MOVEMENT 5
Is mathematical analysis then… only a vain play of the mind?… Far from it; without this language most of the intimate analogies of things would have remained forever unknown to us; and we should forever have been ignorant of the internal harmony of the world…
—Henri Poincaré (1854–1912), “The Value of Science,” The Popular Science Monthly
10
Scientific subjects do not progress necessarily on the lines of direct usefulness. Very many of the applications of the theories of pure mathematics have come many years, sometimes centuries, after the actual discoveries themselves. The weapons were at hand, but the men were not ready to use them.
—Andrew Russell Forsyth (1858–1942), Discussion on the Teaching of Mathematics Which Took Place on September 14th at a Joint Meeting of Two Sections
To divide a cube into two other cubes, a fourth power, or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.
—Pierre de Fermat (1601–1665), written in the margin of his copy of Diophantus’ Arithmetica
We next pivot our attention to a recurring and fascinating property that forms an important layer in the historical tapestry of algebra. We set the stage by returning to computing course averages from Chapter 8 and consider a fourth scenario with the following parameter settings:
Scenario 
Weights/Contributions 
Parameter Values 
Course Average = ax + by + cz 
4 
20% from homework 
a = 20% 
0.20x + 0.60y + 0.20z 
If one of the students has an overall course average of exactly 80% (with no rounding off), can we determine exactly how the student’s scores divvy among the three categories of assignments; that is, can we say with absolute certainty what their average scores are on their homework, tests, and final exam, respectively?
Put another way, is there only one way in which a combination of the three average scores can unite to give an overall course average of exactly 80%?
Clearly, if the student had a straight 80% average across the board on every one of their homework assignments, as well as on all of their tests and their final, this would certainly combine to yield an 80% overall course average. This can be quickly verified: For x = 80, y = 80, and z = 80, we have
course average = 0.20(80) + 0.60(80) + 0.20(80) = 16 + 48 + 16 = 80.
There are other possibilities, however. A few of these are indicated here:
Averages 
Variable Settings 
Course Average = 0.20x + 0.60y + 0.20z 
Course Average 
Homework average = 70 
x = 70 
0.20(70) + 0.60(80) + 0.20(90) = 14 + 48 + 18 = 80 
80 
Homework average = 60 
x = 60 
0.20(60) + 0.60(80) + 0.20(100) = 12 + 48 + 20 = 80 
80 
Homework average = 95 
x = 95 
0.20(95) + 0.60(70) + 0.20(95) = 19 + 42 + 19 = 80 
80 
Homework average = 80 
x = 80 
0.20(80) + 0.60(90) + 0.20(50) = 16 + 54 + 10 = 80 
80 
Homework average = 0 
x = 0 
0.20(0) + 0.60(100) + 0.20(100) = 0 + 60 + 20 = 80 
80 
Thus, the answer to the question of whether or not we can determine with certitude what the average scores for the student are respectively on homework, tests, and the final exam—if we only know the overall average is exactly an 80—is negative; there actually turn out to be quite a lot of possibilities.
Mathematicians call situations where we can’t pin down or determine a single set of values that yield a certain result—due to the existence of many possibilities—indeterminate situations. We have an indeterminate situation here due to the existence of several sets of different average scores for homework, tests, and the final exam that blend together to yield the same overall course average of exactly 80.
Translating the preceding situations into algebraic form shows that they are tantamount to being solutions to the equation 0.20x + 0.60y + 0.20z = 80. The six different combinations of x, y, and z presented satisfy the equation, but there are many more. Because each of these combinations qualify to be called solutions, equations like this one are themselves called indeterminate equations.
We have yet another example of a class of equations that is unlike nearly all of the equations we have discussed throughout the book. The equations we have primarily focused on heretofore—the conditional ones and not identities—generally yielded a single, definitive solution.
One item that distinguishes our indeterminate equation from the earlier equations is that it has three variables. The other equations we solved had only one variable, or ultimately depended upon a single variable. The fact that the number of variables—three—exceeds the number of equations—one—is a significant feature. The extra variables here provide a flexibility and independence that allows many different combinations to work.
Like most things algebraic, indeterminacy is a general phenomenon that often crops up.
THE MAN FROM ALEXANDRIA
Way back sometime in the midst of the Roman era, the mathematician Diophantus did intriguing work with indeterminate situations. Hailing from Alexandria, Egypt, he is one of the shadowiest of the famous mathematicians from antiquity.
And the mystery starts with trying to figure out when he was even alive. The only definitive estimates pin his heyday to within a range of 500 years (150 BCE–350 CE). Another less definitive source places him as being alive around 250 CE, which is the date that many mathematical historians had accepted for a time and some still do.^{1} His ethnicity is also uncertain.^{2}
The amazing creativity of his work combined with its apparent isolation—from any similar known work so far—make it hard even today for mathematical historians to give a complete measure of his contribution to the algebraic narrative. The work in his famous book, Arithmetica, appears to have come out of nowhere, although it is viewed as being unlikely that all of it is due to him alone.^{3}
Just as with Euclid’s famous work, The Elements, and AlKhwarizmi’s famous text on algebra, the book is most likely a combination of earlier work by others as well as original work by the author himself, an extremely capable commentator and expert at mathematical logistics who added in his own explanations, outlook, organization, and perhaps problems and techniques, too.
In the modern viewpoint, indeterminate equations that involve several variables or unknowns raised to positive integer values where we search for integer solutions (or more generally, rational number solutions) are now called Diophantine equations, and the study of such equations is called Diophantine analysis. It is an extremely rich field.
For our purposes here, because we round off averages in each assignment category, the scores for homework, tests, and the final exam are restricted to nonnegative integer values from 0 to 100. This means that looking at the equation 0.20x + 0.60y + 0.20z = 80 this way qualifies it as a Diophantinelike equation. Mathematicians call such equations where the variables are all raised to the first power linear Diophantine equations.^{4} To safely remain in the Diophantine framework, we must consider only those values (of x, y, and z) that combine to be exactly 80, and not those that combine to yield values that round off to 80 (see previous endnote).
Let’s think about this. As there are many solutions that satisfy this equation, together they form a numerical symphony—a collective of solutions if you will. Now the question becomes, is there an algebraic way to capture this ensemble?
Without going into details, algebraic maneuvers for this type of equation show that we can find combinations that work from the following relationship:
final exam score = 400 – (homework average) – 3(test average)
or
z = 400 – x – 3y.
Here, x, y, and z are again nonnegative integers that must be greater than or equal to 0 and less than or equal to 100. Recall that the lower and upper bounds on the values are due to the fact that they represent percentages out of a total with the best score corresponding to 100 (getting everything correct) and the worst corresponding to 0 (getting everything wrong by bad work or through omission).
To see how the relationship z = 400 – x – 3y works, we are free to pick allowable average scores for homework (x) and tests (y) and then plug them into the formula to see what the final exam score (z) would need to be to make the overall course average work out to be exactly 80.
Let’s say that a person got a 100 on all of the homework and a 68 average on the tests; what would the final exam score need to be to make it work? Plugging x = 100 and y = 68 into the formula yields
z = 400 – 100 – 3(68) = 400 – 100 – 204 = 300 – 204 = 96.
The combination of x = 100, y = 68, and z = 96 should result in a course average of 80. Verification yields
0.20(100) + 0.60(68) + 0.20(96) = 20 + 40.8 + 19.2 = 20 + 60 = 80.
If a student had an average of 85 on their homework and an 84 on their tests, then we would have x = 85 and y = 84. To find what final exam score is needed, we plug these values into z = 400 – x – 3y and obtain
z = 400 – 85 – 3(84) = 400 – 85 – 252 = 315 – 252 = 63.
The combination of x = 85, y = 84, and z = 63 should result in an average of 80. Verification yields
0.20(85) + 0.60(84) + 0.20(63) = 17 + 50.4 + 12.6 = 17 + 63 = 80.
You can verify that the five solutions in the table are all related through the formula, as well. Many more combinations can be generated, as we did with the last two examples and as you may do by further experimentation.
TWO MORE INDETERMINATE SITUATIONS
COMBINATIONS OF DAYS
Let’s next consider the equation 28x + 30y + 31z = 365. Are there any positive integers that satisfy this equation? Because we are looking for integer solutions, this becomes a Diophantine situation. We will confine our attention to only positive integers, meaning that 0 is not an allowable value for x, y, or z.
Before seeing if there are any solutions, let’s first see if we can give this equation an interpretation. If we look closely at the equation, we can observe that the values or coefficients attached to each variable correspond to the number of days for some months out of the year. So, if we let x = number of months with 28 days, y = number of months with 30 days, and z = number of months with 31 days, then the question becomes, are there any combinations of 28day, 30day, and 31day months such that they add up to exactly 365 days for the year?
We know that there is at least one set of values that work, namely the situation that we use in our current calendar (for nonleap years), as shown in the following table:
Number of Days 
Month Names 
Number of Months 
Unknown Value 
28 
February 
1 
x = 1 
30 
April, June, September, November 
4 
y = 4 
31 
January, March, May, July, August, October, December 
7 
z = 7 
Placing these x, y, and z values into the expression on the lefthand side yields
28(1) + 30(4) + 31(7) = 28 + 120 + 217 = 148 + 217 = 365.
Thus, x = 1, y = 4, and z = 7 qualifies as one solution to this Diophantine equation.
Are there any other 28day, 30day, and 31day month configurations that combine to equal 365 days in a year? It turns out that, unlike the situation with course grades, there is only one other combination that works—that includes each length of month at least once—namely, the one with x = 2, y = 1, and z = 9. This corresponds to two months that are 28 days long, one month that is 30 days long, and nine months that are 31 days in length. Plugging these values into the lefthand side of the equation gives 28(2) + 30(1) + 31(9), which becomes after multiplication 56 + 30 + 279 = 86 + 279 = 365.
Interestingly, this second situation still corresponds to a 12month year, which we didn’t insist upon. There are only two 12month combinations that work. It turns out, in fact, that even if we allow for more or less than 12 months in a year, no other combination of 28, 30, and 31day months will work for a 365day year.
As with much of this chapter, the details demonstrating this fact are beyond the scope of this book.^{5}
ADULT AND CHILDREN’S MUSEUM TICKETS
Consider a Natural History Museum in which admission tickets for adults cost $12 each and for children $8 each. If a total of $400 is available for purchasing tickets, are there any combinations of adults and children that can attend where all of the money is exactly spent with nothing left over? We assume that at least one adult and one child must be in attendance for any of the combinations.
Proceeding algebraically, we need to represent the unknown number of adults and number of children each by a specific letter. We let x = number of adults and y = number of children. We can treat the setup similarly to what we did with coins as packaged currency in Chapter 5, where here each adult can be considered to have a value of $12, whereas each child has a value of $8. The dollar value of the varying combinations of adults and children who attend the museum can be represented by
12(number of adults) + 8(number of children),
which translates to
12x + 8y.
If 5 adults and 14 children attend, the cost would be 12(5) + 8(14) = 60 + 112 or $172. This amount, of course, doesn’t exactly exhaust the $400.
To algebraically set up the situation of interest, we must represent the condition “total adult cost + total child cost = 400” by an equation. This readily translates to 12x + 8y = 400.
Because the number of adults and children must be equal to positive integers (we can’t have 3.7 adults or 7.4 children), we are looking at this equation through the Diophantine lens.
We list a couple of solution scenarios in the following table:
Number Going to the Museum 
Variable Settings 
Cost = 12x + 8y 
Total Cost 
Adults = 18 
x = 18 
12(18) + 8(23) = 216 + 184 = 400 
$400 
Adults = 8 
x = 8 
12(8) + 8(38) = 96 + 304 = 400 
$400 
An algebraic formula exists that can be used to generate all of the possible combinations of adults and children that work. In words and symbols, we have
number of children = 50 – 1.5(number of adults)
or
y = 50 – 1.5x.
A demonstration of how this formula can be obtained by solving for y in the equation 12x + 8y = 400 is shown in Appendix 5.
There are restrictions on the x, namely that it can’t be a number that causes y to have a negative or fractional value. This means that x (the number of adults) can’t be larger than 32. If there were 33 adults, then y would be a fraction, and 34 adults gives a value of –1 for y. It also must be an even number as only an even number multiplied by 1.5 will yield an integer.
On the other side of the equation, the number of children (y) can be either even or odd as the two examples in the table show, and it ranges from a low of 2 to a high of 47. It turns out that there are a total of 16 positive integer solutions to this equation or a total of 16 different adultchild combinations that work.
If we form the doublet (number of adults, number of children) = (x, y), then the solution combinations that work are these 16 doublets:
(2, 47); (4, 44); (6, 41); (8, 38); (10, 35); (12, 32); (14, 29); (16, 26); (18, 23); (20, 20); (22, 17); (24, 14); (26, 11); (28, 8); (30, 5); (32, 2).
Many more scenarios open up, of course, if we remove the criteria that the budgeted $400 must be exactly spent and simply say that we can spend no more than $400.
A SECONDDEGREE EQUATION
Let’s now examine an issue involving those special positive integers that are perfect squares. A perfect square integer is one that is equal to another integer multiplied by itself (another integer squared).^{6} Thus, 9 is a perfect square because it is equal to the integer 3 times itself (3 × 3 or 3^{2}), and so is 16, which equals 4 times itself (4 × 4 or 4^{2}). The number 12, however, is not a perfect square as no integer times itself equals 12.
The integers between 1 and 200 that are perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, and 196. Adding some of these perfect squares together in pairs reveals the following:
Label 
Perfect Pair 
Sum 
A 
1, 4 
1 + 4 = 5 
B 
4, 9 
4 + 9 = 13 
C 
9, 16 
9 + 16 = 25 
D 
25, 36 
25 + 36 = 61 
E 
36, 64 
36 + 64 = 100 
F 
49, 81 
49 + 81 = 130 
G 
25, 144 
25 + 144 = 169 
H 
16, 169 
16 + 169 = 185 
Five of the sums (A, B, D, F, H) add up to values that are not in the list of perfect squares, but three of them (C, E, G) do add up to values that are themselves perfect squares. So, in these three cases we have the situation where one perfect square plus a second perfect square adds up to yield a third perfect square.
Are there other triplets of three positive integers [such as (9, 16, 25), (36, 64, 100), and (25, 144, 169)] that coordinate this way, that is, three perfect squares such that the largest is equal to the other two added together? This question smells a bit like the other problems involving indeterminacy where we sought to find sets of three integer average scores, sets of three types of months, or combinations of adults and children that blended together, respectively, to yield a course average of 80, a total of 365 days, or a sum of $400 spent.
The difference this time is that instead of trying to find a group of three numbers that give the same value—same course average, same number of days, or same amount of money—we are trying to find a group of three integers—each with the special property that they are perfect squares—such that they satisfy the same relationship with each other: that two of them add up to give the third. Put another way, in the present case there is not a stable value like the course average of 80 (or 365 days or $400), but rather the stability is encoded in the property of the numbers being perfect squares and the relationship they satisfy.
That is, the numbers (9, 16, 25) are all different from (36, 64, 100) and (25, 144, 169), yet in each of the three cases, the property that each number is a perfect square such that the smaller two add up to the third, is satisfied.
There are many equivalent ways to phrase this inquiry. We list a few here:
1. Find positive integers that are perfect squares that can be split into a sum of two other positive integers that are also themselves perfect squares.
2. Find sets of three perfect squares (positive integers) such that one of them is the sum of the other two.
3. Find situations that allow us “to divide a square into two other squares” (Fermat’s language—see this chapter’s epigraphs).
We can try to find answers to this inquiry by trial and error, but that is a daunting process because the majority of sums of two perfect squares don’t add up to a third perfect square. The 14 perfect squares that are less than 200 yield 105 unique pairs of two numbers added together, where unique means that we only count once the sums that are identical when the numbers are reversed. For example, this means that we would only count 1 + 4 and 4 + 1 as one unique pair instead of two pairs because the numbers involved in both sums are the same and addition is commutative. Similarly, the four pairs 25 + 36, 36 + 25, 49 + 64, and 64 + 49 would only count as two unique pairs.
Out of the 105 unique pairs, only four (9 + 16, 36 + 64, 25 + 144, and 81 + 144) add up to a perfect square: 25, 100, 169, and 225, respectively. This amounts to less than 4% of the combinations of the first 14 perfect squares meeting the criteria.
There are many more perfect squares than these first 14, so the difficulty of finding pairs that give us what we want only magnifies.
Can algebra help us out here? And if so, how do we even begin to deploy it in this case?
The situation here involves three special numbers, two of which add together to yield the third, larger number. In the algebraic mindset, we can treat each of these numbers as a variable or unknown. Doing so means that we would then represent each by a letter. Let’s choose the usual suspects: x, y, and z. Continuing, we will let x = the first perfect square integer, y = the second perfect square integer, and z = the third perfect square integer, such that x + y = z.
This representation is problematic, however: Before we can engage it, we must make sure that x, y, and z each have the special property of being perfect square integers. How can we be sure? This is really tantamount to the trialanderror case because we must already know this information about the three numbers before we even engage the algebraic equation; but if we are required in every case to know the information up front, then the algebra is not really helping us.
We need a different plan. As a start, let’s rewrite the perfect squares less than 200 in exponential form as follows:
• Perfect square form: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196.
• Exponential form^{7}: 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}, 7^{2}, 8^{2}, 9^{2}, 10^{2}, 11^{2}, 12^{2}, 13^{2}, 14^{2}.
The following table shows how the four situations where two perfect squares add to a third perfect square translate into exponential form:
Perfect Square Form 
Exponential Form 
9 + 16 =25 
3^{2} + 4^{2} = 5^{2} 
36 + 64 = 100 
6^{2} + 8^{2} = 10^{2} 
25 + 144 = 169 
5^{2} + 12^{2} = 13^{2} 
81 + 144 = 225 
9^{2} + 12^{2} = 15^{2} 
Now, if we look at the equation x^{2} + y^{2} = z^{2}, then the bases of the exponents in the second column of the previous table can be viewed as being integer solutions to this equation, as demonstrated in the following:
Exponential Form 
x 
y 
z 
Integer Solution to x^{2} + y^{2} = z^{2}? 
3^{2} + 4^{2} = 5^{2} 
3 
4 
5 
YES 
6^{2} + 8^{2} = 10^{2} 
6 
8 
10 
YES 
5^{2} + 12^{2} = 13^{2} 
5 
12 
13 
YES 
9^{2} + 12^{2} = 15^{2} 
9 
12 
15 
YES 
3^{2} + 5^{2} = 6^{2} 
3 
5 
6 
NO 
Note that the bottom row of values x = 3, y = 5, and z = 6 (which upon substitution into x^{2} + y^{2} = z^{2} give 3^{2} + 5^{2} = 6^{2}) do not represent a solution of the equation, because the lefthand side gives 9 + 25 = 34 but the righthand side gives 36; and these numbers are not equal. Try a few random values for x, y, and z yourself, and you will see that most of them won’t be solutions.
If we use the exponential form for our new game plan, we can restate all three of the perfect square questions as: “Find three positive integers (x, y, and z) such that they satisfy the equation x^{2} + y^{2} = z^{2}.”
In this form, it becomes more transparent what we seek, and we are free to choose various integer inputs of x, y, and z and place them into the equation to see if they work. That is, this equation now involves the squares of numbers, meaning that no matter what integer values of x, y, and z we initially pick, once we place them into the equation, we get the values x^{2}, y^{2}, and z^{2}, and then we know that we are dealing with three perfect squares.
Consequently, perfect squares are automatically generated from the inputs as we engage the equation. The only thing left to decide is whether or not they satisfy the equation. This is different from using the equation x + y = z whose form does not make transparent what we seek, moreover requiring us to preselect perfect squares before engaging the equation. The automatic generation of perfect squares in the equation x^{2} + y^{2} = z^{2} makes it the more natural and operational setting for the question of sums of perfect squares.
Now that we are seeking positive integer solutions, we are in the Diophantine outlook and this equation becomes a Diophantine equation, this time of the second degree. The second degree comes from the fact that the variables are squared or raised to the second power. In this degree language, the linear Diophantine equations discussed in the previous two sections can be thought of as being Diophantine equations of first degree, because the variables in each are raised only to the first power.
Before looking at possible solutions to the equation involving perfect squares, let’s first see if we can give it a more familiar interpretation, similarly to what we did in the last section with the equation 28x + 30y + 31z = 365.
It turns out that we can by relating it to one of the most famous mathematical results of all time: the Pythagorean Theorem. The Pythagorean Theorem expresses a relationship between the sides of a right triangle, where a right triangle is one that contains a 90degree angle (also called a right angle).
The Pythagorean Theorem: For a right triangle with legs x and y and hypotenuse z, the following relationship holds: x^{2} + y^{2} = z^{2}.
The question about the sum of perfect squares can translate to: “Find right triangles such that the lengths of all three of the sides are positive integers.” The lengths of any three sides that satisfy this condition are called Pythagorean triples or triplets. We have mentioned four so far. Placing these in geometric form yields the following right triangles:
Pythagorean triplets and their triangles
Fortunately, it turns out that there are algebraic ways to generate Pythagorean triples quite easily and to our heart’s content.
We will use new varying quantities—designated by m and n—to state the relationships (once again, the demonstration on how these are obtained is beyond the scope of our discussion):
x = m^{2} – n^{2}, y = 2mn, z = m^{2} + n^{2}.
Let’s see how this works. If we choose m = 2 and n = 1, this will yield the following:
This produces the Pythagorean triple (3, 4, 5), which we already know. Let’s now choose m = 7, n = 5, and find x, y, and z:
This yields the Pythagorean triple (24, 70, 74), which has not been mentioned. Note that 24^{2} + 70^{2} = 74^{2}, which as perfect squares is 576 + 4900 = 5476.
The following table gives a list of several more Pythagorean triples that can be generated from values for m and n,^{8} where it is always the case that x^{2} + y^{2} = z^{2}:
m, n 
Pythagorean Triple x, y, z 
Perfect Square Form x^{2}, y^{2}, z^{2} 
2, 1 
3, 4, 5 
9, 16, 25 
3, 2 
5, 12, 13 
25, 144, 169 
3, 1 
6, 8, 10 
36, 64, 100 
4, 3 
7, 24, 25 
49, 576, 625 
4, 1 
15, 8, 17 
225, 64, 289 
5, 4 
9, 40, 41 
81, 1600, 1681 
6, 5 
11, 60, 61 
121, 3600, 3721 
6, 1 
35, 12, 37 
1225, 144, 1369 
7, 6 
13, 84, 85 
169, 7056, 7225 
8, 7 
15, 112, 113 
225, 12544, 12769 
8, 1 
63, 16, 65 
3969, 256, 4225 
9, 8 
17, 144, 145 
289, 20736, 21025 
10, 9 
19, 180, 181 
361, 32400, 32761 
10, 1 
99, 20, 101 
9801, 400, 10201 
5, 2 
21, 20, 29 
441, 400, 841 
14, 1 
195, 28, 197 
38025, 784, 38809 
15, 14 
29, 420, 421 
841, 176400, 177241 
20, 19 
39, 760, 761 
1521, 577600, 579121 
20, 1 
399, 40, 401 
159201, 1600, 160801 
The variables m and n cannot generate all of the Pythagorean triples out there. For instance, no combination of integer values for m and n can generate (9, 12, 15). However, this particular triplet is an integer multiple of a more basic triplet that can be generated by m and n, namely (3, 4, 5). By integer multiple, we mean that (9, 12, 15) can be obtained from (3, 4, 5) by multiplying each term by the same integer, in this case 3: that is, 3 × (3, 4, 5) = (3 × 3, 3 × 4, 3 × 5) = (9, 12, 15). It turns out to be the case, in general, that if m and n cannot generate a certain triple, then that triplet will turn out to be an integer multiple of a more basic one that can be generated by m and n.
Thus, algebra once again is able to decisively help us in answering the question of finding sets of three perfect squares such that one of them is the sum of the other two.
Through the m and n formulas, we are able to generate with ease and precision as many sets of three numbers that work as we like. Remember, there are a sea of possibilities to choose from, and the overwhelming majority of those choices, if made by guesswork, will not work.
Geometry serves as a critical aid here, too, by providing us with conceptual insight through recognizing the problem as one that can be related to the sides of right triangles, which in turn allows us to involve the Pythagorean Theorem.
In the sense of “The StarSpangled Banner” metaphor, we can think of the right triangle interpretation and the perfect squares interpretation as different renditions of the algebraic song given by the equation x^{2} + y^{2} = z^{2}: renditions that may assist us—through the shuttling back and forth between viewpoints—in better understanding the problem and the equation.
A BABYLONIAN SURPRISE
The different renditions of the equation x^{2} + y^{2} = z^{2} are also able to assist us in providing interesting commentary on an important historical question: Were others aware of the Pythagorean Theorem before the time of Pythagoras (ca. 570–490 BC)?
Surprisingly, through the analysis of several clay/cuneiform tablets from the days of ancient Babylon (ca. 1800 BC) over 1000 years before the time of Pythagoras, it appears that the Babylonians were already familiar with the result. One of the cuneiform documents that gave twentiethcentury mathematicians some of that initial confidence is called Plimpton 322. It has a table that, among other things, lists two of the three sides of 15 right triangles that have integervalued lengths. The sides included on the tablet are the shortest side of the right triangle and the longest side. In terms of our list in the table of m and n values, these would correspond to the short leg (x) and the diagonal side or hypotenuse (z). For example, the Pythagorean triplets (3, 4, 5) and (7, 24, 25) would appear on Plimpton 322 as showing only (3, 5) and (7, 25), respectively. The side whose length is in the middle would not have its value included.
There is a great deal of mystery and disagreement among mathematical historians surrounding the tablet because none of the smaller Pythagorean triples, such as those from our table giving m and n values, are included. The list of values included on Plimpton 322 are given in the following table using modernday numerals:
List of Ratios 
Short Leg (x) 
Hypotenuse (z) 
Entry # 
m 
n 
119 
169 
1 
12 
5 

3367 
4825 
2 
64 
27 

4601 
6649 
3 
75 
32 

12,709 
18,541 
4 
125 
54 

65 
97 
5 
9 
4 

319 
481 
6 
20 
9 

2291 
3541 
7 
54 
25 

Plimpton values not reproduced here 
799 
1249 
8 
32 
15 
481 
769 
9 
25 
12 

4961 
8161 
10 
81 
40 

45 
75 
11 
– 
– 

1679 
2929 
12 
48 
25 

161 
289 
13 
15 
8 

1771 
3229 
14 
50 
27 

56 
106 
15 
9 
5 
Values on Plimpton 322 tablet (in our modernday numerals)
Note that the shaded cells listing the accompanying m and n values are not on the Plimpton 322 tablet. In the last row, the sides generated from m = 9 and n = 5 are not what mathematicians call a primitive triple. Also note that the tablet was written in cuneiform using base 60 or sexagesimal notation.^{9}
Actual Plimpton 322 clay tablet [Photo from Mathematical Cuneiform Texts (Neugebauer and Sachs 1945), permission provided courtesy of the American Oriental Society]
Drawing of Plimpton 322 [Image from (Robinson 2002), provided courtesy of Eleanor Robson; the column headings in box were added by the author]
Decimal cuneiform relationships
The smallest of the fifteen triplets included on Plimpton 322 (entry 11) corresponds to a right triangle with sides x = 45, y = 60, and z = 75. On the table, only 45 and 75 are included. This triplet cannot be generated by m and n values, but is 15 times the triplet (3, 4, 5), which can be generated by the parameters m = 2 and n = 1: x = 15 × 3, y = 15 × 4, and z = 15 × 5.
The largest value on the Plimpton 322 tablet (entry 4) corresponds to the triplet (12709, 13500, 18541), where only the values 12,709 and 18,541 are included. These values give the following relatively large right triangle:
These numbers satisfy the relationship x^{2} + y^{2} = z^{2}: 12709^{2} + 13500^{2} = 18541^{2}, which in perfect squares form is 161518681 + 182250000 = 343768681. Note that these x, y, and z values can be generated from the m and n formulas by setting m = 125 and n = 54.
Why the Babylonians chose such large and sporadic values to list in the Plimpton 322 tablet has mystified mathematicians since the 1940s, when the fact that the rows in the tablet list two of the values in 15 different Pythagorean triplets was discussed by the Austrian mathematical historian Otto Neugebauer and American Assyriologist Abraham Sachs.^{10} It is still an unsettled question as to what purposes the Babylonians had in mind in composing Plimpton 322.
In 2001, the English mathematical historian Eleanor Robson offered an explanation, suggesting that the values in the tablet were used for educational purposes.^{11} For a 2002 article on this theme, she was awarded the Paul R. Halmos–Lester R. Ford Award for expository excellence from the Mathematical Association of America in 2003.^{12}
In 2017 and 2021, Plimpton 322 made a splash in the news once again as mathematicians in Australia resurrected earlier claims that the values in the tablet were used for making trigonometrical calculations, but other scholars and historians, including Robson, question this interpretation.^{13}
Robson’s consistent stance has been that we in our era must be careful not to read too much into the historical record based on our modernday viewpoints, and that the intentions of the ancient authors were often very different from ours. Many other historians share this perspective as well.
To get a sense of the historians’ reasons for urging caution, let’s go back to “The StarSpangled Banner” (or algebraic songs) metaphor from Chapter 7, where, if you recall, the equation 16x + 10 = 106 has multiple interpretations. Now, let’s imagine that each of the five interpretations were in vogue during a different period of history, with the one concerning finding an unknown number (problem 1) being the outlook of ancient mathematicians in 1800 BCE and the one involving power tools (problem 4) being our modern outlook. Because they both ultimately involve the same equation, if after translating an ancient text, we found the equivalent of the modern equation 16x + 10 = 106, we might conclude that the ancient mathematicians knew something about power tools. But this conclusion, of course, would be very wrong.
The simplicity of this analogy makes it a bit of a stretch, but hopefully it illustrates the point that sometimes the very strength of mathematics (the universality of its rules and relationships along with the ensemble of interpretations it allows) can from time to time lead us astray in making accurate assessments of the intentions of the ancients. This sometimes may lead us to underestimate their aims, but it can lead to sometimes overestimating them, as well.
All of this means that more than mathematical formulas are required to really understand what was going on in an earlier era. We must also look at the context of the times, what the authors themselves say, and other documents when available. Given the great length of time from then to now—and the dearth of documents in ancient languages—this can be a monumental, oftentimes impossible, task.
Whatever the reality in terms of the Babylonian authors’ intentions, there is no doubt that the Plimpton 322 tablet is cuneiform and that values on it can be interpreted as being two of the three sides of a right triangle with integer sides. This combined with other cuneiform documents makes this segment of Mesopotamian mathematics interesting in its own right and demonstrates that mathematicians were utilizing the rule associated with Pythagoras of Samos hundreds of years before he flourished. This is something that we have only been made keenly aware of since the middle of the twentieth century.^{14}
FERMAT’S LAST THEOREM
The problems with perfect squares can be naturally extended to other powers. Let’s next consider perfect cubes.
A perfect cube integer is one that is equal to another integer multiplied together three times (another integer cubed). Thus, 8 is a perfect cube because it is equal to the integer 2 multiplied together three times (2 × 2 × 2 or 2^{3}), and so is 125, which equals 5 multiplied together three times (5 × 5 × 5 or 5^{3}). The integers that are perfect cubes from 1 to 200 are 1, 8, 27, 64, and 125.
Now let’s take some perfect cubes and add them together in pairs:
Label 
Pair of Perfect Cubes 
Sum 
A 
1, 8 
1 + 8 = 9 
B 
8, 27 
8 + 27 = 35 
C 
8, 64 
8 + 64 = 72 
D 
27, 64 
27 + 64 = 91 
E 
64, 64 
64 + 64 = 128 
F 
1, 125 
1 + 125 = 126 
G 
27, 125 
27 + 125 = 152 
H 
64, 125 
64 + 125 = 189 
Notice that none of the sums add up to values that are perfect cubes, nor do any of the other pairs of sums from these five numbers (15 unique pairs in total). In fact, it turns out that even if we extend the perfect cubes to include other values such as 216, 343, 512, 729, and 1000, we will never find a sum of two that add up to another perfect cube.
A total shutout from the infinitely many combinations possible.
Put in exponential language, we have that the equation x^{3} + y^{3} = z^{3} has no positive integer solutions at all! And not only is this shutout true for the cubes (or thirddegree equation), but it is also true for the fourthdegree equation x^{4} + y^{4} = z^{4}. Our discussions so far summarize as follows:
• There are infinitely many combinations of three positive integers (x, y, and z) that satisfy the equation x^{1} + y^{1} = z^{1}, or equivalently x + y = z. Examples include (x = 7, y = 10, z = 17) and (x = 432, y = 68, z = 500).
• There are infinitely many combinations of three positive integers (x, y, and z) that satisfy the equation x^{2} + y^{2} = z^{2}. Examples include (x = 3, y = 4, z = 5) and (x = 7, y = 24, z = 25).
• There are no combinations of three positive integers (x, y, and z) that satisfy the equation x^{3} + y^{3} = z^{3}.
• There are no combinations of three positive integers (x, y, and z) that satisfy the equation x^{4} + y^{4} = z^{4}.
It turns out that this total shutout of integer solutions holds not only for the scenarios of third and fourth powers but for all higher integer powers as well. We will use the letter n as a parameter to restate all of these scenarios in one single big algebra equation as follows:
There are no combinations of three positive integers (x, y, and z) that satisfy the equation x^{n} + y^{n} = z^{n} for n ≥ 3 (n is an integer greater than or equal to 3).
or
The equation x^{n} + y^{n} = z^{n} has no positive integer solutions for n ≥ 3 (n is an integer greater than or equal to 3).
Though it is common to read it in the singular, the equation x^{n} + y^{n} = z^{n} really represents infinitely many equations (one for each value of n).
This represents another of the very famous storylines in mathematics: Fermat’s Last Theorem. Pierre de Fermat lived in the 1600s and proved many interesting results about relationships amongst the positive integers. He was a regular correspondent with other mathematicians throughout his life, often sharing the results he proved as challenge problems to them.
Though his efforts in number theory are what he is most famous for, his contributions in analytic geometry, the infinitesimal calculus, physics, and probability are immense and often underrated. In fact, Sir Isaac Newton acknowledged that he received some of his ideas and inspiration for calculus directly from Fermat’s work.^{15} Fermat was truly one of the most remarkable mathematicians of the seventeenth century.
At some point during his research, Fermat jotted down in the margins of his copy of Diophantus’ book, Arithmetica, that he was able to demonstrate why the equations x^{n} + y^{n} = z^{n} all had no solutions for integers n larger than 2, but that he didn’t have enough space in the margins to show it (see this chapter’s epigraphs). And ultimately, he never did show it in his writings.
There are many intriguing items surrounding this story, some mathematical and some historical. Two of the most fascinating are as follows:
1. Why do the equations go from having infinitely many solutions when n = 1 and n = 2 to having no solutions at all for n larger than 2 (n = 3, n = 4, n = 5, and so on)?
That is, what is so special about the two lower powers (especially the squares) and so intractable about higher powers such that the distribution of squares allows for infinitely many solutions, whereas the distribution of cubes, fourth powers, fifth powers, and higher don’t allow any? Why not at least a few solutions for these? One might expect a gradual reduction in the number of solutions as the powers get higher—and the spacing between numbers with that particular power property increases—but why the sharp cliff from infinitely many solutions to none at all, in going from n = 2 to n = 3?
2. Did Fermat actually find a way to demonstrate the general result, and if so why didn’t he share it as a challenge to others or write it down somewhere in his papers?
Mathematicians have found the answer to the first question to be quite intricate and related to a chain of deep results. The last link in that chain was supplied in late 1994 (and published in 1995) by the celebrated mathematician Andrew Wiles (assisted by his former student Richard Taylor), and the proof of Fermat’s Last Theorem is said to come out as a consequence.
Regarding the second question, most mathematicians doubt that Fermat actually found a general answer. They base this in part on two things:
1. No airtight proof was ever found by the many excellent mathematicians who have worked on the theorem from the time of Fermat until the 1990s. This spans a period of over 300 years.
2. The arguments developed by Wiles to prove the theorem were quite long (over 100 journal pages in length) and used complex mathematics that was only developed in the twentieth century and unknown to Fermat.
The fact that Fermat didn’t send it out as a challenge to others also suggests that he may have realized that there were errors in his argument and thus did not in the end have the result that he privately claimed in the margins of his copy of Arithmetica.
CONCLUSION
Over the years, many have questioned the worth of pursuing scientific knowledge out of sheer curiosity and imagination alone, only later to have new emergent outlooks and technologies slam the door on that skepticism.
Electricity and magnetism present a case in point. The sophisticated electrified world that we inhabit today had its origins in the curiosity of the early investigators of electricity and magnetism in the eighteenth and nineteenth centuries, men such as CharlesAugustin de Coulomb, Alessandro Volta, AndréMarie Ampère, Hans Christian Ørsted, Michael Faraday, William Thomson (Lord Kelvin), and James Clerk Maxwell. At the time, it was not obvious what these investigations would grow up to be or if they would even grow up to be anything applicable at all. But grow up they did, and now it is hard to imagine a world without the technologies that spewed forth out of this knowledge.
The investigations of physicists into the structure of the atom offer another case, where our modern world of televisions, cell phones, and computers would not be possible without the curiositydriven discoveries of these late nineteenth and early twentiethcentury scientists.
Curiosity is one of scientists’ most valuable tools. And even when discoveries find no known application, they provide an important background canvas for situating those discoveries that do have a direct impact. It is similarly so with mathematicians. Much of the theoretical advance in science and technology has and continues to be underwritten by mathematics that was originally created more out of simple curiosity about the subject itself rather than with an eye to direct utility.
In considering integer solutions to indeterminate equations, mathematicians embarked on a curiosityladen journey that demonstrates some of what can happen when free inquiry interacts with imagination, mathematical analysis, and circumstance. A bit of this is in evidence from our basic considerations here in this chapter. Let’s discuss a few of these.
COMPLEMENTARY PERSPECTIVES
The Diophantine equation x^{2} + y^{2} = z^{2} connects to the Pythagorean Theorem for right triangles. Let’s look at a couple of things that can occur in the mathematical interplay of ideas from such a connection.
If we look at this equation as only involving the sum of two perfect square integers adding up to equal a third perfect square integer, we will be happy when we find values for x, y, and z that work and throw away the ones that don’t.
However, because the equation also connects to right triangles, we can look at the situation geometrically. When x = 1 and y = 1, yielding 1^{2} + 1^{2} = z^{2}, and we discover that we can’t find any integer values for z such that 2 = z^{2}, we can’t simply throw away these values for x and y. There exists a right triangle with side lengths x = 1 and y = 1:
The circumstances now involve an actual hypotenuse, z, that has an actual length. And the question becomes, if this length is not an integer, then what is it? An answer is now demanded. The length, of course, is z = ≈ 1.41421… This length is not an integer, nor is it even the ratio of two integers (common fraction). It presents us with an example of an entirely new category of numbers called irrational numbers. These numbers have infinitely long decimal representations that don’t repeat in a pattern of digits.
So, looking at this problem from the standpoint of integers alone would lead us to say that it has no solution and then be done with it. But looking at it from the other interpretation—as sides of a right triangle—we can’t ignore it. The need to answer the question of what the length of z is eventually leads to the discovery of a whole new class of numbers.
If we look at the equation x^{2} + y^{2} = z^{2} only from the geometrical viewpoint, we may see it as a twodimensional space problem. This could lead us to naturally extend it to three dimensions, which yields x^{3} + y^{3} = z^{3}, and then seek integer solutions there. But because we are in the geometric viewpoint, and given that the geometrical objects we can easily imagine or visualize consist of either zero dimension (points), one dimension (lines, curves, etc.), two dimensions (rectangles, triangles, etc.), or three dimensions (cubes, spheres, etc.), we may stop at this thirddegree equation, thinking that there are no more physical dimensions to consider.
However, if we look at it from the exponential powers viewpoint, there is nothing to keep us from considering what the fourth power or fifth power versions of this equation look like and so on, leading us ultimately to the equation x^{n} + y^{n} = z^{n}.
This type of thing happens frequently in mathematics where one representation or viewpoint of a phenomena—or algebraic expression—opens up a whole world of new possibilities to which a second representation or viewpoint is totally blind. In the cases discussed here, both viewpoints—geometrical and integer—yield insights that the other respective viewpoint doesn’t naturally offer.
CREATION OF NEW MATHEMATICS
Larger problems are often identified from generalizations of questions that arose out of smaller, particular situations. The search then for commonalities in these larger problems can often lead to dramatically new insights—entire new areas of mathematics even. Two examples of this flow from our discussions in this chapter.
FERMAT’S EXTENSION OF THE DIOPHANTINE EQUATION X^{2} + Y^{2} = Z^{2}: When Fermat extended his investigations to higher powers, initiating a search for positive integer solutions to x^{n} + y^{n} = z^{n}, he was led to his conjecture for values of n ≥ 3, which become known as his famous Last Theorem.
The attempts to prove this theorem, or show gaps in proposed proofs offered, ultimately spawned a large body of new mathematics whose importance far surpasses the importance of proving the original question itself.
The ScottishAmerican mathematician Eric Temple Bell had exactly this type of thing in mind when he shared (in a statement he attributed to mathematician Felix Klein): “Choose one definite objective and drive ahead toward it. You may never reach your goal, but you will find something of interest along the way.”^{16}
HILBERT’S EXTENSION OF SOLUTIONS TO DIOPHANTINE EQUATIONS IN GENERAL (TENTH PROBLEM): In looking at the basic Diophantine equations discussed here, there was a great difference in the form of the solutions to linear Diophantine equations (such as 12x + 8y = 400) and the solutions to the seconddegree equation given by the Pythagorean Theorem (x^{2} + y^{2} = z^{2}). There is also wide variation in how the solutions for each were obtained. This is also true of other Diophantine equations such as the following two:
• y^{2} – 17 = x^{3}.
Here are three integer solutions (out of many):
(x = 2, y = 5), (x = 4, y = 9), (x = 52, y = 375).
• x^{4} – x^{2}y^{2} + y^{4} = z^{2}.
Here are three integer solutions (out of many):
(x = 5, y = 5, z = 25), (x = 10, y = 10, z = 100), (x = 75, y = 75, z = 5625).
The techniques for finding the integer solutions to these two equations are different from the ones we have discussed in detail.
However, there are other Diophantine equations out there that don’t have any positive integer solutions at all, such as x^{10} + y^{10} = z^{10}, x^{15} + y^{15} = z^{15}, and 5x^{4} + 7y^{6} = 1.
Thus, Diophantine equations exist that have either no solutions, a small number of solutions, many more (but still finite) solutions, and infinitely many solutions. This means that the question of whether or not a solution exists will vary from Diophantine equation to Diophantine equation. Since the time of Diophantus, mathematicians have come up with various techniques for solving specific Diophantine equations, even determining when many have no solution. As with those we have discussed, the technique used depended on the particular equation involved.
In 1900 during the International Congress of Mathematicians in Paris, the great German mathematician David Hilbert was so bold as to propose that mathematicians find out if it was possible to find one single, allencompassing, universal method that would work on any and all Diophantine equations, in determining whether they had solutions or not.^{17}
During his talk on Wednesday, August 8, Hilbert discussed only ten of his total list of twentythree interesting questions that he thought twentiethcentury mathematicians should pay some attention to. This question concerning Diophantine equations was number ten on his written list and is known as Hilbert’s Tenth Problem. It was not discussed in his talk. Hilbert’s wording of the problem translated to English summarizes as follows:
Given a Diophantine equation with any number of unknown quantities and with… [integer]… numerical coefficients: To devise a process according to which it can be determined by a finite number of operations whether the equation is solvable in… [integers].
This question was eventually answered in the negative in 1970—no single universal method exists—by the Russian mathematician Yuri Matiyasevich. Matiyasevich’s final conquest of the problem depended critically on the work of three American mathematicians: Martin Davis, Hilary Putnam, and Julia Robinson. Interesting new ideas came out of the techniques used to demonstrate this result, as with Fermat’s Last Theorem.
HISTORICAL INTERPLAY
In studying integer solutions to x^{2} + y^{2} = z^{2} from the geometric viewpoint, mathematicians were led to the notion of Pythagorean triples, eventually finding ways to generate them at will. Euclid gives a demonstration of how to do this in his famous book, Elements, while Viète gives a more symbolic demonstration in his work at the end of the sixteenth century.^{18}
As modern linguists began to crack the ancient Cuneiform writing systems of the Sumerians and Babylonians, modern mathematicians—with their crucial assistance—were eventually able to also acquire an appreciation of some of the mathematics that was known to these ancients. This included their work on seconddegree situations and right triangles. This would ultimately lead mathematicians to the shocking realization—with evidence—that the Pythagorean property appears to have been known and used hundreds of years before the time of Pythagoras. Without the imagination of both ancient mathematicians and contemporary scholars, this very real possibility might have never been realized.
Information such as this could then in turn enhance the viewpoints of historians on how knowledge was transmitted between the Middle Eastern civilizations (Mesopotamia and Egypt) and Ancient Greece. Such mutualistic relationships often occur between the sciences and other disciplines (e.g., radiocarbon dating).
ALGEBRAIC INTANGIBLES
There is a certain elegance to the manner in which perspectives and generalizations interact to yield new insights and unexpected possibilities. Such unanticipated mathematical convergences and unifications over the broad plain of human ideas certainly qualify, in the view of some, as being aesthetic on a variety of levels.
As such, some artifacts of these transactional flairs certainly are worthy of display, reflection, and discussion. Such exhibition often occurs when a famous mathematical result is proven, or some new mathematical knowledge is discovered and gets written up in the media. Unfortunately, even after reading such articles, it usually turns out that too little is still understood by the lay reading audience, and much of the technical reading audience as well. There is probably no easy way to improve upon this, but it is the merging of ideas in deep and sometimes surprising ways—such as shown in miniature here—that often creates the buzz in the air among the experts in the know.
The benefits that accrue from such curiosity not only can help mathematicians and scientists in their investigations, but may also assist other learners in their own personal journeys by allowing them to add to their own “stores of conceptual fuel.”
These are but a few of the kinds of intangibles—from learning algebra—that those partial to the subject sometimes have in mind when they proclaim algebra for all.
In Chapter 11, we illustrate how exploring alternate viewpoints for some of the algebra we have already considered can provide additional insights for both mathematics and science.