Systems of Pythagorean Triples Report

SYSTEMS OF PYTHAGOREAN TRIPLES Abstract. This paper explores systems of Pythagorean triples. It describes the generating formulas for primitive Pythagorean triples, determines which numbers can be the sides of primitive right triangles and how many primitive right triangles those numbers can be a side of, and finally explores systems of three and four right triangles that fit together in three dimensions. 1. Introduction Pythagoras lived during the late 6th century B.C.E. His most famous discovery by far is that on a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse. This theorem, which has come to be known as the Pythagorean Theorem, is not only incredibly powerful, but rich with mathematical extensions that still fascinate mathematicians today. Moreover, at times, seemingly unrelated problems can be reduced to problems involving the Pythagorean Theorem and the many discoveries related to it. The motivation for this paper stems from the following problem. Consider the ellipsoid  x 2 a +  y 2 b +  z 2 c =1 with a > b > c > 0 and a, b, c ∈ Z. Note that if x = 0, then (y/b)2 + (z/c)2 = 1 1 2 √ is an ellipse with focal points at (0, ± b2 − c2 , 0). Similarly, if y = 0, then (x/a)2 + (z/c)2 = 1 √ is an ellipse with focal points at (± a2 − c2 , 0, 0) and if z = 0, then (x/a)2 + (y/b)2 = 1 √ is an ellipse with focal points at (± a2 − b2 , 0, 0). We are interested in ellipsoids whose focal points have integer coordinates. In other words, we are interested in the system (1.1) b2 − c2 = n21 , a 2 − b2 = n22 and a2 − c2 = n23 where a, b, c, n1 , n2 , n3 ∈ Z. Rearranging these equations gives us a system of equations, each of the form given in the Pythagorean Theorem: (1.2) n21 + c2 = b2 , n22 + b2 = a2 and n33 + c2 = a2 . Therefore, the motivation for this paper stems from trying to find integral solutions to System 1.2. However, we start with the basics of the Pythagorean Theorem. We begin by looking at generating formulas for Pythagorean triples. From there, we explore the properties of each side of a primitive right triangle, focussing on how it is possible to fit two right triangles together so that they share either a leg or the hypotenuse. We show that with each side of a primitive right triangle, the total number of triangles that can share that side is 2n−1 , where n is the number of distinct primes in the standard prime factorization of the length of that side. We SYSTEMS OF PYTHAGOREAN TRIPLES 3 follow with a proof that the lengths of the two legs of one right triangle cannot be the lengths of the leg and hypotenuse of another right triangle. From there we move into three dimensions where we first consider the Euler Brick, giving a few simple results as well as considering the perfect cuboid. Lastly, we return to System 1.2 and give Euler’s solution to the problem. 2. Generating All Pythagorean Triples When asked to give examples of Pythagorean triples, a typical math student can usually give two or three examples: (3,4,5), (5,12,13) and maybe (15,8,17). Not many students can come up with more triples off the top of their heads. It is therefore desirable to find a way of generating Pythagorean triples that is simpler than the guess and check method. We begin with some basic definitions and introductory lemmas. Definition 2.1. A Pythagorean triple is a triple of positive integers (a, b, c) satisfying a2 + b2 = c2 . We call a triple primitive when the three integers (a, b, c) have no common factor. To describe Pythagorean triples, it is only necessary to find the primitive triples, since any non-primitive triple is just a multiple of a primitive triple, as is proved in Lemma 2.4. The following basic result will be used frequently throughout this paper. Lemma 2.2. For integers a, b and d, if d divides a and d divides b, then d divides any linear combination of a and b. Proof. Suppose d|a and d|b, then a = dx for some integer x, and b = dy for some integer y. Therefore, for any α and β, αa + βb = (αx + βy)d and thus d|(αa + βb). This completes the proof.  Note that if d|a and d|b, then d|(a2 + b2 ) and d|(a2 − b2 ). Also, if a and b are relatively prime and if d|(a + b) and d|(a − b), then either d = 2 or d = 1, since the sum and difference of (a + b) and (a − b) are 2a and 2b respectively. 4 Figure 1. The Smallest Primitive Pythagorean Triple The next lemma is important when dealing with numbers that can be written as the sum of two squares. Lemma 2.3. Suppose x2 = kl where x, k and l are all positive integers. If gcd(k, l) = 1, then both k and l must be squares. Proof. By the unique factorization of the integers, there exist unique primes p1 , p2 , . . . , pm and unique positive integers α1 , α2 , . . . , αm such that αm 1 α2 x = pα 1 p2 . . . pm and 1 2α2 m x2 = p2α . . . p2α m . 1 p2 Since x2 = kl and gcd(k, l) = 1, we know that if pi |k, then pi - l. Thus, k and l are both equal to the product of some of the pi ’s, where each pi is raised to an even power. Thus, the square roots of both k and l are integers, and therefore, both k and l are squares.  SYSTEMS OF PYTHAGOREAN TRIPLES 5 Using mathematical induction, it is easy to show that if x2 = k1 k2 . . . kn where the ki are pairwise relatively prime, then each ki is a perfect square. We now prove the following lemma about the primitivity of Pythagorean triples. Lemma 2.4. For a Pythagorean triple (a, b, c), the following properties are equivalent: (1) a, b, and c have no common factor, i.e., the triple is primitive, (2) a, b, and c are pairwise relatively prime, (3) two of a, b, and c are relatively prime. Proof. We start by showing that (1) implies (2). To do this, we will prove the contrapositive. Let p be prime and suppose that p divides both a and b. Then by Lemma 2.2, p divides a2 + b2 = c2 . But if p divides c2 , then p|c, and thus (1) is violated. Similarly, if p|a and p|c, then by Lemma 2.2, p|(c2 − a2 ) = b2 , and again, (1) is violated. Obviously the same is true if p|b and p|c. Thus, we have shown that (1) implies (2). It is obvious that (2) implies (3). To prove that (3) implies (1), first assume that a and b are relatively prime. Let p be prime and assume that p|c and p|a. Then p|(c2 − a2 ) = b2 which implies that p|b, which is a contradiction. A similar argument can be used for the other cases. This completes the proof.  Theorem 2.5 gives us an easy way to generate primitive Pythagorean triples. This is a standard result in number theory and can be found in an article entitled Pythagorean Triples by Keith Conrad [1]. Theorem 2.5. If (a, b, c) is a primitive Pythagorean triple, then one of a or b is even and the other is odd. Moreover, taking b to be even, a = k 2 − l2 , b = 2kl, c = k 2 + l2 for some integers k and l with k > l > 0, gcd(k, l) = 1, and k 6≡ l mod 2. Conversely, for such integers k and l the above formulas yield a primitive Pythagorean triple. 6 Proof. First, let k and l be positive, relatively prime integers with k > l > 0 and k 6≡ l mod 2. We check that given a = k 2 − l2 , b = 2kl and c = k 2 + l2 that (a, b, c) is a primitive Pythagorean triple. We observe that a2 + b2 = (k 2 − l2 )2 + (2kl)2 = k 4 + 2k 2 l2 + l4 = (k 2 + l2 )2 = c2 , showing that (k 2 − l2 , 2kl, k 2 + l2 ) is a Pythagorean triple. To show that it is primitive, let d be an integer that divides (k 2 − l2 ) and (k 2 + l2 ); by Lemma 2.2, d must also divide the sum and difference of (k 2 − l2 ) and (k 2 + l2 ), which are 2k 2 and 2l2 . Since (k 2 − l2 ) and (k 2 + l2 ) are odd, d must also be odd, and thus d|k 2 and d|l2 . But since gcd(k, l) = 1 and thus gcd(k 2 , l2 ) = 1, we have d = 1 and thus (k 2 − l2 ) and (k 2 + l2 ) are relatively prime. By Lemma 2.4, we can conclude that (k 2 − l2 , 2kl, k 2 + l2 ) is a primitive Pythagorean triple. Next, assume that (a, b, c) is a primitive Pythagorean triple. To show that one of a and b is even and the other is odd, note that if both a and b are even, then they are both divisible by 2, which, by Lemma 2.4, is a contradiction to the fact that (a, b, c) is a primitive Pythagorean triple. Thus, both a and b cannot be even. Next, assume that both a and b are odd. Then a2 ≡ b2 ≡ 1 mod 4, so c2 = a2 + b2 ≡ 2 mod 4. However, 2 is not a square modulo 4, thus a and b cannot both be odd. Therefore, one of a and b is odd and the other is even. Without loss of generality, from now on, we will take b to be even. We rewrite our equation a2 + b2 = c2 as (2.1) b2 = c2 − a2 = (c + a)(c − a). Since b is even and (c + a) and (c − a) are both even, we can divide both sides of equation (2.1) by 4. Doing this yields  2 b (c + a) (c − a) = . 2 2 2 Now we see that the two integers on the right are relatively prime, because by Lemma 2.2 if d divides both (c + a)/2 and (c − a)/2, it must divide their sum and SYSTEMS OF PYTHAGOREAN TRIPLES 7 Table 1. Examples of Pythagorean Triples k 2 3 4 4 l 1 2 1 3 a 3 5 15 7 b 4 12 8 24 c 5 13 17 25 difference, which are c and a respectively. Since c and a are relatively prime, d = 1. Thus gcd((c + a)/2, (c − a)/2) = 1. By Lemma 2.3, since the product of (c + a)/2 and (c − a)/2 is a square, and since gcd((c + a)/2, (c − a)/2) = 1, we know that (c + a)/2 = k 2 and (c − a)/2 = l2 for some relatively prime positive integers k and l. Doing some simple algebra yields c = k 2 + l2 , a = k 2 − l2 . Finally, (b/2)2 = k 2 l2 implies that b = 2kl. Lastly, we must check that k 6≡ l mod 2. Since k and l are relatively prime, they are not both even. If they were both odd, then k 2 + l2 , 2kl, and k 2 − l2 would all be even, violating the primitivity of (a, b, c). This completes the proof.  Example: Let k = 2 and l = 1. Clearly gcd(k, l) = 1 and k 6≡ l mod 2. Therefore, Theorem 2.5 tells us that (22 − 12 , 2 · 2 · 1, 22 + 12 ) is a primitive Pythagorean triple. This, of course, is just the triple (3,4,5). Table 1 has a list of primitive Pythagorean triples along with their generators. It is helpful to categorize primitive Pythagorean triples in a slightly different way. Again, this is a standard result in number theory and can be found in the article Pythagorean Triples by Keith Conrad [1]. Theorem 2.6. Given a primitive Pythagorean triple (a, b, c) with b even, there exist s and t where s > t ≥ 1 are odd integers that are relatively prime such that a = st, b = s2 − t2 s2 + t2 , c= . 2 2 8 Moreover, for such integers s and t the above formulas yield a primitive Pythagorean triple. Proof. From Theorem 2.5 we know that there exist integers k and l such that a = k 2 − l2 = (k + l)(k − l) where k > l > 0, gcd(k, l) = 1, and k 6≡ l (mod 2). If we take s = k + l and t = k − l we have two odd integers that are relatively prime with s > t ≥ 1. From Theorem 2.5 we know that b = 2kl. Solving for k and l in terms of s and t yields k = (s+t)/2 and l = (s − t)/2, which implies that b = (s2 − t2 )/2, and c = k 2 + l2 = (s2 + t2 )/2. This completes the proof.  Example: Let s = 5 and t = 3. Then Theorem 2.6 tells us that (5 · 3, (52 − 32 )/2, (52 + 32 )/2), or rather (15, 8, 17), is a primitive Pythagorean triple. 3. Primitive Pythagorean Triangles that share a Side The obvious geometric interpretation of Pythagorean triples is right triangles. The sides of a right triangle with integer lengths form the terms of a Pythagorean triple, and visa versa. In this section, and for the rest of the paper, we will be considering Pythagorean triples with right triangles in mind. The first question that we address concerns primitive right triangles that share a side where sharing a side means that two right triangles have one side that is the same length, while the other two sides are different lengths. 3.1. Triangles that share the even length side. From Theorem 2.5, we know that every primitive right triangle has one leg that is even, one leg that is odd, and a hypotenuse that is odd. We first consider those triangles that share the even length side. Lemma 3.1. The even side of a primitive right triangle is divisible by 4. SYSTEMS OF PYTHAGOREAN TRIPLES Proof. The result is clear from Theorem 2.5. 9  Theorem 3.2. Given a number b divisible by 4, let n denote the number of distinct primes in the prime factorization of b. Then b is the even length side of exactly 2n−1 primitive right triangles. Proof. From Theorem 2.5, we know that given a primitive Pythagorean triple (a, b, c), then a = k 2 − l2 , b = 2kl, c = k 2 + l2 for some positive integers k and l where gcd(k, l) = 1, k > l and one of k and l is even while the other is odd. Moreover, we know that given k and l that meet these criteria, they generate a primitive Pythagorean triple. Suppose that b is a number divisible by 4, and let n be the number of distinct primes in the prime factorization of b. In other words, let p1 , p2 , . . . , pn be the primes in the prime factorization of b such that αn 1 α2 b = pα 1 p2 · · · pn . Since b is divisible by 4, one of the primes is equal to 2, and moreover, we know that there are at least 2 factors of 2 in the factorization of b. Without loss of generality we can assume that p1 = 2. With this in mind, we see that the number of primitive right triangles that have a leg of length b is equal to the number of ways we can factor b into b = 2kl where k and l are as in Theorem 2.5. Since b is divisible by 4, and thus (b/2) is divisible by 2, we know one of k and l will be even and the other one will be odd. Since gcd(k, l) must equal 1, we are simply interested in the number of ways we can partition the n primes into two sets where the product of all the terms in one set is k and the product of the terms in the other set is l. In other words, we are interested in the total number of subsets of an n element set divided by 2, since each subset corresponds to a value for k or l, and the compliment of the subset corresponds to the other. This gives us (2n /2) = 2n−1 . The reason we divide by 10 2 is that k must be greater than l, and the total number of subsets counts those partitions where l > k. It remains to show that each new choice of k and l will give a distinct primitive right triangle. Let k1 , l1 and k2 , l2 be generators for two different primitive Pythagorean triples with even term b. If k1 = k2 , then l1 must also equal l2 , and we would not have two distinct right triangles. Thus, we can assume without loss of generality that k1 > k2 , meaning that l1 < l2 . It does not take much to see that it necessarily follows that a1 = k12 − l12 > k22 − l22 = a2 . Thus, given two different choices of k and l, we necessarily get a different value for a, the odd length leg of our triangle, which necessitates a different hypotenuse, or c value, as well. Therefore, there are 2n−1 primitive right triangles that have the even side length equal to b. This completes the proof.  Example: Consider an example with the number 12. We factor 12 into 3 · 22 . Theorem 3.2 tells us that there are 22−1 = 2 primitive right triangles that have 12 as the even leg. Let k = 3 and l = 2. Then clearly k and l satisfy the necessary conditions to generate a primitive Pythagorean triple, and indeed, we see that a = 32 − 22 = 5, b = 2 · 3 · 2 = 12, c = 32 + 22 = 13 is a primitive right triangle. However, we could choose k = 6 and l = 1, in which case we get a = 62 − 1 = 35, b = 2 · 6 · 1 = 12, c = 62 + 1 = 37 as our primitive right triangle. 3.2. Triangles that share an odd length side. In the same way that we counted the number of triangles that share an even length leg, we can count the number of triangles that share an odd length leg. Lemma 3.3. Any odd number can be the odd length leg of a primitive Pythagorean triangle. SYSTEMS OF PYTHAGOREAN TRIPLES 11 Figure 2. Pictorial Example of Theorem 3.2 Proof. By Theorem 2.6, the odd length leg of a primitive Pythagorean triangle can be factored into a = st for relatively prime, odd, positive integers s and t where s > t. Thus, given any positive odd integer q, let t = 1, and set s = q. Therefore, a = q.  Theorem 3.4. Given an odd number a, let n denote the number of distinct primes in the prime factorization of a. Then a is the odd length leg of 2n−1 primitive right triangles. Proof. Not surprisingly, the proof of this theorem is similar to the proof given for Theorem 3.2. Because of this, some of the details will be left out. From Theorem 2.6, we know that given a primitive Pythagorean triple (a, b, c), that a = st, b = s2 − t2 s2 + t2 , c= 2 2 for some positive odd integers s and t where gcd(s, t) = 1, s > t. Moreover, we know that given s and t that meet these criteria, they generate a primitive Pythagorean triple. Let a be an odd positive integer, and let n be the number of distinct primes in the prime factorization of a. In a similar fashion as in the proof of Theorem 3.2, we see that the number of primitive right triangles that have a leg of length a is equal 12 to the number of ways we can factor a into a = st where gcd(s, t) = 1 and s > t. Exactly as in the proof of Theorem 3.2, this number is 2n−1 . Checking that each distinct choice of s and t gives a distinct primitive right triangle can be done in the same way as the proof of Theorem 3.2. Thus, given two different choices of s and t, we necessarily get a different value for b, the even length side of our triangle, which necessitates a different hypotenuse, or c value, as well. Therefore, there are 2n−1 primitive right triangles that have the odd length leg equal to a. This completes the proof.  Example: The number 15 factors into 5 · 3. Theorem 3.4 tells us that there are 22−1 = 2 primitive right triangles that have 15 as the odd length leg. Let s = 5 and t = 3. Clearly s and t satisfy the conditions of 2.6, and so a = 5 · 3, b = 52 − 32 52 + 32 = 8, c = = 17 2 2 is a primitive right triangle. However, we could choose s = 15 and t = 1, in which case we get a = 15 · 1, b = 152 − 1 152 + 1 = 112, c = = 113 2 2 as our primitive right triangle. Figure 3. Pictorial Example of Theorem 3.4 3.3. The Hypotenuse of a primitive right triangle. The last side of the triangle we consider is the hypotenuse. Determining which numbers can be the hypotenuse of more than one primitive right triangle is more involved than the proofs of Theorems 3.2 and 3.4. Because of this, we include references to proofs, instead of the proofs themselves, for some of the theorems that follow. SYSTEMS OF PYTHAGOREAN TRIPLES 13 Theorem 3.5 uses results from Quadratic Reciprocity. Specifically, it uses the well known result in number theory that the equation n2 ≡ −1 (mod p) for an integer n has a solution if and only if p ≡ 1 (mod 4). For more information on Quadratic Reciprocity, see Chapter 24 of A Friendly Introduction to Number Theory by Joseph H. Silverman[2]. Theorem 3.5. If c is the hypotenuse of a primitive right triangle, then c is of the form c = pe11 pe22 · · · penn where pi ≡ 1 (mod 4) for 1 ≤ i ≤ n. Proof. Choose k and l as in Theorem 2.5 so that c = k 2 + l2 where gcd(k, l) = 1 and k 6≡ l (mod 2). Since c is odd, c = pe11 pe22 · · · penn where each pi is odd. For each 1 ≤ i ≤ n, k 2 + l2 ≡ 0 (mod pi ) k2 ≡ −l2 (mod pi ) and (kl−1 )2 ≡ −1 (mod pi ). Here, l−1 is the inverse of l under multiplication mod pi , which we are guaranteed exists since gcd(k, l) = 1. By Quadratic Reciprocity, pi ≡ 1 (mod 4). This completes the proof.  Theorem 3.5 has further clarified our understanding of the hypotenuse of primitive right triangles. In the following theorems, we use this result, as well as a result proved by Judith D. Sally and Paul J. Sally Jr. in Chapter 4 of Roots To Research[3] to prove that given a number c of the form c = pe11 pe22 · · · penn where each pi ≡ 1 (mod 4), that c is the hypotenuse of exactly 2n−1 primitive Pythagorean triangles. 14 Definition 3.6. Given a number c = pe11 pe22 · · · penn where each pi ≡ 1 (mod 4), let N(c2 ) denote the total number of ways, both primitive and non-primitive, that c2 can be written as the sum of two squares of positive integers, not counting trivial variations (a2 + b2 does not count separately from b2 + a2 ). Let Np (c2 ) denote the number of primitive ways c2 can be written as the sum of two squares of positive integers. In Theorem 4.12 of Roots To Research[3], Judith Sally and Paul Sally Jr. prove that given a number c = pe11 pe22 · · · penn where each pi ≡ 1 (mod 4), that N (c2 ) = 1 ((2e1 + 1)(2e2 + 1) · · · (2en + 1) − 1) . 2 Our task in the remainder of this section is to use this number N (c2 ) to find Np (c2 ). For the sake of brevity, for the remainder of this section we will assume that c is of the form c = pe11 pe22 · · · penn where each pi ≡ 1 (mod 4). Lemma 3.7. Given c, Np (c2 ) = N (c2 ) − X Np (c/d)2
where the summation runs through each divisor d of c where 1 < d < c. Proof. We must prove that P
Np (c/d)2 is equal to the number of ways c2 can be written as the sum of two squares in a non-primitive way, since by removing the non-primitive solutions we are left with only the primitive solutions. First, note that given a triple a21 + b21 = (c/d1 )2 , multiplying through by d2 gives us a non-primitive solution (a1 d1 )2 + (b1 d1 )2 = c2 that is counted in N (c2 ). This
P proves that each triple in Np (c/d)2 corresponds to exactly one non-primitive triple in N (c2 ). Next, note that given a non-primitive triple a22 + b22 = c2 with d2 the greatest common factor of a2 , b2 and c, that the triple (a2 /d2 )2 + (b2 /d2 )2 = (c/d2 )2 is
P counted in Np (c/d)2 exactly once. This proves that each non-primitive triple
P in N (c2 ) corresponds to exactly one triple in Np (c/d)2 . SYSTEMS OF PYTHAGOREAN TRIPLES 15 Therefore, there is a one to one relationship between the non-primitive triples in

P P N (c2 ) and Np (c/d)2 . Thus, Np (c2 ) = N (c2 ) − Np (c/d)2 as was claimed. This completes the proof.  Lemma 3.7 gives us a formula for finding Np (c2 ), but the formula is recursive. The next two theorems prove that the formula for Np (c2 ) simplifies to 2n−1 where n is the number of distinct primes in the prime factorization of c. Theorem 3.8. Given c = p1 p2 · · · pn , Np (c2 ) = 2n−1 . Proof. We proceed by induction. Let c = p1 where p1 ≡ 1 (mod 4). Then by
P Lemma 3.7, Np (c2 ) = N (c2 ) − Np (c/d)2 for each divisor d of c between 1 and c. But since c = p1 (and thus e1 = 1), Np (c2 ) = N (c2 ) = 1 ((2e1 + 1) − 1) = 1. 2 Since n = 1 in this base case, Np (c2 ) = 1 = 2n−1 . Our induction hypothesis is that Np (p21 p22 · · · p2k ) = 2k−1 for all 1 ≤ k < n. Since each ei = 1, note that X Np (c/d)2
= Np (p21 ) + Np (p22 ) + · · · + Np (p2n )
+ Np (p21 p22 ) + Np (p21 p23 ) + · · · + Np (p21 p2n ) + · · · + Np (p2n−1 p2n )
+ (3.1) + .. . +
Np (p21 p22 · · · p2n−1 ) + · · · + Np (p22 p23 · · · p2n ) . Since by our induction hypothesis Np (p2i ) = 20 = 1 and Np (p2i p2j p2k ) = 22 = 4 etc., Equation 3.1 can be simplified to X 2 Np (c/d)
      n 0 n 1 n = 2 + 2 + ··· + 2n−2 . 1 2 n−1 16 The binomial theorem tells us that         n 0 n 1 n n n n−1 2 , (1 + 2) = 2 + 2 + ··· + 2 + 0 1 n−1 n n and therefore         n 0 n 1 n (1 + 2)n − 1 n n−1 n−2 2 + 2 + ··· + 2 = − 2 . 1 2 n−1 2 n Therefore, it follows that Np (c2 ) =
(1 + 2)n − 1 X − Np (c/d)2 = 2     (1 + 2)n − 1 (1 + 2)n − 1 n n−1 − − 2 = 2 2 n = 2n−1 . This completes the proof.  The following theorem is the final result for this section. It is proved by induction on the sum of the powers on each prime in the prime factorization of c, and uses Theorem 3.8 as the base case. Theorem 3.9. Suppose c = pe11 pe22 · · · penn where each pi ≡ 1 (mod 4). Then c is the hypotenuse of exactly 2n−1 primitive Pythagorean triangles, or Np (c2 ) = 2n−1 . Proof. We proceed by induction on the sum of the powers on each prime in the prime factorization of c, or rather, we proceed by induction on X ei . 0 l > 0, and one of k and l is even and the other is odd. Solving equations 3.4 and 3.5 for b2 and adding the resulting equations together yields 2b2 = d2 − a2 = (d + a)(d − a). Regardless of whether we are in Case 1 or Case 2, we see that 2b2 = (d + a)(d − a) = 2(k 2 − l2 )2(2kl). Using the fact that b is even, it follows that  2 b = (k 2 − l2 )kl. 2 Since gcd(k, l) = 1, it is clear that gcd(k 2 − l2 , k) = 1 and gcd(k 2 − l2 , l) = 1, since any divisor of k divides k 2 but not l2 , and visa versa for l. Thus, k 2 − l2 , k and l are pairwise relatively prime, and so by Lemma 2.3 each one must itself be a square, since their product is a square. Let k = r2 , l = s2 and k 2 − l2 = t2 and note that t2 = k 2 − l2 = (k + l)(k − l) = (r2 + s2 )(r2 − s2 ). Since one of k and l is even and the other is odd, we know that t2 is odd. Because of this, (k + l) and (k − l) are relatively prime, since any divisor of the two of them would have to divide their sum and difference, which are 2k and 2l respectively. But since t2 is odd, we know that the divisor cannot be 2, thus the divisor would have to divide both k and l, meaning that the divisor is simply 1. It follows that r2 + s2 and r2 − s2 are relatively prime. Thus, by Lemma 2.3, r2 + s2 and r2 − s2 are squares, since their product is t2 . Thus, r2 + s2 = x2 and r2 − s2 = y 2 for some positive integers x and y. Therefore, (y, s, r) and (r, s, x) are two Pythagorean triples where y 2 + s2 = r2 and r2 + s2 = x2 . In other words, the numbers y, s, r and x yield another solution to equations 3.4 and 3.5. However, x2 = s2 + r2 = k + l < 2k ≤ 2kl ≤ d+a < d ≤ d2 . 2 22 Since a, b, c and d was taken at the beginning to be the minimal solution, the fact that x2 < d2 is a contradiction. Therefore, given a Pythagorean triple (a, b, c), it is never the case that there is exists another Pythagorean triple (c, b, d). This completes the proof.  4. Pythagorean Triples in 3-Dimensions 4.1. The Euler Brick. We now move our discussion to considering how we can fit three Pythagorean triangles together in 3-dimensions. Consider a cuboid (right parallelepiped) with side lengths a, b, and c, and face diagonals dab , dac , and dbc , where dab is the face diagonal across the face with sides a and b, and so on, as in Figure 6. Figure 6. Pictorial Example of a Cuboid From the faces of this cuboid, we find three equations: (4.1) a2 + b2 = d2ab (4.2) a2 + c2 = d2ac (4.3) b2 + c2 = d2bc When all six values, a, b, c, dab , dac , and dbc , are integers, the cuboid is called an Euler Brick, and the three equations above are Pythagorean triples. SYSTEMS OF PYTHAGOREAN TRIPLES 23 Euler Bricks have not yet been fully classified, but there are some things we know about them. We call an Euler Brick primitive if all three Pythagorean triples are primitive. Theorem 4.1. There are no primitive Euler Bricks. Proof. We proceed by contradiction. Suppose we have a primitive Euler Brick, and that in Equation 4.1, b is the even term. Then by Theorem 2.5, both a and c must be odd. This is a contradiction because both a and c cannot be odd in Equation 4.2 by Theorem 2.5. Next, let a be the even term in 4.1. Then by Theorem 2.5, both b and c must be odd. This is a contradiction because both b and c cannot be odd in Equation 4.2. Therefore, there are no primitive Euler Bricks. This completes the proof.  In 1740, Nicholas Saunderson found a formula for generating different Euler Bricks, although perhaps not all Euler bricks [4]. Theorem 4.2. If (x, y, z) is a Pythagorean triple, then (a, b, c) = (x(4y 2 − z 2 ), y(4x2 − z 2 ), 4xyz) is an Euler Brick with face diagonals dab = z3 dac = x(4y 2 + z 2 ) dbc = y(4x2 + z 2 ). Proof. The proof is just a matter of computation.  Theorem 4.2 proves that there are infinitely many Euler Bricks. The smallest Euler Brick (measured by the longest side) has sides (a, b, c) = (240, 117, 44) and 24 face diagonals dab = 267, dac = 244 and dbc = 125, and was discovered by Paul Halcke in 1719[4]. If the interior diagonal D of an Euler Brick is also an integer, that is, if D= p a2 + b2 + c2 is an integer, then the Euler Brick is called a Perfect Cuboid. It is not known whether any Perfect Cuboids exist, despite many efforts to prove it impossible and many efforts to find an example of one. 4.2. A Tetrahedron of Right Triangles. Consider the following system of equations: (4.4) n21 + c2 = b2 n22 + b2 = a2 n33 + c2 = a2 . These three equations actually describe the edges of a tetrahedron where three of the faces are right triangles. Moreover, with a little algebra, we see that n21 + n22 = n23 , meaning that the fourth face of the tetrahedron is a right triangle as well. Figure 7 shows how this tetrahedron might look. The question is whether or not it is possible to have a tetrahedron like this with integer sides. In a paper [E753], Euler proved that this is indeed possible. Euler, however, was pursuing a different question. He was looking for three integers a, b and c where all three sums and all three differences are squares. However, a solution to this problem also gives us a solution to our tetrahedron problem. SYSTEMS OF PYTHAGOREAN TRIPLES 25 Figure 7. Pictorial Example of System 4.4 Suppose we have three integers a, b and c such that a + b, a − b, a + c, a − c, b + c, and b − c are all squares. Then, (b + c)(b − c) = b2 − c2 = n21 (a + b)(a − b) = a2 − b2 = n22 (a + c)(a − c) = a2 − c2 = n23 is a solution to System 4.4, since the product of two squares is a square. We will not go through the solution that Euler gives for this problem. The interested reader can read How Euler Did It by Ed Sandifer[5] for a complete explanation of Euler’s solution. The result is stated below. Theorem 4.3. Given positive integers f and g, 26 (4f 2 g 2 ((f 4 + 2f 2 g 2 + 9g 4 ) + (f 4 − 2f 2 g 2 + 9g 4 )))2 a= ·(16f 4 g 4 − (f 4 − 2f 2 g 2 + 9g 4 )(f 4 + 2f 2 g 2 + 9g 4 ))2 b = 2(4f 2 g 2 ((f 4 + 2f 2 g 2 + 9g 4 ) + (f 4 − 2f 2 g 2 + 9g 4 ))) ·(16f 4 g 4 − (f 4 − 2f 2 g 2 + 9g 4 )(f 4 + 2f 2 g 2 + 9g 4 )) c = 2(4f 2 g 2 ((f 4 + 2f 2 g 2 + 9g 4 ) − (f 4 − 2f 2 g 2 + 9g 4 ))) ·(16f 4 g 4 − (f 4 − 2f 2 g 2 + 9g 4 )(f 4 + 2f 2 g 2 + 9g 4 )) is a solution to System 4.4. Proof. The proof is just a matter of calculation. We can factor all the sums and differences to see that they are all indeed squares. That is, a+b= (f − g)2 (f − 3g)2 (f + 3g)2 (f + g)2 (f 2 + 2f g + 3g 2 )2 (f 2 − 2f g + 3g 2 )2 , a−b= (f 2 + 9g 2 )2 (f 2 + g 2 )2 (f 4 − 2f 2 g 2 + 9g 4 )2 , a+c= (f 4 + 2f 3 g + 2f 2 g 2 − 6f g 3 + 9g 4 )2 (f 4 − 2f 3 g + 2f 2 g 2 + 6f g 3 + 9g 4 )2 , a−c= (f 4 − 2f 2 g 2 + 9g 2 )2 (f 2 + 2f g + 3g 2 )2 (f 2 − 2f g + 3g 2 )2 , b+c= 16f 2 g 2 (f 2 − 3g 2 )2 (f 2 + 2f g + 3g 2 )2 (f 2 − 2f g + 3g 2 )2 , and b−c= 16f 2 g 2 (f 2 + 3g 2 )2 (f 4 − 2f 2 g 2 + 9g 4 )2 .  Example: Consider the following example. Let f = 2 and g = 1. It follows from Theorem 4.3 that a = 733, 025, b = 488, 000 and c = 418, 304. Here we see that: a + b = 11052 , a − b = 4952 , SYSTEMS OF PYTHAGOREAN TRIPLES 27 a + c = 10732 , a − c = 5612 , b + c = 9522 , b − c = 2642 . We then use these values to solve for n1 , n2 and n3 in our original equation: n21 = (b + c)(b − c) = 63, 165, 763, 584 = 251, 3282 , n22 = (a + b)(a − b) = 299, 181, 650, 625 = 546, 9752 , and n23 = (a + c)(a − c) = 362, 347, 414, 209 = 601, 9532 . In Euler’s solution, due to the complexity of the problem, he makes assumptions that simplify the problem. Thus, it is not at all clear whether Theorem 4.3 will produce every solution to System 4.4. Moreover, many solutions generated by Theorem 4.3 are trivial. For example, if f = g or f = 3g, then a + b = 0, meaning that n2 = 0, from which it follows that b = a and n1 = n3 . Theorem 4.4. Suppose that a, b, c, n1 , n2 and n3 is a solution set to System 4.4. Then A = a2 + b2 − c2 , B = a2 + c2 − b2 , and C = |b2 + c2 − a2 | and N1 = 2cn2 , N2 = 2an1 , and N3 = 2bn3 is also a solution set. Proof. We need to prove that A2 − B 2 , A2 − C 2 and B 2 − C 2 are all squares. To see that A2 − B 2 is square, note that A − B = 2(b2 − c2 ) = 2n21 , and A + B = 2a2 . Thus, A2 − B 2 = (A + B)(A − B) = 22 a2 n21 = N22 . 28 In a similar fashion, we see that A2 − C 2 = 22 b2 n23 = N32 and B 2 − C 2 = 22 c2 n22 = N12 . This completes the proof.  Example: From the example above, we know that a = 733, 025, b = 488, 000, c = 418, 304, n1 = 251, 328, n2 = 546, 975 and n3 = 601, 953 is a solution set to System 4.4. By Theorem 4.4, A= 733, 0252 + 488, 0002 − 418, 3042 = 600, 491, 414, 209, B= 733, 0252 + 418, 3042 − 488, 0002 = 474, 159, 887, 041, C = |488, 0002 + 418, 3042 − 733, 0252 | = 124, 203, 414, 209, N1 = 2 · 418, 304 · 546, 975 = 457, 603, 660, 800, N2 = 2 · 733, 025 · 251, 328 = 368, 459, 414, 400, and N3 = 2 · 488, 000 · 601, 953 = 587, 506, 128, 000 is also a solution set. The calculations can be verified. 5. Conclusion Theorems 4.3 and 4.4 give us some solutions to our original problem, but perhaps not all solutions. As was mentioned, Euler made many simplifications before coming up with his solutions to System 4.4, and therefore potentially threw out many solutions. However, we have shown that there are, at least, some solutions to the problem we set out to solve. Other questions remain unanswered. It is still unknown whether there are any Perfect Cuboids. The solutions to both the Euler Brick and System 4.4 could potentially be classified in much more elegant terms. We leave these, as well as many other questions, for others to pursue. SYSTEMS OF PYTHAGOREAN TRIPLES 29