Question P.170: Suppose we start with a Pythagorean triple (a, b, c) of posi......
Suppose we start with a Pythagorean triple (a, b, c) of positive integers, that is, positive integers a, b, c such that a² + b² = c² and which can therefore be used as the side lengths of a right triangle. Show that it is not possible to have another Pythagorean triple (b, c, d) with the same integers b and c; that is, show that b² + c² can never be the square of an integer.
Suppose we did have positive integers a, b, c, d such that both a² + b² = c² and b² + c² = d². If b, c had a common factor k, then k² would divide both a² and d², so k would be a factor of a and d as well, and we could replace a, b, c, d by new integers a_1= a/k, b_1= b/k, c_1= c/k, d_1 = d/k and still have a_1^2 +b_1^2 = c_1^2~and~b_1^2+c_1^2 = d_1^2. So we may assume that gcd(b, c) = 1. In particular, b and c cannot both be even. If c is even and b is odd, then a has to be odd, but then a² ≡ 1 (mod 4) and b² ≡ 1 (mod 4), so c² ≡ 1+1 = 2 (mod 4). However, all even squares are 0 (mod 4), contradiction. So c is odd. If b is odd also, then b² ≡ 1 (mod 4), c² ≡ 1 (mod 4), so d² ≡ 1 + 1 = 2 (mod 4), contradiction.
So far we know that gcd(b, c) = 1 and b is even, c is odd (and a is odd). It is known (for a reference, see Problem 118, Solution 2) that under these conditions, the positive integers a, b, c with a² + b² = c² must have the form a = s² − t², b = 2st, c = s² + t² for some positive integers s, t, where gcd(s, t) = 1 and s, t are of opposite parity. Similarly, because b² + c² = d², we have b = 2vw, c = v^2 − w^2, d = v^2 + w^2 for some positive integers v, w, where gcd(v, w) = 1~and~v, w are of opposite parity.
In particular, s^2 + t^2 = v^2 − w^2,~so~v^2 = s^2 + t^2 + w^2. Because s, t are of opposite parity, s²+ t² ≡ 1 (mod 4) and so w cannot be odd lest v^2≡ 2 (mod 4). Consider the equations
s^2 + t^2 = c = v^2 − w^2\\2st = b = 2vw.
If we remove the condition that a > 0, then with respect to these equations, there is no loss of generality in assuming that s is odd and t is even. Let z = gcd(s, v)~and~y = gcd(t, w). Note for later reference that y is even and z is odd. Then from st = vw,
Because gcd(s/z, v/z) = gcd(t/y, w/y) = 1, we can set
x={\frac{s}{z}}={\frac{w}{y}}\,,\\u={\frac{v}{z}}={\frac{t}{y}}\,.Note that x and u are odd and relatively prime. Substituting into
s^2 + t^2 = c = v^2 − w^2,\\we~get\\x^2 z^2 + u^2 y^2 = u^2 z^2 − x^2 y^2,\\or\\x^2(y^2 + z^2) = u^2(z^2 − y^2).Now y² + z² and z² − y² are relatively prime, because any common factor would be an odd factor of both 2y² and 2z². Therefore,
x^2 = z^2 − y^2,~~~~that~is,~~~~ x^2 + y^2 = z^2, ~~~~and\\y^2 + z^2 = u^2.
Observe that u < t < c < d. Thus we would have a new “double Pythagorean triple” whose largest integer is less than the largest integer in the original one. Repeating this process, we arrive at a contradiction, because the sequence of largest (positive) integers cannot decrease indefinitely. We conclude that we can never have had a “double Pythagorean triple” in the first place!