Slot-wise Addition of Pythagorean Triples
Date: 07/17/2003 at 13:28:12 From: Chester Subject: Slot-wise addition of Pythagorean triples Is it possible to have a primitive Pythagorean triple, (a,b,c) such that a^2+b^2 = c^2, and two other Pythagorean triples, not necessarily primitive (x,y,z) and (p,q,r) with the property that a = x+p, b = y+q, and c = z+r? It is very hard to find an example to verify that it is possible, and when dealing with arguments involving even and odd numbers, you see that there is not a direct contradiction, which leaves the question open. The work mainly deals with arguments involved with even and odd numbers, because in a primitive PT, (a,b,c), only one of a,b is even, and the other is odd, leaving c odd.
Date: 07/17/2003 at 18:11:16 From: Doctor Rob Subject: Re: Slot-wise addition of Pythagorean triples Thanks for writing to Ask Dr. Math, Chester. I think that what you ask is impossible. My argument begins like this: x^2 + y^2 = z^2, p^2 + q^2 = r^2, (x+p)^2 + (y+q)^2 = (z+r)^2. Expand the last, and subtract from it the first two. Divide by 2, and the result is p*x + q*y = r*z. Now we know that p, q, r, x, y, and z satisfy p = k*(u^2-v^2), q = k*(2*u*v), r = k*(u^2+v^2), u > v > 0, u + v odd, GCD(u,v) = 1, k >= 1, x = K*(s^2-t^2), y = K*(2*s*t), z = K*(s^2+t^2), s > t > 0, s + t odd, GCD(s,t) = 1, K >= 1. (Alternatively, the formulas for x and y might be swapped, which leads to a separate case.) Substituting, that implies that k*K*[(u^2-v^2)*(s^2-t^2) + (2*u*v)*(2*s*t)] = k*K*(u^2+v^2)*(s^2+t^2), (u^2-v^2)*(s^2-t^2) + (2*u*v)*(2*s*t) = (u^2+v^2)*(s^2+t^2), s^2*u^2 - s^2*v^2 - t^2*u^2 + t^2*v^2 + 4*s*t*u*v = s^2*u^2 + s^2*v^2 + t^2*u^2 + t^2*v^2, 0 = 2*(s^2*v^2 + t^2*u^2 - 2*s*t*u*v), 0 = 2*(s*v - t*u)^2, s*v = t*u. Now notice that s divides t*u, but GCD(s,t) = 1, so s divides u. Write u = d*s, and then v = d*t. Since GCD(u,v) = 1, this implies that d = 1, so s = u and t = v. Now you can easily see that a = x + p = (k+K)*(u^2-v^2), b = y + q = (k+K)*(2*u*v), c = z + r = (k+K)*(u^2+v^2), so GCD(a,b,c) = k + K >= 2, and so (a,b,c) is not a primitive triple. This contradiction eliminates this case. I leave the other case to you. It is simpler, and can be eliminated by parity considerations. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.