Primitive Pythagorean Triple CongruencesDate: 04/17/2003 at 15:09:28 From: Ronnie Subject: Primitive Pythagorean Triples If x,y,z are primitive Pythagorean triples, prove that x+y and x-y are congruent modulo 8 to either 1 or 7. I'm trying to solve the problem in the general form. By looking at known Pythagorean triples, it is easy to see that for the values of x and y, the assertion holds true. I just can't find the relation for the general case. Date: 05/01/2003 at 12:39:05 From: Doctor Nitrogen Subject: Re: Primitive Pythagorean Triples Hi, Ron: For selected integers u, v, the primitive Pythagorean triples (x, y, z) usually have the form x = uv, y = (u^2 - v^2)/2, z = (u^2 + v^2)/2. It can be proved that 4 | y ("4 divides y"), meaning y = 4m for some positive integer m. I will prove this now: As x = uv is odd, y = (u^2 - v^2)/2 is even, since x and y have opposite parity. This means u^2 - v^2 must be of the form u^2 - v^2 = (2^r)s for some integer r > 1 and for some odd integer s. Otherwise if r = 1, (u^2 - v^2)/2 will be equal to s, which is an odd integer, which is impossible as x is odd and x and y must have opposite parity. Therefore r > 1 and y has the form y = (u^2 - v^2)/2 = 2^2(2^r-2)(s) = 4(2^r-2)(s) = 4m for some positive integer m = (2^r-2)(s) Now note that all the possible congruence classes for the integers 4m + x and 4m - x are 4m + x, 4m - x in turn, is congruent to: 0 modulo 8 1 modulo 8 2 modulo 8 3 modulo 8 4 modulo 8 5 modulo 8 6 modulo 8 7 modulo 8 Now since 4m + x and 4m - x are both odd, and since 8 cannot divide an odd integer, the only possible relevant congruences above would be 4m + x and 4m - x are congruent to 1 modulo 8 3 modulo 8 5 modulo 8 7 modulo 8 or involving the set {1, 3, 5, 7} of integers less than 8 and relatively prime to 8. Now since both 4m + x and 4m - x, with y = 4m, are congruent to either 1, 3, 5, or 7 modulo 8, all that is left for you to do is rule out the congruences 3 and 5 modulo 8. - Doctor Nitrogen, The Math Forum http://mathforum.org/dr.math/ |
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