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Math Help - Pythagorean Triple

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    Pythagorean Triple

    Question:
    If x, y, z is a primitive Pythagorean triple, prove that x+y and x-y are congruent modulo 8 to either 1 or 7.
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    Quote Originally Posted by cathwelch View Post
    Question:
    If x, y, z is a primitive Pythagorean triple, prove that x+y and x-y are congruent modulo 8 to either 1 or 7.
    If {x, y, z} is a primitive Pythagorean triple then x and y are of the form m^2-n^2 and 2mn, where gcd(m,n) = 1. Then x\pm y = m^2 \pm2mn - n^2 = (m\pm n)^2 - 2n^2. But the square of an odd number is always congruent to 1 (mod 8), and the square of an even number is congruent to 0 or 4 (mod 8). Since m\pm n is odd, (m\pm n)^2 - 2n^2 will be congruent to 1 (mod 8) if n is even, and congruent to 12=1, or equivalently 7, (mod 8) if n is odd.
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