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    If x^2 + y^2 = z^2 where x,y,z are different integers is the product xyz always a multiple of 60?

    Yes. Solution in next post.

    We wish to prove that given x^2 + y^2 = z^2, the product xyz is always a multiple of 60. If we prove that the product xyz is divisible by 5, 4 and 3 we are done.

    I shall approach this question by using modular arithmetic.
    The possible congruences of a number when taken to modulo 3 is 0, 1 and 2. The square of a number therefore has possible congruences of 0*0 = 0 mod 3, 1*1 = 1 mod 3 and 2*2 = 1 mod 3. To prove that at least one of x,y and z is 0 mod 3, use method of contradiction. Assume z is 1 mod 3. If so, then to maintain a consistent equation, at least one of x and y is 0 mod 3, which leads to a contradiction. Otherwise z is 0 mod 3 and we are done.

    Similary, the possible congruences of a square number taken to mod 4 is 0*0 = 0 mod 4, 1*1 = 1 mod 4, 2*2 = 0 mod 4, 3*3 = 1 mod 4. Again the possible remainders are 0 and 1 mod 4 and it can be argued in similar fashion that at least one of x,y and z is 0 mod 4.

    The possible congruences of a square number taken to mod 5 is 0*0 = 0 mod 5, 1*1 = 1 mod 5, 2*2 = 4 mod 5, 3*3 = 4 mod 5, 4*4 = 1 mod 5. Now, assume that z = 1 mod 5. Again to maintain a consistent equation, either of x and y has to be 0 mod 5. It can be shown that the same case occurs for z = 4 mod 5. If z = 0 mod 5 we are done.

    Therefore, xyz is divisible by 3*4*5 = 60.
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Updated: August 27, 2004
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