-copied Paul's perfect solution:
Odd squares are all = 1 mod 8
The squares of numbers not divisible by 3 are = 1 mod 3.
if ab = 24n - 1 then neither a nor b can be multiples of 2 or 3, since the RHS clearly isn't.
so ab = -1 mod 24
We can safely reduce both a and b mod 24 to find a solution, and then consider:
a*(24-a) = 24a - a^2 = -a^2 mod 24
but a^2 = 1 mod 8 and 1 mod 3 from the above, so a^2 = 1 mod 24 and (a, -a) is a solution mod 24. The sum of the two factors is indeed 0 mod 24.
Adding multiples of 24 to either factor preserves the fact that the product = -1 mod 24, and preserves the fact that the sum = 0 mod 24.
There aren't other solutions because if a*(24-a) = -1 mod 24 and a*(b) = -1 mod 24 then b = 24-a, since a has no factors in common with 24.
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