# Solution to puzzle 103 (Sum of two squares)

Show that a^{2} + b^{2} = 3c^{2} has no solution in positive integers.

Suppose that the greatest common divisor of a, b, c is 1. (If not, we can divide a, b, c by their greatest common divisor.)

A perfect square is congruent to 0 or 1 (modulo 4.)

Hence the equation can be satisfied only if a^{2} = b^{2} = c^{2} = 0 (mod 4.)

Then a, b, c are even; a contradiction.

Therefore a^{2} + b^{2} = 3c^{2} has no solution in positive integers.

## Remarks

Another way to formulate the above argument is to note that if (a, b, c) is an integer solution, then so is (½a, ½b, ½c). It follows that ((½)^{n}a, (½)^{n}b, (½)^{n}c) would be a solution for any integer n, which is impossible, for we could choose n such that 2^{n} > a, b, and c.)

We could also derive a contradiction by noting that, of the set of solutions in positive integers, there is a subset of solutions in which c is minimal. But then, as above, if (a, b, c) is an integer solution, so is (½a, ½b, ½c), with ½c < c. Contradiction.

Source: Traditional

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