Solution to puzzle 103 (Sum of two squares)

Show that a2 + b2 = 3c2 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 a2 = b2 = c2 = 0 (mod 4.)
Then a, b, c are even; a contradiction.

Therefore a2 + b2 = 3c2 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)na, (½)nb, (½)nc) would be a solution for any integer n, which is impossible, for we could choose n such that 2n > 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.