Balti Tee 1994 · Ülesanne 8

Solution: We first show this for odd numbers $a = 2i + 1 \geq 3$. Put $c = 2k + 1$ and $b = 2k$. Then $c^{2} - b^{2} = (2k + 1)^{2} - (2k)^{2} = 4k + 1 = a^{2}$. Now $a = 2i + 1$ and thus $a^{2} = 4i^{2} + 4i + 1$ and $k = i^{2} + i$. Furthermore, $c > b = 2i^{2} + 2i > 2i + 1 = a$.

Since any multiple of a Pythagorean triple (i.e., a triple of integers $(x, y, z)$ such that $x^{2} + y^{2} = z^{2}$) is also a Pythagorean triple, we see that the statement is also true for all even numbers which have an odd factor. Hence only the powers of $2$ remain. But for $8$ we have the triple $(8, 15, 17)$ and hence all higher powers of $2$ are also minimum values of such a triple.