Canadian Mathematical Olympiad

Solution:
Suppose there exist integers $a, b, c$ such that $a^{2} + b^{2} - 8c = 6$.

Then $a^{2} + b^{2} = 8c + 6$.

Consider $a^{2} + b^{2}$ modulo $8$.

The possible quadratic residues modulo $8$ are $0, 1, 4$ (since $0^{2} \equiv 0$, $1^{2} \equiv 1$, $2^{2} \equiv 4$, $3^{2} \equiv 1$, $4^{2} \equiv 0$, $5^{2} \equiv 1$, $6^{2} \equiv 4$, $7^{2} \equiv 1$ modulo $8$).

So $a^{2} + b^{2}$ modulo $8$ can be $0+0=0$, $0+1=1$, $0+4=4$, $1+1=2$, $1+4=5$, $4+4=0$.
Thus, the possible values for $a^{2} + b^{2}$ modulo $8$ are $0, 1, 2, 4, 5$.

But $a^{2} + b^{2} = 8c + 6 \equiv 6 \pmod{8}$.

$6$ is not among the possible values for $a^{2} + b^{2}$ modulo $8$.

Therefore, there are no integers $a, b, c$ for which $a^{2} + b^{2} - 8c = 6$.