23 Find the number of integer pairs $(a, b)$ that satisfy the following conditions: $0 \leqslant a, b \leqslant 36$, and $a^{2}+b^{2}=$ $0(\bmod 37)$.
23. Notice that, $a^{2}+b^{2} \equiv a^{2}-36 b^{2}(\bmod 37)$, so from the condition we have
$$37 \mid a^{2}-36 b^{2},$$

which means
$$37 \mid(a-6 b)(a+6 b),$$

thus
$37 \mid a-6 b$ or $37 \mid a+6 b$.
Therefore, for each $1 \leqslant b \leqslant 36$, there are exactly two $a(a \equiv \pm 6 b(\bmod 37))$ that satisfy the condition, and when $b=0$, from $a^{2}+b^{2} \equiv 0(\bmod 37)$ we know $a=0$. Hence, the number of pairs $(a, b)$ that satisfy the condition is $2 \times 36+1=73$ (pairs).