Find all such integers $x$ that $x \equiv 3(\bmod 7), x^{2} \equiv 44\left(\bmod 7^{2}\right), x^{3} \equiv 111\left(\bmod 7^{3}\right)$.
$x=7 m+3, x^{2} \equiv 49 m^{2}+42 m+9 \equiv 42 m+9 \equiv 44\left(\bmod 7^{2}\right)$, so, $6 m \equiv 5 \equiv 12(\bmod 7)$, that is $m \equiv 2(\bmod 7) . x$ $=7(7 n+2)+3=49 n+17$

$x^{3} \equiv 3 \cdot 17 \cdot 49 n+17^{3} \equiv 111\left(\bmod 77^{3}\right)$, so, $51 n \equiv 0(\bmod 7)$, that is $n=7 k$.

## Answer

$x \equiv 17(\bmod 343)$.