Find the smallest positive integer x where 149|(x^2-69^3) $$
\text{Find the smallest positive integer x where 149|}\left( \text{x}^2-69^3 \right) 
$$
I almost do not spot any clue about this question.
It is from PUMaC-CHINA, 2019.8.17
 A: This follows the comment by  J. W. Tanner and the answer by sirous.
All congruences are mod $149$.
We have $x^2-69^3 \equiv x^2+36 =x^2+6^2 \equiv (x-6j)(x+6j)$, if $j^2\equiv -1$.
Write $149 = 7^2 + 10^2$. Then $10j \equiv 7$ and so $j\equiv 105$.
Thus $x^2-69^3 \equiv (x-34)(x+34)$ and so $34$ is the smallest positive solution.
A: A minute of mental arithmetic via $\sqrt{69^{\large 3}}\equiv 69\sqrt{\color{#90f}{69}}\,$ then negating & pulling out more squares 
$\!\bmod 149\!:\,\  \color{#90f}{69}\equiv -80\equiv \color{#0a0}{-5}\cdot 4^{\large 2}\equiv \color{#0a0}{144}(4^{\large 2})\equiv \color{#0a0}{12^{\large 2}} 4^{\large 2}\equiv 48^{\large 2} $ 
${\rm therefore}\ \ \ \ 69\equiv 48^{\large 2}\,\overset{\large \times\, 69^{\Large 2}}\Longrightarrow\,69^{\large 3}\equiv {\underbrace{(69\cdot 48)}_{\large  23\,\cdot\, \underbrace{3\,\cdot\, 48}_{\LARGE {\bf -}5\ \ }}}^{\large 2}\equiv \color{#c00}{34^{\large 2}}$
A: $69^3=2204 \times 149 +113$
⇒$x^2-69^3 ≡ (x^2-113 )\ mod 149≡ (x^2+36 )\mod 149$
$x^2+36=149 k$
If $k=8$ then:
$x^2=1192 -36=1156=34^2$
⇒ $x= ±34$
