Is there any relation between solution of y for $$y^2 \equiv ax \mod bx \text{ and }y^2 \equiv a \pmod b$$ For example $$y^2 \equiv 30 \mod 70 $$ has solutions $10$ and $60$ but $$y^2 \equiv 3 \pmod 7$$ has no solutions.
I understand we can factorize $x$ and then by CRT we can solve for y. But can we do better than that, I mean without factorising $x$ . If $x$ is large non-prime then can we solve this $$y^2 \equiv ax \mod bx \text{ from solutions of } y^2 \equiv a \pmod b$$
$y^2≡ax \mod bx $ => x divides y
Putting $y=xy'$
$y^2≡ax \mod bx $ <=> $x^2y'^2≡ax \mod bx $
<=> $xy'^2≡a \mod b $
<=> $y'^2≡x^{-1}a \mod b $ // inverse of x in mod b
Now find square root of $x^{-1}a$ in mod b
Thanks to @Roddy MacPhee in comments