verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.
I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity laws to solve this problem. But i get a fraction:
$8x^2 - 2x =  2((2x - \frac{1}{2})^{2} - \frac{1}{4}) $. So the previous statement is equivalent to proving:
$(2x - \frac{1}{2})^2 \equiv 3\frac{1}{2} \cdot \frac{1}{2} \mod 75$. which obviously does not help me further.
Any tips hints on this problem?

Multiply both sides by $8$:
$$(8x-1)^2\equiv 25\pmod{75}$$
So $5\mid 8x-1$. I.e. $3x\equiv 1\equiv 6\pmod{5}\stackrel{:3}\iff x\equiv 2\pmod{5}$. Assuming this,
$$\iff \left(\frac{8x-1}{5}\right)^2\equiv 1\pmod{3}$$
$$\iff \frac{8x-1}{5}\equiv \pm 1\pmod{3}$$
$$\iff 8x-1\equiv \pm 5\equiv \pm 2\pmod{3}$$
$$\iff \left(2x\equiv 0\pmod{3}\, \text{ or }\, 2x\equiv 2\pmod{3}\right)$$
$$\iff \left(x\equiv 0\pmod{3}\, \text{ or }\, x\equiv 1\pmod{3}\right)$$
By Chinese Remainder Theorem we'll surely have a solution. In this case, $x\equiv 12\pmod{15},\, x\equiv 7\pmod{15}$ are all the solutions.
