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?
 A: Hint:
$$
8x^2-2x-3=(4x-3)(2x+1)
$$
Is $2$ invertible mod $75$?

Since a full answer has been given in another post, I will add the following.
Indeed $2\cdot38\equiv1\pmod{75}$, so there are solutions.

Although not required by the question, we can find all the solutions as follows.
$$
\begin{align}
8x^2-2x-3
&=(4x-3)(2x+1)\\
&\equiv8(x-57)(x-37)\pmod{75}
\end{align}
$$
Besides the obvious
$$
x\equiv57\pmod{75}\\
x\equiv37\pmod{75}
$$
we also have
$$
\left.\begin{align}
x&\equiv57\pmod{3}\\
x&\equiv37\pmod{25}
\end{align}\right\}x\equiv12\pmod{75}
$$
$$
\left.\begin{align}
x&\equiv57\pmod{25}\\
x&\equiv37\pmod{3}
\end{align}\right\}x\equiv7\pmod{75}
$$
However, all of the preceding are contained in
$$
\left.\begin{align}
x&\equiv57\pmod{15}\\
x&\equiv37\pmod{5}
\end{align}\right\}x\equiv12\pmod{15}
$$
$$
\left.\begin{align}
x&\equiv57\pmod{5}\\
x&\equiv37\pmod{15}
\end{align}\right\}x\equiv7\pmod{15}
$$

Therefore, all solutions are either $x\equiv7\pmod{15}$ or $x\equiv12\pmod{15}$.

A: 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.
