For what $x$ the expression of the equation has the same remainder after division. For what $x$ does the expression $(50x+7)(3x+11)$ have the same remainder as $(26x − 86)x$ when divided by $111$?
$(50x+7)(3x+11) \equiv y \mod 111$
$(26x − 86)x \equiv y \mod 111$
$(50x+7)(3x+11) - (26x − 86)x \equiv 0 \mod 111 $
$124x^2+657x+77 \equiv 0 \mod 111$
$111 = 3 \times 37$
There is a system

*

*$124x^2+657x+77 \equiv 0 \mod 3$

*$124x^2+657x+77 \equiv 0 \mod 37$
First:
$(11x+12)^2\equiv 1 \mod 3$
And $x\equiv -1 \mod 3$
But I don't know how to solve the second equation and what to do next.
How to combine these two equations and get at what $x$ these expressions have the same remainder from division.

First you have $x= 3y\pm 1$ for some integer $y$.
You can reduce the coeficents mod $37$ so \begin{align}-20x^2+30x+40 &=_{37} -5(4x^2-6x-8)\\
&=_{37} -5(4x^2+68x-8) \\&=_{37}-5 (4x^2+68x +289)+5\\
&=_{37}-5 (2x+17)^2+5\\
\end{align}
So $$(2x-20)^2 =_{37} 1 \implies 2x-20 =_{37}\pm1 $$
So you have 4 possibilites.

*

*First $x=3y+1$ and $2x-20 =_{37} 1$ and thus $ 6y =_{37} 19$ so $$y =_{37}-6\cdot 19 =_{37}-3$$ and finaly $y= 37t-3$ for some integer $t$ which gives $$x=111t-8$$ in first case.

