If $x$ and $y$ are integers such that $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$. 
Given that $x$ and $y$ are integers satisfying $5 \mid x^2 - 2xy - y$ and $5 \mid xy - 2y^2 - x$, prove that $5 \mid 2x^2 + y^2 + 2x + y$.

I have provided a (dumbfounding) solution down below if you want to check out. There should be simpler solutions, I believe so.
 A: We have that $\left\{ \begin{align} 5 &\mid x^2 - 2xy - y\\ 5 &\mid xy - 2y^2 - x \end{align} \right.$$\iff \left\{ \begin{align} 5 &\mid (x^2 - 2xy - y) + (xy - 2y^2 - x)\\ 5 &\mid (x^2 - 2xy - y) - (xy - 2y^2 - x) \end{align} \right.$
$\iff \left\{ \begin{align} 5 &\mid (x + y)(x - 2y - 1)\\ 5 &\mid (x - y)(x - 2y + 1) \end{align} \right.$$\iff \left\{ \begin{align} 5 \mid x + y &\text{ or } 5 \mid x - 2y - 1\\ 5 \mid x - y &\text{ or } 5 \mid x - 2y + 1 \end{align} \right.$
$\implies \left[ \begin{align} 5 \mid x + y &\text{ and } 5 \mid x - y\\ 5 \mid x + y &\text{ and } 5 \mid x - 2y + 1\\ 5 \mid x - y &\text{ and } 5 \mid x - 2y - 1 \end{align} \right.$$\iff \left[ \begin{align} x + y &\equiv x - y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } x \equiv 2y - 1 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } x \equiv 2y + 1 \text{ (mod 5)} \end{align} \right.$
$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } -y \equiv 2y - 1 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } y \equiv 2y + 1 \text{ (mod 5)} \end{align} \right.$
$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text{ (mod 5)}\\ x \equiv -y &\text{ (mod 5) and } y \equiv 2 \text{ (mod 5)}\\ x \equiv y &\text{ (mod 5) and } y \equiv -1 \text{ (mod 5)} \end{align} \right.$$\iff \left[ \begin{align} x &\equiv y \equiv 0 \text { (mod 5)}\\ x \equiv -2 &\text{ (mod 5) and } y \equiv 2 \text{ (mod 5)}\\ x \equiv -1 &\text{ (mod 5) and } y \equiv -1 \text{ (mod 5)} \end{align} \right.$
$\implies \left[ \begin{align} 2x^2 + y^2 + 2x + y \equiv 2 \cdot 0^2 + 0^2 + 2 \cdot 0 + 0 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-2)^2 + 2^2 + 2 \cdot (-2) + 2 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-1)^2 + (-1)^2 + 2 \cdot (-1) + (-1) &\equiv 0 \text{ (mod 5)} \end{align} \right.$
$\iff 5 \mid 2x^2 + y^2 + 2x + y$
A: If $5\mid x^2-2xy-y$ and $5\mid xy-2y^2-x$ then $5$ also divides
$$(3y-1)(x^2-2xy-y)-(3x+2)(xy-2y^2-x)=2x^2+y^2+2x+y.$$
A: 1) When $S=x^2-2xy-y$ and $T = xy-2y^2-x$, then $$ 5|S-T  =
(x-y)(x-2y+1) \ (a)$$
 and $$ 5| S+T =  (x-2y-1)(x+y)\ (b)
$$
Hence I know that $(x-2y+1)-(x-2y-1)=2$, by mathlove's comment. 
Hence at least one of $(x-2y+1),\ (x-2y-1)$ can not be divided by $5$. So by $(a),\ (b)$ we have $5|x^2-y^2$.
2) When $U=2x^2+y^2+2x+y$, then $$ U+S+2T= 3(x^2-y^2) $$
So we complete the proof.
