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.
Best Answer
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.$
$\implies \left[ \begin{align} 2x^2 + y^2 + 2x + y \equiv 2 \cdot 0^2 + 0^2 + 2 \cdot 0 + 0 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-2)^2 + 2^2 + 2 \cdot (-2) + 2 &\equiv 0 \text{ (mod 5)}\\ 2x^2 + y^2 + 2x + y \equiv 2 \cdot (-1)^2 + (-1)^2 + 2 \cdot (-1) + (-1) &\equiv 0 \text{ (mod 5)} \end{align} \right.$
$\iff 5 \mid 2x^2 + y^2 + 2x + y$