There exist infinitely many values $n\in \mathbb{N}$ such that $8n^{2}+5$ is divisible by $7$ and $11$, without using diophantine equations Show that there exist infinitely many values $n\in \mathbb{N}$ such that $8n^2+5$ is divisible by $7$ and $11$, without using diophantine equations.
I have $8n^2+5=8(n^2-9)+77$ and since $7\mid 77$ and $11\mid77$, I am searching for those natural numbers $n$ such that $7\mid(n^{2}-9)$ and $11\mid(n^2-9)$. From the first one I get $n=4+7k$ and the second one $n=8+11l$. Then I have to solve the following diophantine equation $7k-11l=4$,  which I know that got infinitely many solutions.
But I would like to know  a way to show this without the diophantine equations. Is that possible?
 A: Hint $\rm\ 7,11\mid (n\!-\!3)(n\!+\!3)$ $\rm\!\iff\! 77\mid \color{#0a0}{n\!-\!3}\ \ or\ \ 77\mid \color{#0a0}{n\!+\!3}\ \ $ $\rm or \ \begin{align}\rm 7\mid \color{#c00}{n\!-\!3}\\ \rm 11\mid \color{#c00}{n\!+\!3}\end{align}\ \ {\rm or}\  \begin{array}{}\ \ \rm 7\mid\color{#c00}{n\!+\!3}\\ \rm 11\mid\color{#c00}{n\!-\!3}\end{array}\ $
$\!\!\begin{align}
\rm So\ \ \ \  \bmod (7,11)\!:\,\ &\rm n\equiv \color{#0a0}{\pm(3,3)},\, \color{#c00}{\pm(-3,3)}\\[.1em]
\iff \rm \bmod\ \ \ \, 77\!:\quad\  &\rm n\equiv\ \  \color{#0a0}{\pm\! 3},\ \ \ \ \ \ \ \color{#c00}{\pm\! 25}
\end{align}\ $ by CRT.
Remark $ $ Above we computed $\sqrt9 \bmod 77,\,$ i.e. all roots of $\,x^2\equiv 9\pmod{\!77}.\,$ More generally see here and here and here for general methods  to compute modular roots of polynomials using CRT = Chinese Remainder Theorem.
A: If $n$ satisfies a polynomial equation $\mod m$ then the whole residue class of $n$ modulo $m$ satisfies this polynomial equation. you can try all residue classes $0,\ldots,m-1$ to check if there exists a residue class that solves the equation.
In your problem $n=3$ is a solution of $$8n^2+5=0 \mod 77$$ and therefore $$n=\cdots-74,3,80,\ldots$$ are also solutions.
