TITLE: There exist infinitely many values $n\in \mathbb{N}$ such that $8n^{2}+5$ is divisible by $7$ and $11$, without using diophantine equations
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: Hint $\rm\,\ 7,11\:|\:n^2\!-\!9 = (n\!-\!3)(n\!+\!3)\!\iff\! 77\:|\:n\!-\!3\ \ or\ \ 77\:|\:n\!+\!3\ \ or\ \ \begin{eqnarray}7\:|\:n\!&-&\!3\\ \rm 11\:|\:n\!&+&\!3\end{eqnarray}\ \ \ or\ \ \begin{eqnarray}7\:|\:n\!&+&\!3\\ \rm 11\:|\:n\!&-&\!3\end{eqnarray}$
Thus $\rm\,\ mod\ 77\!:\ n \equiv \sqrt{9}\equiv \pm3,\, \pm 25,\ $ i.e $\rm\:\pm(3,3),\, \pm(-3,3)\ mod\ (7,11)\:$