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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?

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Find all $n$ from $0$ to $76,$ inclusive, for which this works. This will be a short finite list. Then, adding a multiple of $77$ to any of these numbers gives another solution. –  Will Jagy Aug 10 '12 at 20:19
Observe that $(8)(3^2)+5=77$. This yields infinitely many solutions. For another bunch, use $(8)(-3)^2+5=77$. There are less crude families. –  André Nicolas Aug 10 '12 at 20:27
I changed "$7|77$" to "$7\mid77$". The latter is considered standard, if I'm not mistaken. –  Michael Hardy Aug 10 '12 at 23:20
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2 Answers

up vote 5 down vote accepted

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 $\rm\:\pm(3,3),\, \pm(-3,3)\ mod\ (7,11)\:$

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If $n$ solves a polynomial equation $\mod m$ then the whole residue class of $n$ modulo $m$ solves 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 soluions

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