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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
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    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

2 Answers2

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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.

Bill Dubuque
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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.

miracle173
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