<s>[INST] When the quotient $\frac {13n^2+n}{2n+2}$ is an integer? Find the nonzero values of integer $n$ for which the quotient $\frac {13n^2+n}{2n+2}$ is an integer?
My Attempt
I assumed $(2n+2 )| (13n^2+n)$ implies existence of integer $k$ such that $$13n^2+n=k(2n+2)$$
$\implies\ 13n^2+(1-2k)n-2k=0$
$\implies\ n=\frac{(2k-1)\pm\sqrt{(1-2k)^2+104}}{26}$
But this is taking me nowhere.
 [/INST] You made a mistake in the last line of your post. The $n^2$ shouldn’t be there. You need: $(1-2k)^2+104=h^2\implies h^2-(1-2k)^2=104\implies (h+1-2k)(h-1+2k)=104=1\cdot 104= 2\cdot 52 = 4\cdot 26= 8\cdot 13= (-1)(-104) = (-2)(-52) = (-4)(-26) = (-8)(-13)$. Those are the possibilities for them.
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