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.

  • 4
    $\begingroup$ If $2n+2 \mid 13n^2+n$ then certainly also $2n+2 \mid 2(13n^2+n) - 13n(2n+2)$ and so we find that $2n+2 \mid -24n$. This should be easier $\endgroup$
    – Mike Daas
    Jun 10, 2020 at 17:10
  • 1
    $\begingroup$ @MikeDaas. Solved. Got $ n=2,3,11$. $\endgroup$
    – Ravindra
    Jun 10, 2020 at 17:19

2 Answers 2


For the sake of future readers, @MikeDaas's method ought to be an answer. It can be simplified to$$2n+2|2(13n^2+n)+(12-13n)(2n+2)=24\implies n+1|12,$$i.e. the nonzero options are $$n\in\{-13,\,-7,\,-5,\,-4,\,-3,\,-2,\,1,\,2,\,3,\,5,\,11\}.$$Now we check which of these actually satisfy $2n+2|13n^2+n$; @Ravi found the positive solutions to be $2,\,3,\,11$, but the negative ones are $-13,\,-5,\,-4,\,-2$.


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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