# 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.

• 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 Jun 10, 2020 at 17:10
• @MikeDaas. Solved. Got $n=2,3,11$. Jun 10, 2020 at 17:19

## 2 Answers

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.