APMO 2013 Problem 2

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Phlembac Adib Hasan
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APMO 2013 Problem 2

Unread post by Phlembac Adib Hasan » Thu May 09, 2013 11:12 am

Determine all positive integers $n$ for which $\dfrac{n^2+1}{[\sqrt{n}]^2+2}$ is an integer.
Here $[r]$ denotes the greatest integer less than or equal to $r$.
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SANZEED
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Re: APMO 2013 Problem 2

Unread post by SANZEED » Thu May 09, 2013 11:47 pm

It's easy to check that $n$ can't be a perfect square. Then let $n=m^{2}+r$, where $1\leq r\leq 2m$.
Then, the expression becomes,
$\displaystyle\frac{(m^{2}+r)^{2}+1}{m^{2}+2}=\displaystyle\frac{m^{2}(m^{2}+2)+2(r-1)(m^{2}+2)+r^{2}-4r+4+1}{m^{2}+2}$
$=m^{2}+2(r-1)+\displaystyle\frac{(r-2)^{2}+1}{m^{2}+2}$ which is an integer by our hypothesis.
So $\displaystyle\frac{(r-2)^{2}+1}{m^{2}+2}$ will be an integer too. Now,$r-2\leq 2m-2$ implies
$\displaystyle\frac{(r-2)^{2}+1}{m^{2}+2}\leq \displaystyle\frac{4m^{2}-8m+4+1}{m^{2}+1}=4-\displaystyle\frac{8m-3}{m^{2}+2}<4$.
Thus $1\leq \displaystyle\frac{(r-2)^{2}+1}{m^{2}+2}\leq 3$.

So we consider $3$ cases:
If $(r-2)^{2}+1=m^{2}+2$, then $(r-2)^{2}=m^{2}+1$, which is impossible unless $m=0$ and $r-2=\pm 1$, so no solution in this case.

If $(r-2)^{2}+1 = 2m^{2}+4$, then $(r-2)^{2}-2m^{2}=3$. Taking modulo $3$ we get that $(r-2)^{2}+m^{2}\equiv 0 \pmod{3}$ which implies $r-2 \equiv m \equiv 0 \pmod{3}$, and this implies $9 \mid (r-2)^{2} - 2m^{2} = 3$, contradiction.

Finally for the remaining case,we can again take modulo $3$ which will eventually show that there are no solutions in the last case too.

So,no such $n$ exists.
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