Rewrite the given equation as: $(m-n)(m+n)=2015=5\times13\times31 \therefore$ the cases are: 1) $m-n=1$ and $m+n=2015$ giving the first solution $(m,n)=(1008,1007)$ 2) $m-n=13$ and $m+n=155$ giving the second solution $(m,n)=(84,71)$ 3) $m-n=5$ and $m+n=403$ giving the third solution $(m,n)=(204,199)$ 4) $m-n=31$ and $m+n=65$ giving the fourth solution $(m,n)=(48,17)$ @below thanks for letting me know

I still believe that this problem is too easy for a mathematic olympiad.

Borbas wrote: Rewrite the given equation as: $(m-n)(m+n)=2015=5\times13\times31 \therefore$ the cases are: 1) $m-n=1$ and $m+n=2015$ giving the first solution $(m,n)=(1008,1007)$ 2) $m-n=13$ and $m+n=31$ giving the second solution $(m,n)=(22,8)$ Your solution is wrong.

The second case is $m-n=13$ and $m+n=155$, with solution $m=84$ and $n=71$. And there are also other cases: -$m-n=5$ and $m+n=403$ with solution $m=204$ and $n=199$ -$m-n=31$ and $m+n=65$ with solution $m=48$ and $n=17$

dangerousliri wrote: Find all couples $(m,n)$ of positive integers such that satisfied $m^2+1=n^2+2016$ . $m^2-n^2=2015 \implies (m+n)(m-n)=2015$ Now, $2015=(65)*(31)$ , $2015=(155)*(13)$ , $2015=(403)*5$ Thus the different pairs $((m+n),(m-n))$ are $(65,31),(155,13),(403,5)$ and solving gives us $(m,n) \in \{(48,17),(84,71),(204,199) \}$