70th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

If $d$ divides $3n^2$, then there exist positive integers $k$ and $m$ such that $3n^2=d\cdot k$ and $n^2+d=m^2$. We substitute to get $n^2+\frac{3n^2}{k}=m^2$, so $(mk{)}^2=n^2(k^2+3k)$. We deduce that $k^2+3k$ is a perfect square. From the inequalities $k^2<k^2+3k<(k+2{)}^2$ we deduce that $k^2+3k=(k+1{)}^2$ which implies $k=1$ and $d=3n^2$ which verifies the problem.

Alternative Solution: Let $d$ be a divisor of $3n^2$ such that $n^2+d=m^2$. We have $m>n$ and $d=(m-n)(m+n)$. Denote $D=(m,n)$, $m=Da$ and $n=Db$. Since $(m-n)(m+n)$ divides $3n^2$, we have $(a-b)(a+b)\mid 3b^2$. The numbers $a$ and $b$ are co-prime, therefore $(a-b)(a+b)\mid 3$. The case $a-b=a+b=1$ implies $b=0$, which is not possible. The only possibility is $a-b=1$ and $a+b=3$, meaning that $a=2$ and $b=1$. Finally we have $m=2n$ and $d=(2n{)}^2-n^2=3n^2$.