For $n>2$ we've $3^{2n}<3^{2n}+3n^2+7<(3^n+1)^2\implies n=2$

Could You please Explain Elaborately... Moreover I also Got the Same answer but there was a fault in my proof so I get a straight 0?

Yes that is the solution I did in the exam. Probably, that is the only comparatively challenging in the paper. Easy to see that $n$ is even. $k^2-3^{2n}=3n^2+7$. Now, $k^2-(3^n)^2\ge 2\cdot 3^n+1$. Since $k\ge 3^n+1\implies k^2-(3^n)^2\ge (3^n+1)^2-(3^n)^2$. However $2\cdot 3^n+1\ge 3n^2+7$ for $n\ge 4$. (Easy to prove by induction.) Only equality holds iff $\boxed{n=2}$.

prat007 wrote: Could you please explain elaborately? He simply used the fact that there cannot be perfect square between $n^2$ and $(n+1)^2$. prat007 wrote: Moreover I also got the same answer but there was a fault in my proof so I get a straight 0? Usually there is no marks in RMO for totally wrong method and correct answer.[Alas it is not MCQ!]

can anybody tell me whether my solution is correct or not $3^{2n}+3n^2+7=k^2$ $3n^2+7=(k-3^n)(k+3^n)$ now since $3n^2+7$ is irreducible over R since its roots are imaginary hence $3n^2+7= k+3^n$ and $1=k-3^n$ thus $6=2.3^n-3n^2$ then by induction i proved that the only solution is n=2

The solution is not true. Is $k+3^n$ a polynomial that you are using the fact of the irreducibility?

Adi2497 wrote: can anybody tell me whether my solution is correct or not $3^{2n}+3n^2+7=k^2$ $3n^2+7=(k-3^n)(k+3^n)$ now since $3n^2+7$ is irreducible over R since its roots are imaginary hence $3n^2+7= k+3^n$ and $1=k-3^n$ thus $6=2.3^n-3n^2$ then by induction i proved that the only solution is n=2 iarnab_kundu wrote: The solution is not true. Is $k+3^n$ a polynomial that you are using the fact of the irreducibility? it is not a polynomial but always gives integer values

That does not matter. Actually one side you are using the fact of an irreducibility of a polynomial and the other side its a mere function. Its not correct. I assure you.