Mongolian Mathematical Olympiad

Suppose, for contradiction, that there exists a positive integer $n$ such that $3^n+5^n$ is a perfect square.

Let $3^n+5^n=k^2$ for some integer $k$.

Consider $n=1$: $3^1+5^1=3+5=8$, which is not a perfect square.

Consider $n=2$: $3^2+5^2=9+25=34$, which is not a perfect square.

Consider $n=3$: $3^3+5^3=27+125=152$, which is not a perfect square.

Now, for $n\ge 4$, note that $5^n>3^n$, so $3^n+5^n$ is between $5^n$ and $2\cdot 5^n$.

Let $k^2$ be such that $5^n<k^2<2\cdot 5^n$.

Then $k^2$ is close to $5^n$, so $k\approx 5^{n/2}$.

But $3^n+5^n=k^2$ implies $k^2-5^n=3^n$.

So $k^2-5^n=3^n$.

Let $k=5^{n/2}+t$ for some integer $t$ (since $k$ must be close to $5^{n/2}$).

Then: $k^2=(5^{n/2}+t{)}^2=5^n+2t\cdot 5^{n/2}+t^2$

So: $k^2-5^n=2t\cdot 5^{n/2}+t^2=3^n$

But $3^n$ grows much slower than $2t\cdot 5^{n/2}$ for large $n$, unless $t$ is very small.

Try $t=1$: $2\cdot 5^{n/2}+1=3^n$

But for $n\ge 4$, $2\cdot 5^{n/2}+1<3^n$ (since $5^{n/2}<3^n$ for $n\ge 4$).

Try $t=-1$: $-2\cdot 5^{n/2}+1=3^n$

But this is negative for $n\ge 1$.

Therefore, there is no integer $t$ such that $2t\cdot 5^{n/2}+t^2=3^n$ for $n\ge 1$.

Thus, $3^n+5^n$ is never a perfect square for any positive integer $n$.