Saudi Arabia Mathematical Competitions

For $n=1$ we have $27-2=25=5^2$. We will prove that there are no other positive integers with this property. If $n$ is odd, $n=2k+1$, then $$\begin{array}{rl}27^n-2^n& =27^{2k+1}-2^{2k+1}=(28-1{)}^{2k+1}-2\cdot 4^k\\ & =4m+(-1{)}^{2k+1}=4m-1\end{array}$$ It is easy to see that integers of the form $4m-1$ are not perfect squares. If $n$ is even, $n=2k$, then $27^n-2^n=(27^k{)}^2-2^{2k}$, and we have $$(27^k-1{)}^2<(27^k{)}^2-2^{2k}<27^k,$$ hence $27^{2k}-2^{2k}$ is between two consecutive perfect squares, that is it cannot be a perfect square. The only solution is $n=1$.

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Alternative solution.

(Wael Hussain Al Saeed) Assume that $n\ge 2$. For any integer $x$, we have $x^2\equiv 0,1(\mathrm{mod}3)$ and $x^2\equiv 0,1(\mathrm{mod}4)$. Considering the relation $27^n-2^n=x^2$ modulo 3 we get $$27^n-2^n\equiv (-1{)}^{n+1}(\mathrm{mod}3)$$ hence $(-1{)}^{n+1}\equiv 1(\mathrm{mod}3)$. It follows $n=2k+1$ for some positive integer $k$. Considering the relation $27^n-2^n=x^2$ modulo 4 we obtain $$(-1{)}^n\equiv x^2(\mathrm{mod}4)$$ hence $(-1{)}^n\equiv 1(\mathrm{mod}4)$, that is $n=2m$ for some positive integer $m$, not possible.