22nd Junior Turkish Mathematical Olympiad

Let $3^a+3^b+3^c=m^2$. Since $m$ is odd we have $m^2\equiv 1(\mathrm{mod}8)$. For each positive integer $k$ we have $3^k\equiv 1$ or $3^k\equiv 3(\mathrm{mod}8)$. Therefore, $3^a\equiv 3^b\equiv 3^c\equiv 1(\mathrm{mod}8)$ and we get that $a$, $b$ and $c$ are odd numbers. Let $a\le b\le c$. Then $3^a(3^{b-a}+3^{c-a}+1)=m^2$. Now since $a$ is odd, we get that $3^{b-a}+3^{c-a}+1\equiv 0(\mathrm{mod}3)$. Thus, $a=b=c$ and we get all solutions: $(a,b,c)=(2k-1,2k-1,2k-1)$, where $k$ is a positive integer.