Romania 15

Find all positive integers $a, b, c$ such that $11^{a}+3^{b}=c^{2}$.
By reasoning modulo 3, we see that $a$ is even. We set $k=a / 2$. We have $3^{b}=\left(c-11^{k}\right)\left(c+11^{k}\right)$. Therefore, for some $u, v$ such that $u \geq v \geq 0$ and $u+v=b, c+11^{k}=3^{u}$ and $c-11^{k}=3^{v}$. Thus, $2 \times 11^{k}=3^{v}\left(3^{u-v}-1\right)$. Then $v=0$, so $2 \times 11^{k}=3^{u}-1$. Let $\omega$ be the order of 3 modulo 11. We have $\omega \mid u$. And if $u>\omega$, by Zsigmondy's theorem, there exists a prime number $p \neq 11$ such that $p \mid 3^{u}$. From $\mathrm{m} \tilde{A}^{a}$ me, $p \neq 2$ because $3^{1}-1$ is a multiple of 2. Therefore, $u=\omega=5$. Thus, $2 \times 11^{k}=242$, which is true only when $k=2$.

Therefore, there is only one solution: $a=4, b=5$ and $c=\frac{3^{5}+1}{2}$.