Balti Tee 1994 · Ülesanne 9

Solution: Considering the equality $2^{a} + 3^{b} = n^{2}$ modulo $3$, it is easy to see that $a$ must be even. Obviously $n$ is odd so we may take $a = 2x$, $n = 2y + 1$ and write the equality as $4^{x} + 3^{b} = (2y + 1)^{2} = 4y^{2} + 4y + 1$. Hence $3^{b} \equiv 1 \pmod{4}$ which implies $b = 2z$ for some positive integer $z$. So we get $4^{x} + 9^{z} = (2y + 1)^{2}$ and $4^{x} = (2y + 1 - 3^{z})(2y + 1 + 3^{z})$. Both factors on the right-hand side are even numbers but at most one of them is divisible by $4$ (since their sum is not divisible by $4$). Hence $2y + 1 - 3^{z} = 2$ and $2y + 1 + 3^{z} = 2^{2x - 1}$. These two equalities yield $2 \cdot 3^{z} = 2^{2x - 1} - 2$ and $3^{z} = 4^{x - 1} - 1$. Clearly $x > 1$ and a simple argument modulo $10$ gives $z = 4d + 1$, $x - 1 = 2e + 1$ for some non-negative integers $d$ and $e$. Substituting, we get $3^{4d + 1} = 4^{2e + 1} - 1$ and $3 \cdot (80 + 1)^{d} = 4^{2e + 1} - 1$. If $d \geq 1$ then $e \geq 1$, a contradiction (expanding the left-hand expression and moving everything to the left we find that all summands but one are divisible by $4^{2}$). Hence $e = d = 0$, $z = 1$, $b = 2$, $x = 2$ and $a = 4$, and we obtain the classical $2^{4} + 3^{2} = 4^{2} + 3^{2} = 5^{2}$.