, 2010

Let $1 + 2^2 + 3^3 + 4^n = x^2$. This implies $32 = x^2 - 4^n$ or, equivalently, $2^5 = (x - 2^n)(x + 2^n)$. As the l.h.s. is a power of 2, the factors in the r.h.s. are of the form $x - 2^n = 2^a$ and $x + 2^n = 2^{5-a}$ where $a$ is 0, 1 or 2. Subtracting the first of the two equalities from the second gives $2^{n+1} = 2^{5-a} - 2^a$. This leads to an integral $n$ only if $a = 2$; then $n = 1$. A check shows that $1 + 2^2 + 3^3 + 4^1 = 6^2$ indeed.

Observe that $1 + 2^2 + 3^3 + 4^n = 32 + 4^n = 2^5 + 2^{2n} = 2^5 \cdot (1 + 2^{2n-5})$. If $n \ge 3$, then $2n - 5 \ge 1$; hence $2^{2n-5}$ is an even integer and $1 + 2^{2n-5}$ is therefore odd. Thus in the prime factorization of the number given in the problem, the exponent of 2 is 5. As this is odd, the number cannot be a perfect square. If $n = 2$ or $n = 1$, then $32 + 4^n = 48$ or $32 + 4^n = 36$, respectively, where only the latter is a perfect square. Consequently, only $n = 1$ is possible.

If $n = 1, 2, 3$, then the given number is 36, 48, 96, respectively, where only the first is a perfect square. If $n \ge 4$, then $2 \cdot 2^n + 1 \ge 2 \cdot 16 + 1 > 32$, implying $(2^n)^2 < (2^n)^2 + 32 < (2^n)^2 + 2 \cdot 2^n + 1 = (2^n + 1)^2$. As the number under question is equal to $(2^n)^2 + 32$, it falls between two consecutive perfect squares, hence cannot be a perfect square itself.