23. (USA 4) Find all natural numbers $n$ for which $2^{8}+2^{11}+2^{n}$ is a perfect square.
23. It is easily verified that no solutions exist for \( n \leq 8 \). Let us now assume that \( n > 8 \). We note that \( 2^{8} + 2^{11} + 2^{n} = 2^{8} \cdot (9 + 2^{n-8}) \). Hence \( 9 + 2^{n-8} \) must also be a square, say \( 9 + 2^{n-8} = x^{2}, x \in \mathbb{N} \), i.e., \( 2^{n-8} = x^{2} - 9 = (x-3)(x+3) \). Thus \( x-3 \) and \( x+3 \) are both powers of 2, which is possible only for \( x=5 \) and \( n=12 \). Hence, \( n=12 \) is the only solution.