We have given $k>1$ primes $p_1 < p_2 < \cdots < p_k$ and a integer $n$ s.t. $(1) \;\; p_1^2 + p_2^2 + \ldots + p_k^2 = 2^n$. Let $S_k$ be the sum on the LHS of equation (1). Hence $2^n \geq 2^2 + 3^2 = 4 + 9 = 13 > 8 = 2^3$, yielding $n>3$. Hence, if $p_1 \neq 2$, then $8 \mid S_k - k = 2^n - k$ by equation (1), which means $8 \mid k$, which implies $k \geq 8$. Assume $p_1=2$. Then according to equation (1) $8 \mid 2^2 + (k - 1) = 4 + k - 1 = k + 3$, yielding $k \geq 5$. Consequently $S_k \geq 2^2 + 3^2 + 5^2 + 7^2 + 11^2 = 4 + 9 + 25 + 49 + 121 = 208$. The fact that $208 + 13^2 - 11^2 = 208 + 169 - 121 = 256 = 2^8$ give us $2^2 + 3^2 + 5^2 + 7^2 + 13^2 = 2^8$. Summa summarum, the smallest value of $k$ satisfying equation (1) is 5.