7 In the subset $S$ of the set $X=\{1,2, \cdots, 50\}$, the sum of the squares of any two elements is not a multiple of 7. Find the maximum value of $|S|$.
7. Divide $X$ into 2 subsets: $A_{1}=\{x \mid x \equiv 0(\bmod 7), x \in X\}, A_{2}=$ $X \backslash A_{1}$. Then $\left|A_{1}\right|=7,\left|A_{2}\right|=43$. Clearly, $S$ can contain at most 1 number from $A_{1}$, so $|S| \leqslant 43+1=44$. On the other hand, for any integer $x$, if $x \equiv 0, \pm 1, \pm 2, \pm 3(\bmod 7)$, then $x^{2} \equiv 0, 1, 4, 2(\bmod 7)$. Therefore, if $x^{2}+y^{2} \equiv 0(\bmod 7)$, then $x^{2} \equiv y^{2} \equiv 0(\bmod 7)$. Thus, let $S=A_{2} \cup\{7\}$, then $S$ meets the requirement, and at this point $|S|=44$. Hence, the maximum value of $|S|$ is 44.