TITLE: Can we prove a stronger claim?
QUESTION [12 upvotes]: Here
it is asked whether for every $k>1$, there is a prime $p$ and $w_1,w_2,\cdots ,w_k>1$, such that $p$ divides $w_1^2+w_2^2+\cdots +w_k^2$, but none of the summands. 

Can we prove there is a solution for every $k$, if the $w$s must be pairwise distinct ? The prime $p$ need not be the same for different $k$s.

REPLY [2 votes]: $$
\text{One more solution . . .}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;
$$
\begin{align*}
\text{Claim:}\;&\text{The prime $p=17$ works for all integers $k > 1$.}\\[8pt]
\text{Case$\;(1)\text{:}\;$}&\text{$k$ is even.}\\[8pt]
&\text{Choose integers}\;x_1,...,x_k,\;\text{all distinct, such that}\\[8pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k}{2}}}\;\text{of them are}\;\equiv 4\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k}{2}}}\;\text{of them are}\;\equiv 1\;(\text{mod}\;17)
\\[8pt]
&\text{Then}\;17\;\text{doesn't divide any of $x_1,...,x_k$}\\[4pt]
&\text{but}\;\;\sum_{i=0}^k x_i^2 \equiv 
\left({\small{\frac{k}{2}}}\right)4^2 
+
\left({\small{\frac{k}{2}}}\right)1^2 
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;
\left({\small{\frac{k}{2}}}\right)(4^2+1^2)\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;
\left({\small{\frac{k}{2}}}\right)(0)\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;0\;(\text{mod}\;17)\\[4pt]
&\text{which proves the claim for case $(1)$.}\\[8pt]
\text{Case$\;(2)\text{:}\;$}&\text{$k$ is odd, $k \ge 3$.}\\[8pt]
&\text{Choose integers}\;x_1,...,x_k,\;\text{all distinct, such that}\\[8pt]
&\qquad{\small\bullet}\;\;\text{one of them is}\;\equiv 6\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k+1}{2}}}\;\text{of them are}\;\equiv 4\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k-3}{2}}}\;\text{of them are}\;\equiv 1\;(\text{mod}\;17)
\\[8pt]
&\text{Then}\;17\;\text{doesn't divide any of $x_1,...,x_k$}\\[4pt]
&\text{but}\;\;\sum_{i=0}^k x_i^2 \equiv 
6^2 + \left({\small{\frac{k+1}{2}}}\right)4^2
+
\left({\small{\frac{k-3}{2}}}\right)1^2
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + \left({\small{\frac{k+1}{2}}}\right)16
+
\left({\small{\frac{k-3}{2}}}\right)1
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + \left({\small{\frac{k+1}{2}}}\right)16
+
\left({\small{\frac{k-3}{2}}}\right)18
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + (8k+8) + (9k-27)
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
17k-17\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;0\;(\text{mod}\;17)\\[4pt]
&\text{which proves the claim for case $(2)$.}\\[8pt]
&\text{Thus, the proof of the claim is complete.}\\[8pt]
\end{align*}