# Primes dividing sums of squares but not dividing any of summands

I am inexperienced when it comes to number theory (even elementary one) so do not have an idea at this moment on how to solve this one.

Let us go through some examples.

For $k=2$ we can have prime that divides $n_1^2 + n_2^2$ but that does not divide neither $n_1$ nor $n_2$. Take $n_1=3$ and $n_2=5$. That prime here is $p=17$.

For $k=3$ we can take as an example $n_1=3$ and $n_2=5$ and $n_3=7$. Then the prime is $p=83$.

For $k=4$ we can give an example $n_1=n_2=3$ and $n_3=n_4=5$. Then the prime is $p=17$.

We will not allow that our prime $p$ be equal to $2$ so we will search for counterexamples that have $p\neq 2$.

Question is:

Is it true that for every $k \in \mathbb N \setminus \{1\}$ there exist natural numbers $w_1,w_2,...w_k$ (all different from $1$) such that there is a prime $p \neq 2$ that divides $\sum_{i=1}^k w_i^2$ but does not divide any of the numbers $w_1,w_2,...w_k$?

I am not sure in my fluency of English, so when I say that "$p$ does not divide any of the numbers $w_1,w_2,...w_k$" I mean that $p$ does not divide $w_1$ and $p$ does not divide $w_2$ and ... and $p$ does not divide $w_k$.

• This seems obvious. Have you tried induction on $k$? – Crostul Oct 1 '17 at 12:15
• @Crostul No, I tend to avoid induction-based proofs, if possible. – user480281 Oct 1 '17 at 12:16
• For when $k$ is not a power of $2$, take $w_i=1~\forall~1\leq i\leq k$ and $p$ an odd prime divisor of $k$. – Prasun Biswas Oct 1 '17 at 12:19
• @PrasunBiswas I will edit to avoid trivial counterexamples. – user480281 Oct 1 '17 at 12:20
• @Peter I was thinking about some variants of that, not exactly of that but of something approximately similar. – user480281 Oct 1 '17 at 13:00

We have $$131=3^2+4^2+5^2+9^2=5^2+5^2+9^2$$ If $k\ne 5$ and $k>2$, there are non-negative integers $a$ and $b$ wirh $3a+4b=k$ Use the summands of the first represenation $b$ times and the summands of the second representaition $a$ times.

Moreover, $131|7^2+9^2+4^2+4^2+10^2$ , so the cases $k=5$ is solved as well.

• But does such representation allow number $1$? I am sure that it is. – user480281 Oct 1 '17 at 12:30
• 17=4^2+1^2? Or I did not understand your answer. – user480281 Oct 1 '17 at 12:33
• Yes, this would be an allowed representation and $17$ does neither divide $1$ nor $4$ – Peter Oct 1 '17 at 12:34
• But I do not allow that summands take value 1, I made an edit to avoid repetitions of 1 and trivial counterexamples. – user480281 Oct 1 '17 at 12:36
• I made an edit a little bit before this answer. – user480281 Oct 1 '17 at 12:36

We will prove a more stronger claim

Question(I): Suppose that the natural number $k \geq 3$ and a prime number $p \geq 7$ are given; then there exist natural numbers $w_1,w_2,...w_k$ (all different from $1$) such that $p$ divides $\sum_{i=1}^k w_i^2$ but does not divide any of the numbers $w_1,w_2,...w_k$?

Before proving this; we will prove another problem:
Question(I): Suppose that the natural number $k \geq 3$ and a prime number $p \geq 7$ are given; then there exist natural numbers $w_1, w_2, ..., w_k$ such that $p$ divides $\sum_{i=1}^k w_i^2$ but does not divide any of the numbers $w_1,w_2,...w_k$?

Lemma: Given a prime number $p \geq 7$ and any integer $a$ such that $p \nmid a$.
Then there are natural numbers $n, m$ such that:

$$p \mid n^2+m^2-a; \qquad \qquad p\nmid n; \qquad \qquad p \nmid m.$$ Proof: See here.

Proof of question(II):

• If $p\nmid k-2$
let $w_1=w_2=...=w_{k-2}=1$; and let $a=-\sum_{i=1}^{k-2}w_i^2$;
now by the above lemma there are natural numbers $n, m$ such that $p\mid n^2+m^2-a$;
now let $w_{k-1}=n$ and $w_{k}=m$; so we have done!

• If $p\nmid k+1$
let $w_1=w_2=...=w_{k-3}=1$, $w_{k-2}=2$; and let $a=-\sum_{i=1}^{k-2}w_i^2$;
now by the above lemma there are natural numbers $n, m$ such that $p\mid n^2+m^2-a$;
now let $w_{k-1}=n$ and $w_{k}=m$; so we have done!

By considering the argument described here question(I) follows immediately from question(II).

At the end for the case $k=2$, you can choose $p \overset{4}{\equiv}1$ arbitrary;
and again there exist integers $m, n$ such that: $$p \mid n^2+m^2; \qquad \qquad p\nmid n; \qquad \qquad p \nmid m.$$