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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 $w_k$.

2 Answers2

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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.

Peter
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  • But does such representation allow number $1$? I am sure that it is. –  Oct 01 '17 at 12:30
  • 17=4^2+1^2? Or I did not understand your answer. –  Oct 01 '17 at 12:33
  • Yes, this would be an allowed representation and $17$ does neither divide $1$ nor $4$ – Peter Oct 01 '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. –  Oct 01 '17 at 12:36
  • I made an edit a little bit before this answer. –  Oct 01 '17 at 12:36
  • But Peter, all the $w_i$ must be different than 1. –  Oct 01 '17 at 12:40
  • Then, my method doesn't work. Sorry, didn't notice that you demand that.. But the prime is arbitary, yes ? – Peter Oct 01 '17 at 12:41
  • Yeah, it doesn´t work but it is a nice idea. I conceived this question after some thinking about sums of squares and although trivial I do not know how to solve it. –  Oct 01 '17 at 12:44
  • Yes, p does not need to be fixed through all counterexamples of this conjecture. –  Oct 01 '17 at 12:45
  • @AntoinedePaladin Edited my answer. – Peter Oct 01 '17 at 13:00
  • What if $\sum w_i^2$ is divisible by 3? Or 5. –  Oct 01 '17 at 13:06
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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: $$p \mid n^2+m^2; \qquad \qquad p\nmid n; \qquad \qquad p \nmid m. $$

Davood
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