20
$\begingroup$

This question on MSE asked the following:

"Given $x,y,z \in \mathbb{N},$ find probability that $x^2+y^2+z^2$ is divisible by $7.$"

The OP did not declare the assumed probability model, and was duly criticised for that. On the other hand, it is only natural to assume that $x$, $y$, $z$ are independently uniformly distributed mod $7$.

A case analysis then shows that that the probability in question is ${1\over7}$. This simple result led me to solve the same problem for the primes $p=3$, $5$, $11$, and $13$. In each case I obtained ${1\over p}$ as result. Further experiments showed that the remainder of $s=x^2+y^2+z^2$ mod $p$ is not uniformly distributed mod $p$, but that in any case the probability of $s=0$ mod $p$ is equal to ${1\over p}$ for all $p\leq107$. This leads to the following

Conjecture. Let the integeres $x$, $y$, $z$ be independently uniformly distributed modulo the prime $p$. Then the probability that $s:=x^2+y^2+z^2$ is divisible by $p$ is equal to ${1\over p}$.

Maybe this well known. Otherwise I'd like to see a proof.

$\endgroup$
2
  • $\begingroup$ If the conjecture is true this probably means that $w:=x^2+y^2+z^2$ is and uniformly distributed random variable in $\Bbb N$. $\endgroup$
    – Masacroso
    Feb 27 '17 at 20:38
  • $\begingroup$ so far, this is part of theorem 4.12 in Charles Small, Arithmetic of Finite Fields. Alright, it is exactly theorem 4.6 on page 91, need notation.... $\endgroup$
    – Will Jagy
    Feb 27 '17 at 20:47
17
$\begingroup$

Found it. Given odd dimension $n$ and quadratic form $$ f = a_1 x_1^2 + a_2 x_2^2 + \cdots + a_n x_n^2, $$ everything in a finite field with odd number of elements $q,$ the count $$ \#\left(f = b\right) \; = \; q^{n-1} + q^{(n-1)/2} \; \; \chi \left( \; (-1)^{(n-1)/2} \; b a_1 a_2 \ldots a_n\right). $$ At the bottom of page 91 Small points out that $$ \#\left(f = 0\right) \; = \; q^{n-1} . $$ When $b \neq 0$ we need to know what $\chi$ means.

Aah. Page 86, very simple. We have finite field $F$ and element $a.$ First $\chi(0) = 0.$ If $a$ is a nonzero square, $\chi(a)=1.$ If $a$ is nonzero and not a square, $\chi(a)=-1.$

Charles Small, Arithmetic of Finite Fields, Theorem 4.6 on page 91,

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.