## Sum of squares modulo a prime

What is the probability that the sum of squares of n randomly chosen numbers from $Z_p$ is a quadratic residue mod p?

That is, let $a_1$,..$a_n$ be chosen at random. Then how often is $\Sigma_i a^2_i$ a quadratic residue?

-
Speaking as a non-expert, I should think that for n much larger than, say, log(p) (or maybe even 4), the probability should approach 1/2. Are you interested in the answer for n << p, or p << n, or is there some other relationship between n and p which would make answering the question easier? Gerhard "Email Me About System Design" Paseman, 2011.07.05 – Gerhard Paseman Jul 5 2011 at 22:34
Also, I assume p is prime from your title, but it would help to call that out in the body of the question. Gerhard "Email Me About System Design" Paseman, 2011.07.05 – Gerhard Paseman Jul 5 2011 at 22:37
Yes, $p$ prime — and odd — is implicit both in the title and in the use of "quadratic residue". – Noam D. Elkies Jul 5 2011 at 22:53
This is my intuition as well. n should be at most polynomial in log(p) and p is meant to be a prime. Is there a standard theorem for this? – unknown (google) Jul 5 2011 at 22:54
As you can see, if $p \rightarrow \infty$ then already $n=3$ is enough to get near-equidistribution (and $n=2$ fails only because 0 is over- or under-represented depending on whether $p$ is $+1$ or $−1 \bmod 4$). – Noam D. Elkies Jul 6 2011 at 2:36

## 4 Answers

Here is a slightly different argument: Let $Q$ be a non degenerate quadratic form over $\mathbb{F}_q$ of rang $n$ and determinant $d$. Let $A(n,d)=|{x\in \mathbb{F}_q^n:Q(x)=0}|$. The claim is that $A(n,d)=q^{n-1}+O(q^{n/2})$. For $n>2$ we can write $Q(X)=Q_0(X_1,\ldots,X_{n-2}) +X_{n-1}X_n$, where $Q_0$ is a form of rank $n-2$ in the variables $X_1,\ldots,X_{n-2}$. This decomposition shows instantly that that $A(n,d)=(2 q-1) A(-d,n-2) +(q-1) (q^{n-2}-A(-d,n-2))$. Proceeding by induction we get the estimate $A(n,d)=q^{n-1}+O(q^{n/2})$. (The error term can be computed exactly using Gauss sums). Applying this to the forms $X_1^2+\cdots +X_n^2-X_{n+1}^2$ and $X_1^2+\cdots+ X_n^2$ we get that the desired probability is $(A(n+1,-1)-A(n,1))/(2 q^n)= (q-1)/(2q) +O(q^{-n/2})$.

-

The probability depends on the parity of $n$ and the residue of $p$ modulo $4$: it can be calculated in a straightforward way using Gauss sums.

Let $n$ be $2k$ or $2k+1$, and let $p\equiv r\pmod{4}$ where $r=\pm 1$. Then, assuming I made no mistake, the probability equals $$\frac{p+1}{2p}+\frac{p-1}{2p}(rp)^{-k}.$$

Note that in my calculation I regarded zero as a quadratic residue. If we exclude zero then the final answer will look slightly different, with a main term $\frac{p-1}{2p}$ as Noam Elkies said.

-
 Hah! A fellow Zero supporter. Welcome, Zero-brother! Gerhard "Sound The Drums And Cannon" Paseman, 2011.07.05 – Gerhard Paseman Jul 5 2011 at 23:59

This probability can be calculated exactly, and indeed it approaches $1/2$ rather quickly — more precisely, for each $p$ it approaches the fraction $(p-1)/(2p)$ of quadratic residues $\bmod p$. This can be proved by elementary means, but perhaps the nicest way to think about it is that if you choose $n$ numbers $a_i$ independently and sum $a_i^2 \bmod p$, the resulting distribution is the $n$-th convolution power of the distribution of a random single square — so its discrete Fourier transform is the $n$-th power of the D.F.T., call it $\gamma$, of the distribution of $a^2 \bmod p$. For this purpose $\gamma$ is normalized so $\gamma(0)=1$. Then for $k \neq 0$ we have $\gamma(k) = (k/p) \gamma(1)$ [where $(\cdot/p)$ is the Legendre symbol], and $$p \gamma(1) = \sum_{a \bmod p} \exp(2\pi i a^2/p),$$ which is a Gauss sum and is thus a square root of $\pm p$. It follows that $|\gamma(k)| = p^{-1/2}$, from which we soon see that each value of the convolution approaches $1/p$ at the exponential rate $p^{-n/2}$, and the probability you asked for approaches $(p-1)/(2p)$ at the same rate.

As noted above, this result, and indeed the exact probability, can be obtained by elementary means, yielding a (known but not well-known) alternative proof of Quadratic Reciprocity(!). But that's probably too far afield for the present purpose.

-
From your answer I infer 0 does not count as a quadratic residue mod p. (I definitely agree it is not as interesting as a quadratic residue, but should it be really ostracized from the set of squares?) Gerhard "Supports The Rights of Zero" Paseman, 2011.07.05 – Gerhard Paseman Jul 5 2011 at 23:12
While 0 is indeed a square, it does not count as a "quadratic residue" (nor as a "quadratic nonresidue" — though this term should really have been "nonquadratic residue"). Properly 0 should count as half square and half non-square (think about the number of square roots), and then the limit would really be 1/2. – Noam D. Elkies Jul 5 2011 at 23:55
For quadratic reciprocity: let $n$ be an odd prime prime $l \neq p$, and $N$ the number of solutions of $\sum_{i=1}^l a_i^2 = 1 \bmod p$. Cyclic permutation of the coordinates has two fixed points or none according as $N \equiv 2$ or $0 \bmod l$. From the formula for $N$ it soon follows that $(l/p) \equiv {p^*}^{(l-1)/2} \bmod l$ where $p^* = \gamma(1)^2 = p$ or $-p$ according as $p \equiv 1$ or $-1 \bmod 4$. By Legendre's formula this means $(l/p) = (p^*/l)$. Exercise: modify this to determine $(2/p)$ from the count of solutions of $x_1^2 + x_2^2 \equiv 1 \bmod p$. [Original source?] – Noam D. Elkies Jul 6 2011 at 0:01
:-) If no reference turns up here I'll post a reference request as a question... It worked quite well for my one previous M.O. query (on the integral for $\frac{22}{7} - \pi$). – Noam D. Elkies Jul 6 2011 at 3:17
This idea can also be used to get at a few memorable cases of higher reciprocity. For example: if $p \equiv 1 \bmod 3$, and you already know that the number of solutions of $x^3 + y^3 = 1 \bmod p$ is $p - 2 - a$ where $4p = a^2 + 27 b^2$ for some integers $a,b$, then it follows at once that $2$ is a cubic residue iff $a$ is even, that is, iff $p$ itself can be written as $a^2 + 27b^2$. [NB the count is $p-2-a$, not the familiar $p+1-a$, because for this purpose we must exclude the three points at infinity.] – Noam D. Elkies Jul 6 2011 at 14:41
show 6 more comments

.5

Let me atone for giving too few details by giving too many.

Let $$S=\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}\sum_{m=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p}$$ The innermost sum is $p$ if $a_1^2+\cdots+a_n^2-t^2\equiv0\pmod p$ and zero otherwise, so $S$ counts $2p$ whenever $a_1^2+\cdots+a_n^2$ is a (nonzero) quadratic residue, $p$ whenever it's zero. On the other hand, $$S=\sum_{m=0}^{p-1}\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p}$$ so $$S=p^{n+1}+\sum_{m=1}^{p-1}\sum_{a_1=0}^{p-1}\dots\sum_{a_n=0}^{p-1}\sum_{t=0}^{p-1}e^{2\pi im(a_1^2+\cdots+a_n^2-t^2)/p}$$ so $$S=p^{n+1}+\sum_{m=1}^{p-1}\left(\left(\sum_{a_1=0}^{p-1}e^{2\pi ima_1^2/p}\right)\cdots\left(\sum_{a_n=0}^{p-1}e^{2\pi ima_n^2/p}\right)\left(\sum_{t=0}^{p-1}e^{2\pi imt^2/p}\right)\right)$$ Each of those inner sums is a Gauss sum and known to equal $\sqrt p$ in modulus (more detail: the sum is ${m\overwithdelims()p}\sqrt{{-1\overwithdelims()p}p}$), so $|S-p^{n+1}|\le(p-1)p^{(n+1)/2}$. For $n\gt1$, the main term beats the error term, and you get a good estimate.

-
If you're going for the shortest answer, I recommend changing character sets. Gerhard "Brevity Is The Soul" Paseman, 2011.07.05 – Gerhard Paseman Jul 5 2011 at 22:48
Also $\frac12$ is not quite right if $p$ is fixed and only $n$ grows. – Noam D. Elkies Jul 5 2011 at 23:00
Or "Levity is the role of wit"... – Noam D. Elkies Jul 5 2011 at 23:57
The answer is worth the question. +1 – Wadim Zudilin Jul 6 2011 at 2:55