# Sums of two same powers modulo $p$

It is related to this question.

Question. Suppose that $p$ is a prime, $p-1$ is divisible by $q^2$ for some $q$. Is it true that every number modulo $p$ is a sum of two $q$-th powers.

If $q=2$, then (*) is true. Indeed in that case $p\equiv 1 \mod 4$. Take any number $t \mod p$. WLOG we can assume that $t$ is odd (the product of two sums of two squares is a sum of two squares). If $t\equiv 1 \mod 4$, then consider the arithmetic progression $4np+t, n\ge 0$. By Dirichlet, it contains a prime $p'\equiv 1\mod 4$. By Fermat, $p'$ is a sum of two squares, hence $t$ is a sum of two squares modulo $p$. If $p\equiv -1 \mod 4$, then consider the arithmetic progression $8np+t-2p, n\ge 0$. By Dirichlet, it contains a prime number $p'$ of the form $4k+1$ (since $p\equiv 1\mod 4$) and we are done.

That question may be easier than the question cited above. Or it may be a known open problem.

Update Several simplifications of the argument for $q=2$ were proposed. Although the statement is not true when $(p-1)/q^2$ is small, it is true when this quotient is large enough. I think this answers my question almost completely. Thanks to everybody who gave an answer or a comment.

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What do you mean by odd in $\mathbb{Z}/(p)$? You can simplify by taking just the first progression, and then saying every $k$ in $\mathbb{Z}/(p)$ is equivalent to $t \mod p$, where $t \equiv 1 \mod 4$. I think you need $q$ to be prime. It doesn't work for $q=4$: $6,7,10,11$ are not sums of fourth powers $\mod 17$ (the fourth powers are $0, \pm1, \pm4$) –  Zack Wolske Apr 1 '12 at 18:37
@Zack: Thank you. No I did not assume that $q$ is prime. So you gave a counterexample to my statement. If you upgrade it to an answer, I will accept it. –  Mark Sapir Apr 1 '12 at 18:42
The curve $x^q+y^q=a$ has points over $\mathbb{F}_p$ for $a\ne 0$ as soon as $p > q^4$ or thereabouts, by the Weil bound. –  Felipe Voloch Apr 1 '12 at 18:59
For $q=2$ there is a much simpler proof that every element of $\mathbf{F}_p$ is a sum of two squares, namely by cardinality reasons. –  François Brunault Apr 1 '12 at 19:21

Suppose $p=q^2 + 1$. Then the set of $q$-th powers, including 0, has size $q+1$. The number of elements given by a sum of pairs of these is at most $(q+1) + \frac{(q+1)(q)}{2} = \frac{q^2 + 3q + 2}{2} < q^2 + 1$ whenever $q > 3$. So, for example, $6,7,10,11$ are not expressible as a sum of two fourth powers in $\mathbb{Z}/(17)$.

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The literature on Waring's problem in Finite fields may be relevant. Theorem 6 at http://holdenlee.files.wordpress.com/2010/12/finite-field-waring.pdf states:

Theorem Let $d| p-1$ be fixed and let $A$ be the set of $d$th powers in $\mathbb{F}_{p}$, then for all sufficiently large $p$ the set $A$ is a basis of order 2.

In fact, the theorem holds for a sparse subset of $A$ (see the link). Of course, this only works asymptotically with $d$ fixed, which doesn't settle the question.

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@Mark: Thank you! So the problem has been considered before (perhaps not exactly in the same formulation). It is quite possible that it is all that is known about it. –  Mark Sapir Apr 1 '12 at 18:23
@Mark: It looks like Theorem 5.1 answers my question for sufficiently large (but not too large $p$. But I do not understand something. What if $p\equiv 3 \mod 4$, $d=2$. Then the set of squares is not a basis of order 2. Right? –  Mark Sapir Apr 1 '12 at 18:32
You can modify your above argument to work for $p \equiv 3$, since $p$ is odd, everything is equivalent ($\mod p$) to some $t \equiv 1 \mod 4$. –  Zack Wolske Apr 1 '12 at 18:46
@Zack: You are correct. Thanks again. –  Mark Sapir Apr 1 '12 at 18:56