Let $F$ be a finite field of $q = p^d$ elements, and suppose $\alpha\in F^\times$ is a square which generates $F$ over $\mathbb{F}_p$. I'm interested in understanding to what extent it is possible that $\alpha,\alpha+1,\ldots,\alpha+(p-1)$ are all squares.

This is trivially true for $p = 2$, and false if $d = 1$, and $p \ge 3$.

Can it happen for arbitrarily large primes $p$ that you can find $\alpha$ such that $\alpha,\alpha+1,\ldots,\alpha+(p-1)$ are all squares?

  • 1
    $\begingroup$ Can you please clarify the final question? You have a question: does there exist $\alpha$ such that...in $F$ and you asking whether there is an affirmative answer depending on just $p$ or $d$ also? $\endgroup$
    – tkr
    May 24 at 1:42
  • $\begingroup$ @tkr I think I've clarified now $\endgroup$ May 24 at 2:03

1 Answer 1


The displayed bound in the last paragraph of Section 2 here, with $r = p$ and $\varepsilon_i = 1$ for $i = 1,\ldots,p$, shows that there exists $\alpha \in F^\times$ such that $\alpha, \ldots,\alpha+p-1$ are all squares in $F^\times$ unless the count $N_{p^d}$ there is $0$, which would imply $$ \frac{p^d}{2^p} < (p-1)p^{d/2} + \frac{p}{2}, $$ and for fixed $p$ this is false for large enough $d$. How large?

Setting $x = p^{d/2}$, the above inequality says $$ x^2 < (p-1)2^px + p2^{p-1}, $$ so $$ x^2 - (p-1)2^px - p2^{p-1} < 0. $$ The largest real root of $X^2 - (p-1)2^pX - p2^{p-1}$ is $$ \frac{(p-1)2^p + \sqrt{(p-1)^22^{2p} - 4(p-1)p2^{2p-1}}}{2} < \frac{(p-1)2^p + (p-1)2^p}{2} = (p-1)2^p. $$ A negative value of that quadratic polynomial occurs only when $X$ is less than the larger real root, so necessarily $$ x = p^{d/2} < (p-1)2^p. $$ Thus $$ d < 2\log_p((p-1)2^p) = 2\log_p(p-1) + 2p\log_p(2) < 2 + 2p, $$ so when $d \geq 2p+3$ there is at least one $\alpha \in F^\times$ such that $\alpha, \alpha+1, \ldots, \alpha+p-1$ are all squares in $F^\times$.

Note. I have not taken into account here the requirement from your first paragraph that $F = \mathbf F_p(\alpha)$. The number of $\alpha$ just with the quadratic residue property is $p^d/2^p + O_p(p^{d/2})$. Count the number of generators of $F$ over $\mathbf F_p$ as a field extension and I suspect you'll probably be able to show by contradiction that for all large $d$ (depending on $p$) there will be an $\alpha$ fitting all the conditions you want.

  • $\begingroup$ Wow your theorem is remarkably relevant! Though, in the theorem you seem to require $p > r$, whereas in my case I want $p = r$. Does that make a difference? $\endgroup$ May 24 at 4:28
  • 1
    $\begingroup$ It's not "my" theorem; all this stuff was known decades ago. You are mixing up the meaning of $p$ in what I wrote and in what you wrote. I wrote most of Section $2$ over $\mathbf F_p$, so $p > r$ simply means more numbers are in the field than the list of conditions: to have $r$ different conditions you need at least $r$ numbers in your field (or $r+1$ if you want $r$ conditions on nonzero numbers in the field). So when working over $\mathbf F_q$ for $q = p^d$ you want $r = p$ and $q > r = p$, which holds when $d \geq 2$. $\endgroup$
    – KCd
    May 24 at 5:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .