# In a finite field, can all of the $\mathbb{F}_p$-translates of a square also be square?

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?

• 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?
– tkr
May 24 at 1:42
• @tkr I think I've clarified now May 24 at 2:03

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.
• 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? May 24 at 4:28
• 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 $d \geq 2$.