2
$\begingroup$

Let $\lfloor x\rfloor$ be the floor function.

QUESTION: Does the determinant
$$D_p=\det\left[\left\lfloor\frac{i^2+j^2}p\right\rfloor\right]_{1\le i,j\le(p-1)/2}$$ vanish for each prime $p>7$ with $p\equiv3\pmod4$?

My comptation suggests that the answer should be positive. Any ideas?

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Your definition of $D_p$ does not require $p$ to be prime, but just odd and at least 3. The following more specific suggestion is based on data up to 400: for odd $n \geq 3$, $D_n \not= 0$ if and only if $n=7$, $n=25$, or or $n$ is a prime that's $1 \bmod 4$ other than $13$ ($D_7 = -1$, $D_{25} = 9$, and $D_{13} = 0$). $\endgroup$
    – KConrad
    Commented Nov 19, 2018 at 16:58
  • $\begingroup$ @KConrad: I've added a few words on this case into my answer. Generally, the only interesting case in this setup is $p\equiv1\pmod4$, and I think it should contain more counterexamples. Can you check that for larger $p$? $\endgroup$
    – Ilya Bogdanov
    Commented Nov 19, 2018 at 22:40
  • $\begingroup$ @IlyaBogdanov I asked Alvaro Lozano-Robledo to check primes $p \equiv 1 \bmod 4$ up to 2000 and he didn't find any example with $D_p = 0$ other than $p = 13$. $\endgroup$
    – KConrad
    Commented Nov 20, 2018 at 4:38
  • $\begingroup$ I've checked now up to 5000, and the only example is $p=13$. $\endgroup$
    – Álvaro Lozano-Robledo
    Commented Nov 20, 2018 at 5:53

1 Answer 1

Reset to default
3
$\begingroup$

Let $p$ be large enough. Then there are two pairs of consecutive squares $a$, $a+1$ and $b$, $b+1$ modulo $p$ (otherwise the parities of sqiares modulo $p$ cannot alternate more than constant times, but among them there are $1,4,9,16,\dots$). Since $-(a+1)$ is not a square, the difference between the rows corresponding to $a$ and $a+1$ has all its entries equal (namely, if $i^2\equiv a$ and $j^2\equiv b$, this common value is $\frac{i^2-j^2+1}p$). The same holds for the rows corresponding to $b$ and $b+1$. Therefore, these four (or three, if $a+1=b$ or vice versa) rows are linearly dependent.

Small values of $p$ can be checked manually. The above arguments, however, may also be relevant, if such two pairs exist.

[ADDENDUM] THis addresses the generalization suggested by @KConrad, on the values of $D_n$ for odd $N\geq 3$. Sorry for being sketchy.

If $n$ is divisible by at least two distinct primes, then any square $\mod n$ coprime with $n$ appears at least twice among the $i^2$; the difference of such two rows is a constant row again. Hence, in this case $D_n=0$. (But what if we compose such matrix of representatives of all distinct squares modulo $n$???)

If $n=p^k$ with $p\geq 7$ and $k\geq 2$, then the rows for $i=p^{k-1}$, $i=2p^{k-1}$, and $i=3p^{k-1}$ are dependent for the same reason; act similarly for $3^k$ and $5^k$ with $k\geq 4$.

Essentially, the remaining cases are $p\equiv1\pmod 4$, and they seem to be a bit more interesting. Here is what I can say on them so far.

Arrange the rows in the increasing order according to $i^2\mod p$, and the columns --- in decreasing order according to $j^2\mod p$. Let $a$ and $b$ be the squares corresponding to the $i$th and the $(i+1)$st rows of the new matrix. Then the difference of these two rows is constant, apart from the column corresponding to $-b$ (whose entry is by $1$ larger). This shows that the rank of the matrix is at least $\frac{p-1}2-1$, and the degeneracy may happen if the first row can be expressed via the differences mentioned above.

I have some aditional thoughts on this case, but it still seems that this case is rare but should not appear only for $p=13$...

Improve this answer
$\endgroup$
1

You must log in to answer this question.

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