5
$\begingroup$

Let $p$ be an odd prime. We know that the polynomial $x^{p-1}-1$ splits into linear factors modulo $p$. If $p$ is of the form $4k+1$ then we can write $$x^{p-1}-1=x^{4k}-1=(x^{2k}+1)(x^{2k}-1).$$ The theorem of Lagrange tells us that any polynomial congruence of degree $n$ mod $p$ has at most $n$ solutions. Hence we can deduce from this factorization that $-1$ is a quadratic residue modulo $p$. Similarly if $p$ is of the form $3k+1$ we can write $4(x^{p-1}-1)=4(x^{3k}-1)=(x^k-1)((2x^{k}+1)^2+3)$ and deduce that $-3$ is a quadratic residue mod $p$.

Can we prove in this fashion that $-2$ is a quadratic residue mod $p$ if $p$ is of the form $8k+1$ or $8k+3$?

Note that I am interested only in this specific method. I know how to prove this using different means.

$\endgroup$
0
$\begingroup$

In the case $p = 8k+1$, one can write $$x^{p-1} - 1 = (x^{4k}-1)(x^{4k}+1) = (x^{4k}-1)((x^{2k}+1)^2+2(x^{2k})) \,.$$

Now for any $x$ such that $(x^{2k}+1)^2+2(x^{2k}) = 0$, we see that $y = (x^{2k}+1)/x^k$ satisfies $y^2 = -2$.

I think that this method breaks down in cases of the form $p= 8k+3$. For $k=1$, the resulting polynomial to factor is $x^{10}-1$. However, the corresponding cyclotomic extension field $\Bbb Q(e^{i\pi/5})$ does not include $\Bbb Q(\sqrt{-2})$ as a subfield (its Galois group is cyclic of order 4, with $\Bbb Q(\sqrt{5})$ as the only quadratic subfield), which I think implies that there's no comparable decomposition for $x^{10}-1$.

Note: while looking at that example, I did find that if $p$ is of the form $5k+1$ then you can factor $$4(x^{p-1}-1) = 4x^{5k}-4 = (x^k-1)(4x^{4k}+4x^{3k}+4x^{2k}+4x+4) =(x^k-1)((2x^{2k}+x^k+2)^2 - 5x^{2k})\,. $$ Similar reasoning to the above shows that in this case 5 is a quadratic residue mod $p$.

$\endgroup$
1
  • 1
    $\begingroup$ Do you mean $(x^{4k}-1)((x^{2k}-1)^2+2(x^{2k}))$? $\endgroup$ – cansomeonehelpmeout Jan 5 '18 at 20:39
0
$\begingroup$

Just a sketch for the moment, I have the strong feeling the following lines can be greatly simplified.

If $p\equiv 1\pmod{8}$ in $\mathbb{F}_p^*$ there is an element of order $8$, since $\mathbb{F}_p^*$ is a cyclic group of order $p-1$. In particular $\Phi_8(x)=x^4+1$ completely factors over $\mathbb{F}_p$. If $\alpha\in\mathbb{F}_p$ is a root of $\Phi_8$ we have $$ 0=\alpha^2+\frac{1}{\alpha^2}=\left(\alpha+\frac{1}{\alpha}\right)^2-2 $$ hence $\left(\frac{2}{p}\right)=1$. Since $p\equiv 1\pmod{4}$ we also have $\left(\frac{-1}{p}\right)=1$ and $-2$ is a quadratic residue $\pmod{p}$ since the Legendre symbol is multiplicative. In general, the degree of the splitting field of $\Phi_8$ over $\mathbb{F}_p$ is given by the least $k$ such that $8\mid (p^k-1)$. In particular $\Phi_8$ is never irreducible over $\mathbb{F}_p$, since $n^2\equiv 1\pmod{8}$ holds for any odd $n$. If $p$ is an odd prime $\not\equiv 1\pmod{8}$, by denoting as $i$ and $\sqrt{2}$ the elements of $\mathbb{F}_p$ or $\mathbb{F}_p^2$ fulfilling $\beta^2+1=0$ and $\beta^2-2=0$ we have that $\Phi_8$ factors over $\mathbb{F}_p$ as the product of two quadratic, irreducible polynomials. The complex roots of $\Phi_8$ are given by $\frac{\pm 1\pm i}{\sqrt{2}}$ and they stay the same with the previous assumption. The irreducible factor of $\Phi_8$ vanishing at $\frac{1+i}{\sqrt{2}}$ may only have the following forms:

$$ x^2-\sqrt{2}x+1,\qquad x^2-i,\qquad x^2-i\sqrt{2}x-1.$$ You just have to show these forms are respectively associated with $p=8k+7,8k+5,8k+3$ to have that the $p=8k+7$ case is the only case in which $2$ is a quadratic residue but $p\not\equiv 1\pmod{8}$. And this is simple since by Frobenius automorphism the conjugated root of $\frac{1+i}{\sqrt{2}}$ is $\left(\frac{1+i}{\sqrt{2}}\right)^p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.