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The answer is yes. Since $p\equiv 1\pmod{4}$, and $m$ is odd, we have $m^2-p^k\equiv 0\pmod{4}$. The only integers that are not differences of squares are exactly those that are $2\pmod{4}$.

As I mentioned in another answer, using GCDs and fractions is usually a bad idea. Here is how I would interpret all of this material. Let $N$ be an odd perfect number. Write its prime factorization …

(This is an extended comment.) There couldn't be anything special about base 10, could there? Notation: Given two positive integers $m,n$, let $m\oplus n$ be the integer that results from prepending …

It is rare that complicating an expression leads to an insight. There are, of course, important counter-examples to this principle. But generally speaking, one seeks to simplify an expression, rathe …

Suppose that $p$ is a prime number that divides $2^{2n+1}-1$. This means that $2^{2n+1}\equiv 1\pmod{p}$. Consequently, the order of $2$ modulo $p$ must be an odd number dividing $2n+1$. On the othe …

Let's use some standard bounds on the $n$th prime, as found in this paper by Pierre Dusart. We have $$ p_{2n} \leq 2n[\ln(2n)+\ln(\ln(2n))-0.9484] = 2n\left[\ln(n)+\ln\left(\frac{2}{e^{0.9484}}\ln(2n) …

Doing an internet search I found the paper Groups of small order as Galois groups over $\mathbb{Q}$ by Jack Sonn, from 1989. Theorem 1 of that paper asserts that every group of order less than 672 is …

This is just a long comment that might be helpful: Treat $a$ as a parameter, and treat $x$ as a variable. The Diophantine expression $$ \exists x\ ((x^2<a) \land (a<(x+1)^2)) $$ defines $a$ as a nons …

The answer to your new question is still no. I mentioned this problem (or rather, the group-theoretic reformulation given by Will Sawin) to my colleague Steve Humphries, and he found the following tw …

Let $\zeta$ be the Riemann zeta-function, and let $t> 0$. I'm interested in the growth rate of $$ \left|\frac{\zeta'}{\zeta}\left(-\frac{1}{2}+it\right)\right| $$ as $t\to\infty$. It is easy to find …

The answer to the question posed by Aaron Meyerowitz to darij grinberg in the comments is unfortunately negative, even in the integers, by taking $a=c=u=v=w=0$ but $b=d=1$. However, it has a positive …

First, every odd prime divides some Mersenne number. (In fact, every odd integer divides infinitely many Mersenne numbers.) So that is no restriction at all. Second, the claim from Wikipedia is corr …

There are two main sources of repeated logs. (These sources can be further refined into natural subcategories, but I'll only mention a couple of those subcategories.) Those two main sources are: Typ …

Factorization in the ring $\mathbb{Z}[x]/(x^2+1)\mathbb{Z}[x]\cong \mathbb{Z}[i]$ is well known. For instance, $5$ and $13$ (and any prime $\equiv 1\pmod{4}$) are no longer prime. The factorization o …

This is impossible from standard results about cyclotomic polynomials and their factors. Note that $\frac{q^{p-1}-1}{p-1}$ is just $$ \prod_{d|(p-1), d>1}\Phi_d(q), $$ where $\Phi_d(x)$ is the $d$th …