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This question is an offshoot of this closely related MO question.

Here, we consider the Diophantine equation $$m^2 - p^k = 2^r t,$$ where $r \geq 2$ and $\gcd(2,t)=1$.

Furthermore, we place the following restrictions: $$p \equiv k \equiv 1 \pmod 4$$ $$2^r \neq t$$ $$2^r t > \frac{3373m^2}{3375} \approx {10}^{750}$$ $$\min(t,2^r) < m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r) \text{ based on mathlove's answer }$$here.

Basically, we are trying to check whether there is a solution to the said Diophantine equation satisfying $p^k < m$, and all of the other constraints above. (Thus, we would like to determine whether there is a counterexample to our conjecture that $m < p^k$.)

From this answer to a closely related MSE question, we know that $$m = 9, p = 5, k = 1, r = 2, t = 19$$ is a counterexample to $m < p^k$, except that it does not satisfy the third restriction above, which must necessarily be satisfied by an odd perfect number $p^k m^2$ with special prime $p$.

We tried to search for other counterexamples via Sage Cell Server, but it is currently unable to do this search in the range that we require.

Could somebody out there with more computing power, and more adept programming skills in e.g. Python, lend us a hand please?

Jose Arnaldo Bebita Dris
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  • Why not replace $2^rt$ with $4s$ having just a single variable? – Max Alekseyev Jun 24 '21 at 14:35
  • If you just take $k=2j$, $m=2^r+p^j$, $t=m+p^j$ and assume that $2^r > p^{2j}$, then this is always satisfied. – Mike Bennett Jun 24 '21 at 22:33
  • @MaxAlekseyev: When $p^k m^2$ is an odd perfect number with special prime $p$ satisfying $\sigma(m^2)/p^k$ is a square, then $\sigma(m^2) \equiv 1 \pmod 4$ and $k = 1$. But $\sigma(m^2) \equiv 1 \pmod 4$ is equivalent to $p \equiv k \pmod 8$. Therefore, if $\sigma(m^2)/p^k$ is a square, then $m^2 - p^k \equiv 0 \pmod 8$. Which is the reason why we put $2^r t$ for $m^2 - p^k$ there. But we do get your point. – Jose Arnaldo Bebita Dris Jun 25 '21 at 03:44
  • @MikeBennett: For the particular problem that we are considering (the problem of determining whether $m < p^k$ holds in general for an odd perfect number $p^k m^2$ with special prime $p$), we know that $p \equiv k \equiv 1 \pmod 4$ holds. So $k = 2j$ is not possible, and your argument breaks down. – Jose Arnaldo Bebita Dris Jun 25 '21 at 03:46
  • Then state that in the question. – Mike Bennett Jun 25 '21 at 04:23
  • Okay doing so now. – Jose Arnaldo Bebita Dris Jun 25 '21 at 04:24

1 Answers1

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I think that $$m=10^{375}+1,p=5,k=1,r=2,t=25\cdot 10^{748}+5\cdot 10^{374}-1\tag1$$ is a solution.

Proof :

When $m=10^{375}+1,p=5,k=1$, we have $$m^2-p^k=(10^{375}+1)^2-5\equiv 1-5\equiv 4\pmod 8$$ from which $r=2$ and $$t=\frac{(10^{375}+1)^2-5}{2^r}=25\cdot 10^{748}+5\cdot 10^{374}-1$$ follow.

Now, (1) satisfies

  • $r \geq 2$

  • $\gcd(2,t)=1$

  • $2^r \neq t$

  • $p \equiv k \equiv 1 \pmod 4$

  • $p^k\lt m$

  • $\dfrac{3373m^2}{3375} \approx {10}^{750}$

Now, (1) satisfies $2^r t > \dfrac{3373m^2}{3375}$ since $$2^rt-\frac{3373m^2}{3375}=m^2-5-\frac{3373m^2}{3375}=\frac{2m^2-5\times 3375}{3375}\gt 0$$

(1) satisfies $\min(t,2^r) < m$ since $$m-\min(t,2^r)=m-2^r=m-4\gt 0$$

(1) satisfies $m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2}$ since $$\begin{align}m < \dfrac{1+\sqrt{1+2^{r+2}t}}{2}&\iff 2m-1\lt \sqrt{1+2^{r+2}t} \\\\&\iff 2m-1\lt \sqrt{1+4(m^2-5)} \\\\&\iff (2m-1)^2\lt 1+4(m^2-5) \\\\&\iff m\gt 5\end{align}$$ which does hold.

(1) satisfies $\dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r)$ since $$\begin{align}\dfrac{1+\sqrt{1+2^{r+2}t}}{2} < \max(t,2^r)&\iff \dfrac{1+\sqrt{1+16t}}{2} < t \\\\&\iff \sqrt{1+16t}\lt 2t-1 \\\\&\iff 1+16t\lt (2t-1)^2 \\\\&\iff t\gt 5\end{align}$$ which does hold.

So, it follows that $(1)$ is a solution.

mathlove
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