Is $\frac{(p-1)^p+1}{p^2}$ squarefree for all primes $p \geq 7$? I did some small testing and it seems to hold up to $p \leq 47$. Also, note that the above expression is indeed an integer by binomial expansion.
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2Very interesting. The probability that a random integer is square-free is about $.6$, so these fifteen data points might seem very convincing, but there is a bias that increases that probability significantly for these numbers, because are not divisible by any prime p or below. (An earlier version of this comment overstated the probability.) So my guess is we should keep looking for a counterexample. The computation gets very slow though! – hunter Jan 07 '24 at 03:36
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2If $q^2$ divides $(p-1)^p+1$ then $q\equiv\pm1\pmod{p}$. – Servaes Jan 07 '24 at 03:38
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13This question should not be getting close votes. It's an interesting question, it's not homework (in any reasonable class), OP has shown work, it's not at all clear what the answer is. – hunter Jan 07 '24 at 03:48
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@Servaes Would you mind explaining why that's true? – Command Master Jan 07 '24 at 03:54
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@Servaes Just thinking... if $q \equiv \pm 1 \pmod p$, then $q$ is even (since $p$ is odd by hypothesis). Consequently $q^2$ is even and there is only one even prime... – Marco Ripà Jan 07 '24 at 03:55
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3@MarcoRipà That's not true - $q$ can be $2p+1$, for example, which is odd. – Command Master Jan 07 '24 at 03:56
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4@Servaes I think even $q\equiv 1\pmod p$ is necessarily true, since the order of $p-1$ mod $q$ must be exactly $2p$ (unless $q=p$ which I haven't bothered to rule out, presumably by working out the $p$-adic expansion of $(p-1)^p$). – Erick Wong Jan 07 '24 at 04:02
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9If, heuristically, we look at the "expectation" of the number of primes squared dividing such $p$'s, every $q$ only has $O(\frac{\log(q)}{\log(\log(q))})$ possible $p$'s it might divide, so the total number is $\sum_{q}{O(\frac{\log(q)}{\log(\log(q)) q^2})}$, which converges, so we should only expect a limited number of counterexamples, if there are any. I can confirm with a computer there aren't for for $q\leq10^7$, so the remaining "probability" is $\approx 9 \cdot 10^{-6}$. – Command Master Jan 07 '24 at 04:06
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3This situation reminds me of this preprint of mine about an old open problem in recreational mathematics (https://arxiv.org/abs/2205.10163)... very very low chances to get a counterexample, but very hard to prove that there can't be any. I will look forward if this problem will be solved since I think that it won't be an easy task. – Marco Ripà Jan 07 '24 at 04:10
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2@ErickWong I implicitly assumed $q\neq p$ because we are looking at $\frac{(p-1)^p+1}{p^2}$, which is easily shown to be coprime to $p$. – Servaes Jan 07 '24 at 04:51
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1This could be on par with the famous question whether the Mersenne numbers with prime exponent and the Fermat numbers are all squarefree , with the difference that in this case we have a huge restriction , namely that a prime disproving this must be a Wieferich prime. But here , it appears to be similar difficult to find a counterexample , if it exists. – Peter Jan 10 '24 at 06:56
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Intriguingly, $(p - 1)^p + 1 = np^2$. So, is $\frac{(p - 1)^p + 1}{p^2}$ "like" $(p - 1)^p + 1$, in any way? Closer $p^2$ is to $1$, the better the chances that $\frac{(p - 1)^p + 1}{p^2}$ isn't square free. Ergo, the conjecture, if it is that $\frac{(p - 1)^p + 1}{p^2}$ is square-free, is likely true. – Hudjefa Jan 11 '24 at 12:34
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Too, $(p - 1)^p + 1 = \left(2n\right)^p + 1$. So, is $\frac{\left(2n\right)^p + 1}{p^2}$ square-free? – Hudjefa Jan 11 '24 at 12:43
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It has to be that $\left(2n\right)^p + 1 = p^2q^2k$ – Hudjefa Jan 11 '24 at 12:48
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1According to my calculations , $q$ must exceed $10^{10}$ , so finding a counterexample will be very difficult. – Peter Jan 11 '24 at 22:02
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1Not sure if this is helpful, but two tidbits: As $(p-1)^p + 1 = m p^3 + p^2$ for some integer $m$, we can write $N = \frac{(p-1)^p + 1}{p^2} = mp + 1 = ks^2$, where $k$ is square free. In this format the goal is to prove there are only solutions with $s=1$, making $N$ square free. The other tidbit (from numerical tests), it appears $N \equiv 1 \text{ or } 7 \pmod{24}$. That restricts $s \equiv 1 \pmod{24}$, and $k \equiv 1 \text{ or } 5 \pmod{24}$. – PPenguin Jan 12 '24 at 01:14
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1It seems that if $q$ is some prime factor of $(p-1)^p + 1$ other than $p$ then $q\equiv 1 \left(\mod p\right)$ – Juan Moreno Jan 13 '24 at 21:26
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1@JuanMoreno Yes, this is true because by using primitive roots of $p^2$. – FAN WENDI HCI Jan 14 '24 at 01:19
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1Interestingly, it seems that $(p-1)^{2k+1} +1$, where $k>0$ and $2k+1<p$, is always square free, so it seems that the property you have identified could be a particular case of a more general one – Juan Moreno Jan 15 '24 at 16:46
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What is implied from knowing the answer to this question? – I Zuka I Jan 16 '24 at 08:07
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Why the restriction $p\geq7$? – mr_e_man Aug 02 '24 at 07:09
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@mr_e_man I think it's because for $p=3$ the expression is equal to $1$. – User Aug 02 '24 at 08:07
1 Answers
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This is a long comment, not an answer, so, apologies for posting!
I think ABC conjecture is, to some extent, supportive of this observation. I may be wrong, please point out my mistake!
ABC conjecture says that for any given $\varepsilon>0$, there exists finitely many coprime triples $(a,b,c)\in\Bbb N^3$ such that $a+b=c$ and $c>d^{1+\varepsilon}$, where $d=\mathrm{rad}(abc)$ is the product of only the distinct prime factors of $d$, without multiplicities. Take $a=(p-1)^p$ and $b=1$ and hence, $c=(p-1)^p+1$. If $c/p^2$ is squarefree, $\mathrm{rad}(c)=c/p$. Therefore, $\mathrm{rad}(abc)=\mathrm{rad}(p-1)c/p$. ABC conjecture implies that there are only finitely many primes $p$ such that $c>(\mathrm{rad}(p-1)c/p)^{1+\varepsilon}$. If we take $q=\mathrm{rad}(p-1),$ $\varepsilon=1$, we get $$1>q^2c/p^2=\frac{q^2((p-1)^p+1)}{p^2}=q^2\left(\left(1-\frac{1}{p}\right)^2(p-1)^{p-2}+\frac{1}{p^2}\right).$$
We note that if $c$ is not squarefree, then $\mathrm{rad}(c)$ must be at most $c/3p$. I am not sure if we can prove, at least, there are only finitely many such primes using ABC.
James Moriarty
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I think this ABC is not enough to show anything about 'squarefree-ness' but can potentially prove that $((p-1)^p+1)/p$ can't have large power of a prime. – James Moriarty Mar 19 '25 at 11:59