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Let $p\ge 5$ be a prime number,show that neither of the two numbers $$\dfrac{p^p-1}{p-1}; \dfrac{p^p+1}{p+1}$$ can be a prime power

I know that $(p-1)|p^p-1$ and $(p+1)|p^p+1$

I have read some similar problems:

Prove that $\frac{a^n-1}{b^n-1}$ and $\frac{a^{n+1}-1}{b^{n+1}-1}$ can't both be prime.

Choosing $a$ s.t. $\frac{a^k - 1}{a-1}$ is not a prime power

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  • $\begingroup$ For the first number, you can see this question $\endgroup$ – GAVD Sep 26 '17 at 10:01
  • $\begingroup$ In general, $\dfrac{a^{2k+1}\pm1}{a\pm1}=m^n$ seems to admit $(a=3,~k=2)$ as the only solution. $\endgroup$ – Lucian Sep 29 '17 at 15:56
  • $\begingroup$ @GottfriedHelms In order to avoid trivialities such as you mention, a working definition is this: A natural number is prime if it has exactly two different divisors. $\endgroup$ – uniquesolution Oct 4 '17 at 19:57
  • $\begingroup$ Maybe relevant: "If p is a prime, then p^p-1 has at least a prime factor that is congruent to 1 modulo p." oeis.org/A212552 $\endgroup$ – Hyperplane Oct 24 '17 at 14:38
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If we have a solution in integers to $$ \frac{x^n-1}{x-1} = y^q, $$ with $n \geq 3$, $q \geq 2$ prime, $x, y > 1$ and $$ (x,y,n,q) \neq (18,7,3,3), $$ then a result of Bugeaud, Mignotte and Roy (Pacific J. Math. 2000) ensures that there exists a prime divisor of $x$ which is congruent to $1$ modulo $q$. If we assume that $x=n=p$, this implies that $p \equiv 1 \mod{q}$, say $p = aq+1$ and so we can rewrite the equation $$ \frac{p^p-1}{p-1}=y^q $$ as $$ p (p^a)^q - (p-1) y^q = 1. $$ An old result of someone or other implies that the equation $$ p u^q - (p-1) y^q = 1 $$ has only the solution $(u,y) =(1,1)$ in positive integers and so $a=0$, a contradiction.

A similar argument works for showing that $\frac{p^p+1}{p+1}$ isn't a perfect power (replacing $x$ by $-x$), after a bit of work.

These results I'm citing require a fair bit of machinery, and hence this proof is pretty far from elementary.

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COMMENT. Something that has been happening in recent months is that MSE beginners post difficult problems that most likely do so with all deliberation. I want here to report some facts mentioned by L. J. Mordell in relation to the proposed problem.

(Note that $x ^ p$ is more general than $p ^ p$).

Euler had already shown that the equation $y^2-1=x^3$ has the only solutions $x=0,-1,2$. For the prime $p\gt3$, Nagell proved that if $$y^2-1=x^p$$ then one has $$p\equiv 1\pmod8\text{ and } y\equiv 0\pmod p$$ furthermore $$y\pm1=2x_1^p\text { and }y\mp2{p-1}x_2^p$$ so that $$x_1^p-2^{p-2}x_2^p=\pm 1\\\text { and }p\text{ divides }\frac{x^p+1}{x+1}\text { but } p^2\text {does not divide }\frac{x^p+1}{x+1}$$ Furthermore if $x_1+y_1\sqrt p$ is the fundamental unit of $\mathbb Q(\sqrt p)$ then $x_1+y_1\equiv 1\pmod 8$ this condition being equivalent to the fact that $2$ is a biquadratic residue of $p$.

T. Nagell. Sur l’impossibilité de l’équation indéterminée $z^p+1=y^2$. Norsk Mat. Forenings Skrifter, $\mathbf 1$ (1921),Nr. 4.

T. Nagell. Sur une equation à deux indéterminées.Norsk Vid. Selsk Forh. $\mathbf 7$ (1934) p. 136-139.

Mordell adds “We conclude this section by mentioning the impossibility of the equations $$y^3=x^p+1, \space |x|\gt 1,\space\space \text{ Nagell }\\ y^3=x^p-1, \space |x|\gt 2,\space\space \text{ Nagell }\\y^4=x^p+1\text{ Selberg }$$ The last result is now a special case of Chao Ko’s theorem.”

Chao Ko. On the diophantine equation $x^2=y^n+1,\space xy\ne 0$. Scientia Sinica (Notes),$\mathbf {14}$ (1964),p.457-460.

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