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

  • $\begingroup$ 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

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.


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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