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The following equation

$$ \sigma(P^n)=\sigma(Q^m) $$

has only the trivial solution

$$ P=Q, m=n, $$

for $P,Q$ irreducible polynomials of degree $>1$ in the ring $$ GF(2)[t] $$

where for any $A \in GF(2)[t]$

$$ \sigma(A) = \sum_{d \mid A} d $$

Question: What happens over the integers. More precisely:

What are the solutions (besides $p=q,m=n$) (or what can be said about the solutions, besides $m \equiv n \pmod{2}$ ) of the equation

$$ \sigma(p^n) = \sigma(q^m) $$

for odd prime numbers $p,q$ and positive integers $m,n$


$$ \sigma(n) = \sum_{0<d, d \mid n} d. $$

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In other words, $(p^r-1)/(p-1)=(q^s-1)/(q-1)$. Have you looked in Guy, Unsolved Problems In Number Theory? I'm pretty sure this equation is discussed there. – Gerry Myerson Apr 15 '11 at 23:29
The equation does not seems to be in my copy of Guy's book (my copy buyed august 1996). The equation does not seems to appear in classical Mordell's book Diophantine Equations. – Luis H Gallardo Apr 16 '11 at 0:06
MR0118701 (22 no.9472) Makowski, A. Schinzel, A. Sur l'\'equation ind\'etermin\'ee de R. Goormaghtigh. (French) Mathesis 68 1959 128–142. R. Goormaghtigh conjectured that the only solution of $N=1+x+\cdots+x^m=1+y+\cdots+y^n$, $m>n>1$, $y>x>1$, are $N=31$, $y=5$, $x=2$, $m=4$, $n=2$ and $N=8191$, $y=90$, $x=2$, $m=12$, $n=2$. The authors prove several special cases of this conjecture. Reviewed by P. Erd\H{o}s. – Luis H Gallardo Apr 16 '11 at 0:35
I detected the following misprint in my copy of Guy's book: In Page 158, line 12 R. Goormachtigh should be R. Goormaghtigh. Google was unable to find something on the former; I have not tried the latter. – Luis H Gallardo Apr 16 '11 at 0:36
There is a little bit of information about the question at and the most recent (3rd) edition of Guy's book has a brief discussion in Problem B25 on pages 124-125. I'm not aware of anything more recent but I could have missed something. Might be worth searching arXiv for Goormaghtigh, or searching MathSci if you have access. – Gerry Myerson Apr 16 '11 at 6:03

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