# On a variant of the equation $\sigma(x)=2^n$

It is a nice exercise to prove that the only solutions (positive integers $x$) of the equation on the title are products of Mersenne primes; with all exponents equal to $1$.

Question: It is true that the only solutions $A \in GF(2)[x]$ to the equation $$\sigma(A) = x^a(x+1)^b$$ are products of distinct Mersenne irreducible polynomials $M$ where this means $$M = x^c(x+1)^d+1$$ and $M \in GF(2)[x]$ is irreducible.

Trivial example: $$\sigma(x^2+x+1)=x(x+1).$$ As usual $\sigma(n)$ is the sum of all positive divisors of the positive integer $n$ and $\sigma(A)$ is the sum of all divisors (including $1$ and $A$) of the polynomial $A$ in $GF(2)[x])$.

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Is this exercise in a book somewhere, so I can refresh my memory on how to go about solving it? –  Robert K Apr 23 '11 at 20:04
Just try an induction on $\omega(x)$ –  Luis H Gallardo Apr 24 '11 at 1:21
In the OEIS page that I included in the question you may find a link to a proof, but it is amusing to try it yourself! –  Luis H Gallardo Apr 24 '11 at 9:17