MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $1$.

((see also: A046528 in the OEIS))

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 $GF(2)[x])$.

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.