# Binary neighbors perfect polynomials, instead of consecutive

Inspired by Andrei's nice solution of 52609, (namely consecutive perfect polynomials): Denote by $A$ the full ring of polynomials in one variable $t$ over the finite field with $2$ elements. For any polynomial $P \in A$ define $$\sigma(P) = \sum_{d \mid A} d.$$ A polynomial $P \in A$ is called perfect if $$P = \sigma(P).$$ If $P$ has no roots in $GF(2)$ then $P$ is odd, otherwise it is even.

Andrei proved:

$P$ is odd perfect iff $P$ is perfect and $P$ is a square in $A,$ and deduced:

$P$ and $P+1$ cannot both be perfect.

Observe that one of $P,P+1$ is odd while the other is even.

Take now a polynomial $P \in A.$ What is the next polynomial of the same type? i.e., having the same parity. Seems that the following definition is appropriate for this:

Call neighbors two polynomials $P,Q \in A$ if $\deg(P)>2$ and ($Q=P+t(t+1)$ or $P=Q+t(t+1)$).

Question: There are neighbors polynomials $P,Q \in A$ such that $P$ and $Q$ are both perfect ?

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Google says: voisins --> neighbors, not neighbords. So, I corrected the post accordingly. Is perhaps a too literal translation. – Luis H Gallardo Jan 20 '11 at 18:17
It is not the same thing but (seems to me) that there is a conjecture Gary Mullen ?) claiming infinity of irreducible polynomials $Q$ and $Q+t^2+t$ over a finite field with an even cardinal. – Luis H Gallardo Jan 26 '11 at 21:51