Denote by $A$ the full ring of polynomials in one variable $t$ over the finite field with $q$ elements.
For any monic polynomial $P \in A$ define
$$
\sigma(P) = \sum\_{d \mid P, d\, \text{monic}} d.
$$
A monic polynomial $P \in A$ is called `perfect` if
$$
P = \sigma(P).
$$


Let $q=2.$
Two polynomials $P,Q \in A$ are called `consecutive` if $\deg(P)>0$ and ($Q=P+1$ or $P=Q+1$).


In AMM problem 10771 $[1999,166]$ proposed by Florian Luca and resolved by Francis B. Coghlan
it is proved the inexistence of consecutive perfect numbers.


Question: Let $q=2.$ Are there consecutive polynomials $P,Q \in A$ such that $P$ and $Q$ are both perfect.


EDIT. Changed incorrect $A$ by correct $P$ in the definition as observed by Valerio.



Can two Consecutive Polynomials both be perfect ?
The answer is no. This follows from the following claims (which I will prove below):


**1.** If $P$ is perfect and $P(0)\ne 0$ or $P(1)\ne 0$ then $P$ is a square.


**2.** A perfect polynomial which is a square has no roots in $\mathbb{F}\_2$.


Assume that $P$ and $P+1$ are both perfect. From Claim 1, either $P=Q^2$ is a square, or $P+1$ is a square. Consequently they are both squares (since $Q^2+1=(Q+1)^2$). From Claim 2 we get that $P$ and $P+1$ have no roots in $\mathbb{F}\_2$, which is clearly absurd.


**Proof of Claim 1.** Let $P=P\_1^{a\_1}\ldots P\_n^{a\_n}$ be the prime decomposition of $P$ and assume that $P(0)=1$. Then $P\_i(0)=1$. Since $P$ is perfect one has
$$(\*)\qquad\qquad P=\prod\_{i=1}^n(1+P\_i+\ldots + P\_i^{a\_i}),$$
so $1=P(0)=\prod\_{i=1}^n(1+a\_i)$, proving that $a\_i$ are all even. If $P(1)=1$, apply the previous argument to $P(X+1)$.


**Proof of Claim 2.** The hypothesis implies that (\*) holds and all $a\_i$ are even. Then $1+P\_i(0)+\ldots + P\_i^{a\_i}(0)=1$ for all $i$, so $P(0)=1$. The proof that $P(1)=1$ is similar.


