Can two Consecutive Polynomials both be perfect ? 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.
 A: 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.
