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.

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Is there any particular reason you restrict to $q=2$? – Thomas Bloom Jan 20 '11 at 12:56
I think that in the definition of $\sigma(P)$ you want $d$ to divide $P$ – Valerio Talamanca Jan 20 '11 at 13:14
@Valerio: Of course !. I picked the definition from another post... – Luis H Gallardo Jan 20 '11 at 14:33
@Thomas: The situation is already complicated enough for $q=2$ so... – Luis H Gallardo Jan 20 '11 at 14:34

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.

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@Andrei: It is also a perfect solution ! – Luis H Gallardo Jan 20 '11 at 15:47
Thank you, I enjoyed this problem! – Andrei Moroianu Jan 20 '11 at 22:08