Polynomials over $\mathbb{F}_2$ without multiplicity 1 factors Let $f \in \mathbb{F}_2[T]$ such that both $f$ and $f + 1$ have the property that every irreducible factor in the unique factorization domain $\mathbb{F}_2[T]$ appears with multiplicity at least $2$. Does it follow that $f$ must be the square of a polynomial? And if this is true, can this statement by generalized?
 A: I think the implication is true:
Write $f=\prod\limits_{i=1}^nf_i^{r_i}$ for the irreducible factors $f_i$ of $f$ and $f+1=\prod\limits_{j=1}^m\tilde{f}_j^{s_j}$, respectively. Then, as in the comment above, one has that $\prod\limits_{i=1}^nf_i^{r_i-1}$ and $\prod\limits_{j=1}^m\tilde{f}_j^{s_j-1}$ divide the derivation $f'=(f+1)'$ and thus, as $f$ and $f+1$ are of course prime to each other (and thus $f_i$ and $\tilde{f}_j$, for all $i,j$ and finally their products are prime to each other) that 
$$\prod\limits_{i=1}^nf_i^{r_i-1}\cdot \prod\limits_{j=1}^m\tilde{f}_j^{s_j-1}\text{ divides }f'.$$
As by assumption $r_i,s_j\geq 2$ for all $i,j$, that is $r_i-1\geq r_i/2$ resp. $s_j-1\geq s_j/2$, the degree of the left hand side, which is $$\sum_{i=1}^n(r_i-1)\text{deg} (f_i)+\sum_{j=1}^m(s_j-1)\text{deg}(\tilde{f}_j)\geq 1/2 \left(\sum_{i=1}^nr_i\text{deg}(f_i)+\sum_{j=1}^ms_j\text{deg}(\tilde{f}_j)\right) =\text{deg} (f)$$ is at least the degree of $f$, a contradiction. The only remaining possibility is $f'=0$, that means $f(T)=g(T^2)$ for some $g\in\mathbb{F}_2[T]$, and thus $f(T)=g(T^2)=g(T)^2$ is a square. Concerning the possibility of generalizations -- the first thing that comes to mind is that one can replace 2 by any prime $p\in\mathbb{N}$ everywhere in the above statement (except of course that the multiplicities of the irreducible factors are still assumed to be at least $2$ (and not $p$)) to get the implication "$f$ is of the form $g^p$" instead of "$f$ is a square".
Kind regards!
A: This follows from the Mason--Stothers theorem, which is the polynomial analogue of the $ABC$ conjecture.  We need to be careful about stating the theorem in a way that works in characteristic $p$, as follows.
Theorem (Mason, Stothers):  If $k$ is a field and $a(x)$, $b(x)$, and $c(x)$ are nonzero polynomials in $k[x]$ such that (i) $a(x) + b(x) = c(x)$, (ii)  $\gcd(a(x),b(x)) = 1$, and (iii) at least one of the derivatives $a'(x)$, $b'(x)$, or $c'(x)$ is not $0$, then 
$$
\max(\deg a(x), \deg b(x), \deg c(x)) \leq \deg({\rm rad}(a(x)b(x)c(x)) - 1.
$$
In characteristic $0$ we can rewrite (iii) in the equivalent form that at least one of $a(x)$, $b(x)$, or $c(x)$ is not constant, but in positive characteristic it is stronger to say at least one of the polynomials has a nonzero derivative  than to say at least one of the polynomials is not constant.
Corollary: Let $k$ be a field with positive characteristic. If $f(x) \in k[x]$ and $\alpha \in k^\times$ have the property that all irreducible factors of $f(x)$ and $f(x)+\alpha$ have multiplicity at least $2$, then $f(x) \in k[x^p]$, where $p$ is the characteristic of $k$.
(Your question uses $k = {\mathbf F}_2$ and $\alpha = 1$. This corollary does not replace your $2$ by $p$ everywhere: the multiplicities still only have to be at least $2$ even when $k$ has characteristic $p$.) 
Proof: We will show by contradiction that $f'(x) = 0$, which implies (and is in fact equivalent to saying) $f(x)$ lies in $k[x^p]$.
Assuming $f'(x) \not= 0$, set $a(x) = f(x)$, $b(x) = \alpha$, and $c(x) = f(x)+\alpha$. These three polynomials fit the conditions of the Mason--Stothers theorem. Therefore
$$
\max(\deg f,\deg(f+\alpha)) \leq \deg({\rm rad}(f(f+\alpha))-1.
$$
Therefore
$$
\deg f + \deg(f + \alpha) \leq 2\deg({\rm rad}(f(f+\alpha))-2.
$$
By hypothesis the irreducible factorizations of $f$ and $f+\alpha$ are $f = u\pi_1^{e_1}\cdots \pi_r^{e_r}$ and $f+\alpha = v\widetilde{\pi}_1^{d_1}\cdots \widetilde{\pi}_s^{d_s}$, where $u, v \in k^\times$, $\pi_i$ and $\widetilde{\pi}_j$ over all $i$ and $j$ are distinct monic irreducibles in $k[x]$, $e_i \geq 2$ for all $i$, and $d_j \geq 2$ for all $j$. In the above inequality, the left side 
is 
$$
\sum_i e_i\deg \pi_i + \sum_j d_j\deg \widetilde{\pi}_j \geq 2\left(\sum_i \deg \pi_i + \sum_j \deg \widetilde{\pi}_j\right).
$$
From the definition of the radical and the relative primality of $f$ and $f+\alpha$, 
$$
\deg({\rm rad}(f(f+\alpha)) = \sum_i \deg \pi_i + \sum_j \deg \widetilde{\pi}_j.
$$
Therefore
$$
2\left(\sum_i \deg \pi_i + \sum_j \deg \widetilde{\pi}_j\right) \leq 2\left(\sum_i \deg \pi_i + \sum_j \deg \widetilde{\pi}_j\right) - 2, 
$$
which implies $0 \leq -2$. That's a contradiction. Thus $f'(x) = 0$.
QED
That proof is similar to the method used to prove Catalan's conjecture in $k[x]$ from the Mason--Stothers theorem, which is how I came up with the above proof.
Corollary: Let $k$ be a finite field. If $f(x) \in k[x]$ and $\alpha \in k^\times$ have the property that all irreducible factors of $f(x)$ and $f(x)+\alpha$ have multiplicity at least $2$, then $f(x)$ is a $p$th power in $k[x]$, where $p$ is the characteristic of $k$.
Proof: Since $k$ is finite we have $k = k^p$, so $k[x^p] = k[x]^p$. Now use the previous corollary.
QED
