I have a very hard problem: Prove that, if $2^{2^j} a + 1$ divides $c^{2^j}+1$ for fixed integers $a,c$ and all nonnegative integers $j$, then $a=1$ and $c=2^l$ for some odd positive integer $l$, or else $a=0$.
Here is my progress on the problem so far:
Let $p$ be a prime divisor of $2^{2^j}a+1$. We have $p\mid 2^{2^j}+1\mid c^{2^j}+1$ for all $j\ge 0$, hence $c^{2^j}\equiv -1\pmod p$. Squaring both sides leads to $c^{2^{j+1}}\equiv 1\pmod p$. Thus, it is obvious that the multiplicative order of $c$ modulo $p$ is $2^{j+1}$. By Fermat's Little Theorem, we also know that $c^{p-1}\equiv 1\pmod p$, since obviously $\gcd(c,p)=1$. It follows that $2^{j+1}\mid p-1$ for all prime divisors $p$ of $2^{2^j} a + 1$. This means that $p=2^{j+1}k+1$ for all prime divisors $p$ of $2^{2^j} a+1$, which by the Euler-Lucas theorem is precicely a property of the Fermat numbers, i.e. $F_j=2^{2^j}+1$.
This is where I reached. I'm hoping that I can somehow conclude from this that $a=1$, i.e. $2^{2^j} a + 1$ is indeed the $j$'th fermat number (as Gilles Bonnet pointed out), but it seems difficult.
Added later: What if $2^n a+1$ divides $c^n+1$ for all nonnegative integers $n$, instead of just powers of 2? Does the problem become significantly easier?
Best Answer
As Elaqqad said in his comment, we can use Hadamard Quotient Theorem. For example, in this article (in french), it is proved the following result (see page 190).
Theorem : if $u$ and $v$ are in $\mathbb{Z}\backslash\{-1;0;1\}$ and $P_0$, $P_1$, $Q_0$, $Q_1$ are polynomials in $\mathbb{Z}[X]\backslash\{0\}$ such that for all integer $n>0$, $$P_1(n)u^n+P_0(n)\,|\,Q_1(n)v^n+Q_0(n)$$ then we can find $t\in\mathbb{N}^*$ such that $v=u^t$ and $Q_0P_1^t=(-1)^{t+1}Q_1P_0^t$.
So it is easy to deduce that if $a\not=0$ is such that $2^na+1\,|\,b^n+1$ for all integer $n>0$, then $a=1$ and $b=2^{2k+1}$ for some $k\geq0$.