 Solution

For $b>1$, Euler's theorem allows $d=\phi (b)>0$, since $a,b$ are coprime integers. At $b=1$ every congruence holds and one may take $d=1$. Let $y$ be the multiplicative order of $a$ modulo $b$. Write a nonnegative exponent $x=qy+r$ with $0\le r<y$ using the Euclidean division. Then $a^x\equiv a^r(\mathrm{mod}b)$, so $a^x\equiv 1$ implies $r=0$ by minimality of $y$. Thus $y\mid x$. Negative exponents, if included, give the same conclusion using the inverse residue of $a$.

A prime divisor of a Fermat number $F_n=2^{2^n}+1$ is odd, and satisfies $2^{2^n}\equiv -1(\mathrm{mod}p)$. Therefore $2^{2^{n+1}}\equiv 1(\mathrm{mod}p)$, so its multiplicative order divides $2^{n+1}$. It does not divide $2^n$, since $-1\ne 1$ modulo the odd prime. Every divisor of $2^{n+1}$ is a power of two, henceBy Fermat's little theorem and the divisibility just proved, $2^{n+1}\mid p-1$, giving $\boxed{p\equiv 1(\mathrm{mod}{2}^{n+1})}$.