Solution:

Consider the sequence $\{b_{n}\}$ defined by $b_{n}=a_{n}+d$ for all $n$, so that $b_{n+2}=b_{n+1}+b_{n}$ for all $n$. This sequence is determined by its first two terms $b_{1}$ and $b_{2}$, and the same holds true if we reduce the sequence $\bmod a_{1}$. Taking remainders $\bmod a_{1}$, pairs of consecutive terms will repeat themselves, and so the sequence $\{b_{n}\}$ is periodic mod $a_{1}$, i.e., there exists a positive integer $\ell$ for which $b_{k+\ell} \equiv b_{\ell}\pmod{a_{1}}$ for all $k$, and so $a_{\ell+1} \equiv a_{1} \pmod{a_{1}}$. Thus, $a_{1} \mid a_{\ell+1}$. From (i), we must have $\left|a_{1}\right|=\left|a_{\ell+1}\right|$, and in fact, $\left|a_{1}\right|=\left|a_{k \ell+1}\right|$ for all $k$. In particular, $a_{k \ell+1}$, and thus $b_{k \ell+1}$, assumes at most two distinct values.

It suffices to show that $\{b_{n}\}$ is constant. Consider the characteristic polynomial of the recurrence defining $\{b_{n}\}$, $P(x)=x^{2}-x-1$. Let $\varphi$ and $\psi$ be the distinct roots of $P$, with $\varphi>\psi$. Note that in fact $\varphi>1$ while $0>\psi>-1$. There exists a unique pair of constants $c_{1}, c_{2}$, dependent on the values of $b_{1}$ and $b_{2}$, satisfying the system
$$\begin{aligned} & b_{1}=c_{1}+c_{2} \\ & b_{2}=c_{1} \varphi+c_{2} \psi \end{aligned}$$
It can be proved easily by induction that $b_{n}=c_{1} \varphi^{n-1}+c_{2} \psi^{n-1}$ for all $n \geq 1$. From this, we get that $\left|b_{n}\right| \geq\left|c_{1}\right| \varphi^{n-1}-\left|c_{2}\right|$. If $c_{1} \neq 0$, then $\left|b_{n}\right|$ eventually grows without bound, which contradicts our previous assertion that $\left|b_{k \ell+1}\right|$ assumes at most two values. Thus, $c_{1}=0$, and consequently, $b_{n+1}=\psi b_{n}$. However, $b_{n+1}$ and $b_{n}$ are integers while $\psi$ is irrational. This forces us to conclude that $b_{n+1}=b_{n}=0$, and so $c_{2}=0$ as well. Thus, $b_{n}=0$ for all $n$, and $a_{n}=a_{1}$ (with $d=-a_{1}$ ).