17. (a) Let $n=\sum_{i=1}^{k} 2^{a_{i}}$, so that $\alpha(n)=k$. Then $$n^{2}=\sum_{i} 2^{2 a_{i}}+\sum_{i<j} 2^{a_{i}+a_{j}}.$$ It is easy to see that $\alpha(n^2) \geq \alpha(n)$. For $n=2^a$, we have $\alpha(n^2) = \alpha(n) = 1$. For $n=2^a + 2^b$ with $a > b$, we have $\alpha(n^2) = \alpha(n) + 1$. For $n=2^a + 2^b + 2^c$ with $a > b > c$, we have $\alpha(n^2) = \alpha(n) + 2$. In general, for $n=\sum_{i=1}^{k} 2^{a_{i}}$ with $a_1 > a_2 > \cdots > a_k$, we have $\alpha(n^2) = \alpha(n) + (k-1)$.

(b) For $m \geq 1$, let $n_{m}=\sum_{i=0}^{2^{m}-1} 2^{i(i+1) / 2}$. It is easy to see that $\alpha(n_{m})=2^{m}-m$. On the other hand, squaring and simplifying yields $n_{m}^{2}=1+\sum_{i=0}^{2^{m}-2} 2^{i(i+3) / 2}$. Therefore, $\alpha(n_{m}^{2})=2^{m}-m+1$. It follows that $$\lim _{m \rightarrow \infty} \frac{\alpha(n_{m}^{2})}{\alpha(n_{m})}=\lim _{m \rightarrow \infty} \frac{2^{m}-m+1}{2^{m}-m}=1.$$

(c) Let $\gamma \in [0,1]$ be a constant to be chosen later, and let $N_{i}=2^{m_{i}} n_{i}-1$ where $m_{i}>\alpha(n_{i})$ is such that $m_{i} / \alpha(n_{i}) \rightarrow \theta$ as $i \rightarrow \infty$. Then $\alpha(N_{i})=\alpha(n_{i})+m_{i}-1$, whereas $N_{i}^{2}=2^{2 m_{i}} n_{i}^{2}-2^{m_{i}+1} n_{i}+1$ and $\alpha(N_{i}^{2})=\alpha(n_{i}^{2})-\alpha(n_{i})+m_{i}$. It follows that $$\lim _{i \rightarrow \infty} \frac{\alpha(N_{i}^{2})}{\alpha(N_{i})}=\lim _{i \rightarrow \infty} \frac{\alpha(n_{i}^{2})+(\theta-1) \alpha(n_{i})}{(1+\theta) \alpha(n_{i})}=\frac{\theta-1}{\theta+1}$$ which is equal to $\gamma \in[0,1]$ for $\theta=\frac{1+\gamma}{1-\gamma}$ (for $\gamma=1$ we set $m_{i} / \alpha(n_{i}) \rightarrow \infty$).

(d) Let be given a sequence $\left(n_{i}\right)_{i=1}^{\infty}$ with $\alpha(n_{i}^{2}) / \alpha(n_{i}) \rightarrow \gamma$. Taking $m_{i}>\alpha(n_{i})$ and $N_{i}=2^{m_{i}} n_{i}+1$ we easily find that $\alpha(N_{i})=\alpha(n_{i})+1$ and $\alpha(N_{i}^{2})=\alpha(n_{i}^{2})+\alpha(n_{i})+1$. Hence $\alpha(N_{i}^{2}) / \alpha(N_{i})=\gamma+1$. Continuing this procedure we can construct a sequence $t_{i}$ such that $\alpha(t_{i}^{2}) / \alpha(t_{i})=\gamma+k$ for an arbitrary $k \in \mathbb{N}$.