a) We prove that $ \alpha(n^2)\leq\alpha(n)(\alpha(n)+1)/2$. Let $ k=\alpha(n)$; $ n=2^{u_1}+\cdots{2^{u_k}}$. Thus $ n^2=\sum_{i=1}^{k}{2^{2u_i}}+\sum_{i<j}2^{u_i+u_j+1}$ and $ \alpha(n^2)\leq{k+(_2^k)}=k(k+1)/2$. b) If $ n=2^u+1$, $ u\geq{2}$, then the above inequality is an equality because $ \alpha(n)=2$ and $ \alpha(n^2)=3$.

(c) there exists a sequence $ (n_i )^{\infty}_1$ such that $ \frac {\alpha ( n^2_i )}{\alpha (n_i }$ goes to zero as $ i$ goes to $ \infty.$ $ n_i=2^{i+1}+(2^{i}-1)$ for $ i>3$ then $ \alpha(n_i)=1+(i)=i+1$ $ n_i=2^{i+1}+2^{i}-1=3.2^{i}-1$ $ n^2_i=(3.2^{i}-1)^2=9.2^{2i}-6.2^i+1$ $ n^2_i=2^{2i+3}+2^{2i}-3 2^{i+1}+1$ $ n^2_i=2^{2i+3}+2^{2i}-2^{i+3}+2^{i+1}+1$ $ n^2_i=2^{2i+3}+2^{i+3} (2^{i-3}-1) +2^{i+1}+1$ then $ \alpha(n^2_i)=1+(i-3)+1+1=i$

Hi mszew, $ \alpha(n_i^2)/\alpha(n_i)$ goes to $ 1$.

Here's a solution for part (c). Let $k$ be a sufficiently large positive integer, and $a > 2^k \cdot 10^{10^{10^{10}}}$ be an even larger positive integer. Let $n= 2^{a-1} - 2^{a-2} - 2^{a-4} - \cdots - 2^{a-2^k}.$ Observe that $\alpha (n) = 2^k - k.$ On the other hand, we can easily find that $$n^2 = 2^{2(a-2^k)} + 2 \sum_{1\le i,j \le k, i \neq j} 2^{a-2^i} \cdot 2^{a-2^j},$$ which implies that $\alpha(n^2) \le 1 + \binom{k}{2}$ since it's a sum of that many powers of two. From here, we are clearly done since $$\lim_{k \rightarrow \infty} \frac{1 + \binom{k}{2}}{2^k-k} = 0.$$ $\square$

The part a as stated here is false and there's a typo. In IMO compendium it stated the correct inequality that is $ \alpha(n^2) \leq \frac{1}{2} \alpha(n)(\alpha(n) + 1)$ and not $ \alpha(n) (n^2 ) \leq \frac{1}{2} \alpha(n)(\alpha(n) + 1)$