For which values of $a$ is the solution for $x^2 - y^2 = a^3$ unique? 
For which values of $a$ is the solution unique? $$x^2-y^2=a^3$$

I'm not sure how to do this, so I've been looking at this guy's solution.
$x^2 - y^2 = a^3$ is factored into $(x-y)(x+y) = a^3$. He says that the parity of both $x-y$ and $x+y$ is the same because when you add them together you get $2x$. Then he says let $a = {p_1}^{k_1} ....{p_r}^{k_r}$ be the prime factorization. Then he says that $x^2 - y^2 = a^3$ has exactly $2(3k_1 + 1) ... (3k_r + 1)$ solutions for $(x,y)$. I don't understand how he knows that or where he got that from. Can someone explain this to me, thank you.
 A: Lets focus on the case when $a>0$. 
Let $a=p^{k_1}_{1}...p^{k_r}_{r}$ be the prime factorisation of $a$ then the number of distinct positive divisors of $a$ is equal to $\tau(a)=(k_1+1)...(k_r+1)$.
Therefore the number of distinct positive divisors of $a^3$ is $\tau(a^3)=(3k_1+1)...(3k_r+1)$. As said above $$x^2-y^2=a^3\Rightarrow (x-y)(x+y)=a^3$$ 
$x-y$ and $x+y$ have the same parity either they are both even or both odd. The existence of at least a solution is clear as $a$ and $a^2$ have the same parity and one could conjecture $x-y=a$ and $x+y=a^2$ which would yield $a+2y=a^2\Rightarrow y=\frac{a(a-1)}{2}\in \mathbb{N}$ if $a\in\mathbb{N}$. Now your question is why do we have that the total number of possible solutions is $2{\tau(a^3)}$? Lets consider the case when $a$ is an odd positive integer. Notice that if $d$ is a divisor of $a^3$ so is $a^3/d$. Now because $a$ is odd, $d$ and $a^3/d$ must necessarily have the same parity. They are both odd. Therefore the pair $(d,a^3/d)$ or $(a^3/d,d)$ could be a potential solution to $(x-y,x+y)$. How many such pairs do we have? We have exactly $2\cdot\tau{(a^3)}$. 
Now consider when $a$ is positive even integer, then $a=2^{k_1}...p^{k_r}_r$. Notice that in this case $(2^{3k_1},a^3/2^{3k_1})$,$(a^3/2^{3k_1},2^{3k_1})$,$(1,a^3)$ and $(a^3,1)$ can not be candidates for $(x-y,x+y)$ because $2^{3k_1}$ and $a^3/2^{3k_1}$ or $1$ and $a^3$have different parities. So in this case we have $2(\tau(a^3)-2)$ possible solutions.
