Integer values of difference between cube and sqaure I am trying to find square integer values for
$k = a^3-b^2$ and  $\gcd(a,b) = 1$
i.e. the values of and b for which k is a perfect square. 
 A: All right, I think I remember how to derive this.  First, you'll want the formula for Pythagorean triples
$$(m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$$
Now multiply both sides by $m^2+n^2$.
$$(m^2+n^2)^3=(m^2+n^2)[(m^2-n^2)^2+(2mn)^2]$$
Finally, from here, you'll want to use Brahmagupta's Identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2$$
to yield equations in the proper form with $a=m^2+n^2$.
A: If $a$ is such that its  prime divisors congruent to $3$ módulo $4$ appear with an even exponent, then the same is true of $a^3$. Fermat's theorem on sums of two squares implies that $a^3$ is the sum of two squares.
Edit.
The above answer does not take into account the requirement $\gcd(a,b)=1$. Let $a=p$ be a prime such that
$$p\equiv1\mod 4.$$
The $p$ can be written in a unique way as
$$
p=b^2+c^2,\quad b,c\in\mathbb{N},\quad b\,c\ne0.
$$
The formula for the number of representations of an integer as a sum of two squares shows that $p^3$ can be written as a sum of two squares in two different ways. One of them is
$$
p^3=(b\,p)^2+(c\,p)^2.
$$
Let the other one be $p^3=d^2+e^2$. Then we must have $\gcd(p,d)=\gcd(p,e)=1$. For instance
$$\begin{align*}
5^3&=2^2+11^2\\
13^3&=9^2+46^2\\
17^3&=47^2+52^2
\end{align*}$$
A similar analysis can be carried out for other integers $a$.
