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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.

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    I wrote up the details as Theorem 8.4 in http://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/Zinotes.pdf.2012-12-18

2 Answers 2

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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 $a$.

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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 $a=m^2+n^2$.