When does $\sqrt{4x^3+4y^3+1}$ a natural number?

Question

If $x$ and $y$ is a natural number, how can we find $y$ in terms of $x$ so the $\sqrt{4x^3+4y^3+1}$ is a natural number?

Answer 1

For one family of solutions set $y=x^2$. Then

$z=\sqrt{4x^3+4y^3+1}=\sqrt{4x^3+4x^6+1}=\sqrt{(2x^3+1)^2}=2x^3+1$

This gives:

$(x,y,z) = (1,1,3), \space (2,4,17), \space (3,9,55), \dots$

There may, of course, be other solutions.

Answer 2

$4x^3+4y^3=k^2-1=(k-1)(k+1)$

Since LHS is even RHS is a product of two consecutive even numbers, let $k=2m+1$ we have:

$4x^3+4y^3=2m(2m+2)=4m(m+1)$

$x^3+y^3=m(m+1)$

So whenever $x^3+y^3=m(m+1)$ and $k=2m+1$,then $4x^3+4y^3+1=k^2$

For example:

$m=1$ ⇒ $1(1+1)=2=1^3+1^3$⇒$k=2+1=3$

$m=8$ ⇒ $8(8+1)=72=4^3+2^3$ ⇒ $k=2.8+1=17$

$m=90$ ⇒ $90(90+1)=8190=19^3+11^3$ ⇒ $k=2.90+1=181$.