Finding the smallest integer solution to $N=x_1^2+x_2^2-y^3$ for any given $N$, with $x_1>0$, $x_2>0$ and $y>0$

Question

Finding the smallest integer solution to $N=x_1^2+x_2^2-y^3$ for any given $N$, with $x_1>0$, $x_2>0$ and $y>0$

I’ll provisionally define smallest as the minimum $y$ for a given $N$, but I am open to suggestions. Here’s an example where defining the smallest as minimum $x_1x_2y$ would change the result: $$57=3^2+7^2-1^3=1^2+8^2-2^3$$

The answer I accepted to this question Conjecture: $N=x_1^2+x_2^2-y^3$ has integer solutions for all $N$, with $x_1>0$,$x_2>0$ and $y>0$ provides a parametric solution (Andrew Adler) for all $N$ (except for a few trivial cases).

$$2a+1=\left(a^3-3a^2+a\right)^2+\left(a^2-a-1\right)^2-\left(a^2-2a\right)^3$$

$$4a+2=\left(2a^3-2a^2-a\right)^2+\left(2a^3-4a^2-a+1\right)^2-\left(2a^2-2a-1\right)^3$$

$$8a+4=\left(a^3+a+2\right)^2+\left(a^2-2a-1\right)^2-\left(a^2+1\right)^3$$

$$16a+8=\left(2a^3-8a^2+4a+2\right)^2+\left(2a^3-4a^2-2\right)^2-\left(2a^2-4a\right)^3$$

$$16a=\left(a^3+7a-2\right)^2+\left(a^2+2a+11\right)^2-\left(a^2+5\right)^3$$

However, the answers produced by these ingenious formulae grow ever further from the smallest solution as $a$ increases.

$$57=19628^2+755^2-728^3$$

Example Smallest Solutions

$$-5=7^2+17^2-7^3$$ $$-4=4^2+14^2-6^3$$ $$-3=1^2+2^2-2^3$$ $$-2=3^2+4^2-3^3$$ $$-1=1^2+5^2-3^3$$ $$0=2^2+2^2-2^3$$ $$1=1^2+1^2-1^3$$ $$2=1^2+3^2-2^3$$ $$3=8^2+8^2-5^3$$ $$4=1^2+2^2-1^3$$ $$5=2^2+3^2-2^3$$

My question Is there a method, other than brute force, to find the smallest solution for given $N$?

If not, is there a method of reducing a known solution to a smaller solution?

Update 23 February 2018

When $N$ is square, $$N=a^2=1^2+a^2-1^3$$

Answer 1

There are some infinite subsets of the integers that can be addressed using Fermat's two-square theorem:

If $N$ is of the form $4n$ and $N+1$ is a prime, then there exist $x, y$ such that $x^2+y^2=N+1$ and therefore:

$$N = x^2+y^2-1^3$$

Using the Brahmagupta-Fibonacci identity, this can be extended to cases where $N+1$ is a product of primes of the form $4n+1$.

Also, if $N$ is of the form $8n+2$ and $(N+8)/2 = 4(n+1)+1$ is a prime, then there exist $x, y$ such that $x^2+y^2 = (N+8)/2$. Hence:

$$N+8 = (x+y)^2 + (x-y)^2$$

$$N = (x+y)^2 + (x-y)^2 - 2^3$$

and this is the smallest solution since $N+1^3$ is of the form $8n+3$ so cannot be a sum of two squares. Again, this extends to cases where $(N+8)/2$ is a product of primes of the form $4n+1$.