Integer solutions for eighth power Diophantine equation

Question

I'm trying to solve the equation $x^8+y^8=n$ for integers $x$ and $y$ and a given $n$ which is quite large (about 160 decimal digits). There's a hint that $x$ and $y$ may be not too different.

I tried $x=int(\sqrt[8]{\frac{n}{2}})$ and then increasing $x$ by one in each step, checking whether $n-x^8$ is itself an eighth power. Did that in Python, had no luck after letting the program run for some time.

Any ideas how to speed up the search?

Answer 1

As suggested by Somos you can try first Gaussian integer factorization.

There are various resources on the net, that can help you performing this task. This URL for instance works with numbers as large as your $n$ : https://www.alpertron.com.ar/GAUSSIAN.HTM

It happens in this case that the returned numbers are perfect $4^{th}$ powers thus a solution is $$99778214590404624163^{\ 8}+96305429813873971636^{\ 8}$$

Answer 2

Solving in pari/gp:

? n=17223507887602526420151651536619310810581211968884712600471622453277672311605232817186376770608196483963283605920702226380449555021499188148868557483176337161377;
?
? Q=bnfinit(x^8+1);
?
? N=bnfisnorm(Q, n)
%1 = [Mod(99778214590404624163*x^3 - 3472784776530652527*x^2 + 3472784776530652527*x + 96305429813873971636, x^8 + 1), 1]
?
? f=lift(N[1]);
?
? x=polcoeff(f, 0)
%2 = 96305429813873971636
?
? y=polcoeff(f, 3)
%3 = 99778214590404624163
?
? x^8+y^8==n
%4 = 1