I was reading this piece of conjecture for Generalized Taxicab Number: https://arxiv.org/pdf/1901.09053.pdf
I was wondering for T(2,5,n) which we still haven't found any pairs, can we prove that it actually doesn't have a pair which the numbers are similar. For example, a^5+b^5 <> c^5+d^5, in which c^5=2*d^5. Simply put, can we prove that for any c, d of positive integers, we c^5<>2*d^5.
The ideas i have are using Fermat's infinite descent method, but can we prove that there are infinite integer numbers which can form another equation like c^5=2*d^5, so that c^5<>2*d^5? I can prove that it will, but i'm not sure it will be integers.
If we can, then with Fermat's last theorem, a^n+b^n<>c^n, we can easily see that a^5+b^5<>c^5, means a^5+b^5<>d^5+d^5. So we can never find n=2 ways in that the numbers are similar. After which we can focus on only different numbers, and assume c<d so we can again apply FIDM to prove that we can't find another pair.
I was wondering for T(2,5,n) which we still haven't found any pairs, can we prove that it actually doesn't have a pair which the numbers are similar. For example, a^5+b^5 <> c^5+d^5, in which c^5=2*d^5. Simply put, can we prove that for any c, d of positive integers, we c^5<>2*d^5.
The ideas i have are using Fermat's infinite descent method, but can we prove that there are infinite integer numbers which can form another equation like c^5=2*d^5, so that c^5<>2*d^5? I can prove that it will, but i'm not sure it will be integers.
If we can, then with Fermat's last theorem, a^n+b^n<>c^n, we can easily see that a^5+b^5<>c^5, means a^5+b^5<>d^5+d^5. So we can never find n=2 ways in that the numbers are similar. After which we can focus on only different numbers, and assume c<d so we can again apply FIDM to prove that we can't find another pair.
