A Diophantine Equation.

[phoenix1729]

How to solve the Diophantine equation $a^{2n}+b^2=c^{2n}$, where $n>1$ ^ $a,b,c\ne0$?

[Buenchico]

Welcome to The AnswerBank, Phoenix1729.

Despite Diophantine equations being the subject of my university thesis (nearly half a century ago!) though, I'm afraid that I can't help you with a solution and I doubt that many others here can either. (Sorry!).

The purpose of my post though is to direct you to another forum, where you're more likely to be able to find a useful answer, which is the Mathematics section of the excellent Stack Exchange:
https://math.stackexchange.com/

Good luck in your quest!

[ClareTG0ld]

I'm not sure I share Buenchico's pessimism about what help on this problem can be found here. Still, I am not going to solve this problem tonight.

I will, however, note that this is equivalent to investigating Pythagorean triples: let a^n = A and c^n = C, then your equation reduces to

A^2 +b^2 = C^2 .

It so happens that all integer solutions of this equation can be written in either the form

A = k(m^2 - p^2), b = 2kmp , C = k(m^2 + p^2) ,

or the form

A = 2kmp, b = k(m^2 - p^2) , C = k(m^2 + p^2) .

for any set of integers (k,m,p). As an example, k = p = 1, m = 2 gives the solution (A=3 , b=4 , C=5) or (A=4, b=3, C=5).

This makes the problem equivalent to solving

c^n = k(m^2 + p^2) and {a^n = k(m^2 - p^2) or  2kmp }

for the same n.

At this point I've not been able to progress any further, but my exploration of "small" triples (A,b,C) is leading me to suspect that there are few solutions, if any, for n>1. But at least these formulas are a useful starting point: in particular, it's probably useful to start with the smaller question, of what solutions exist specifically for

c^n = k(m^2 + p^2) ;

where n>1 etc. One (trivial) set of solutions is when n=2 and k=1, so that c,m,p are themselves a Pythagorean triple; can a = m^2-p^2, or a = 2mp, in this case also be an exact square?