Beal's Conjecture Counterexample

 I'm a math geek. Every once in a while I like to attack big problems as an intellectual workout. The problem I picked this time was known as Beal's conjecture. Why should you care? Aside from the sheer beauty of formulas and ramifications to fields like math and cryptography/cryptocurrency, it has a million-dollar prize. https://www.bealconjecture.com/

Everyone is familiar with the Pythagorean Theorem: A^2+B^2=C^2, but Fermat claimed (recently proven after 357 years of attempts) that this formula can't work for any power over two. 

A^n+B^n =C^n has no integer solutions. In the process of trying to prove Fermat's Last Theorem, a mathematician by the name of Andrew Wiles determined that in the general case A^x + B^y = C^z has no solutions for x, y, and z above 3 unless A,B, and C have a common factor. The American Mathematical Society has offered the big prize to anyone who can prove or disprove the conjecture.

I found how I could take any prime and use modulo arithmetic rules to construct examples where that prime is the common factor.

2^3 + 2^3 = 2^4

3^6 + 3^6*2^3 = 3^8

(5^2*41)^10 + (5^2*41*2)^10 = (5^2*41)^11

7^3 + 7^4 = 7^3*2^3

11^5*3^5 + 11^5*3^5*2^5 = 11^6*3^6  ... ad infinitum.

If there had been even one prime I couldn't do this with, it might have been a chink in the armor. No such luck. After playing with this for a week, I became convinced it's true, and nobody will be able to find a counterexample. Why? Because in an attempt to disprove it, I made the rules even weaker. I got rid of the common exponent condition and just looked for a counterexample where A, B, and C have no factors in common but all the exponents are above three. A^x + p1^m*p2^n*... = C^z

My reasoning for this was that if I generated a few dozen weaker solutions, I could then generalize a pattern for the counterexamples, and I could finagle one that also worked in the stronger Beal case. But that's never going to happen. In the looser construct, I found exactly one solution:

271^3 + 73^3*3^5*2^3 = 919^3

This example cannot be converted to the Beal case without increasing the power of 3 to a multiple of three. However, to do that, I would theoretically have to multiply every term in the equation by common factors, violating our rules and thus proving Beal's conjecture.

Of course, to make this an official proof I would have to show conclusively that this is the only solution, not just assert that it was the only one that doesn't overflow the integer variables on my computer and I didn't make a mistake in my programming. This is close enough for me. It adds to the body of knowledge without me having to lock myself in a room alone for months. These sorts of problems will eat you alive if you let them.