Wednesday, September 4, 2013

Properties of A^x + B^y = C^z with Jonathan Neal


During our study of the Beal Conjecture with my friend and fellow mathematics graduate Jonathan Beal, we looked at the equation

A^x + B^y = C^z

where A, B, C, x, y, and z are integers. As a result A^x, B^y, and C^z are integers.

Let A^x be even (where A is a multiple of 2). Let B = p*m where p is a prime number. Then B^y = p^y * m^y.

The only even prime is 2. All other prime numbers (3, 5, 7, etc. ) are odd.

So if A^x is even and
p = 2 and m is even: p^y is even, m^y is even, B^y is even, and C^z is even.
p = 2 and m is odd: p^y is even, m^y is odd, B^y is even, and C^z is even.
p ≠ 2 and m is even: p^y is odd, m^y is even, B^y is even, and C^z is even.
p ≠ 2 and m is odd: p^y is odd, m^y is odd, B^y is odd, and C^z is odd

Assuming A^x is odd and
p = 2 and m is even: p^y is even, m^y is even, B^y is even, and C^z is odd.
p = 2 and m is odd: p^y is even, m^y is odd, B^y is even, and C^z is odd.
p ≠ 2 and m is even: p^y is odd, m^y is even, B^y is even, and C^z is odd.
p ≠ 2 and m is odd: p^y is odd, m^y is odd, B^y is odd, and C^z is even.

Eddie


This blog is property of Edward Shore. 2013




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