3.1.1 Andrew Beal’s conjecture

This proof is also applicable for Andrew Beal’s conjecture.
For a case where exponent Nc for c represents any number greater than  2 and exponents for a and b  have common odd divisor N  greater than 2 we can express this as:

and the proof results directly from equation (3.3), which has no solution for natural numbers  s = b^Nb + a^Na  and  x  other than zero.

We can see that according to Beal’s conjecture all exponents for  a, b and  c  can be different but for a and  b have common odd divisor.

We should clarify now the case when  x=(a+b) – c=0
We have then for (3.2) c=s, b=s – a   and

thus

The left side of the equation contains only one component:  s to the power of  Nc. The expression on the right side in square brackets can be divisible by  s
only when  N is divisible by  s. However, it already cannot be divisible by  s² since the sum of two last components in square brackets is not divisible by  s²- see below:

The lack of divisibility of these last two components by  s² eliminates purposefulness of further exponent increase for  s as this being a divisor for  N. The same applies for second possible divisor of  N, which would have to be divisible by  s and then an exponent could be increased. Max exponent for  s as being a divisor of  N could be  N−1, but then expression in square brackets would be negative. It would become negative already at lower values of exponent for s.
The equation (3.8) would be then as shown below:

We can conclude then that  Nc=2 and N=s.

The equality:  s²= N^N=(N − a)^N+a^N, must be verified and it is only fulfilled for N=3.
For N>3, N² is definitely smaller than  (N − a)^N+ a^N.

This explains the single case:
 1³ + 2³ =3² as well as its equivalent case:  3³ + 6³ =3⁵.