Is Proteus' Theorem true?

[syndixxx]

There is the so-called “Proteus theorem or heteroscaling theorem,” which was probably first voiced (as a separate statement) by Ib Petrov (Ib is a real name, he was surprised himself) in his article of the same name. Petrov Ib. "Proteus' Theorem or Heteroscaling Theorem", self-published, 2024. — 6 с. (publication in Russian).

Actually the theorem goes like this:

Theorem: For integers a, b and c, where c > 1, a > 1 and b > 1, the equations a^b+b^a=c^a and a^b+b^a=c^b do not have complete solutions.

In essence, this is a special case of Diophantine equations. And it seems that the proof given by the author is understandable (I would say even on an intuitive level), but somehow it seems to me not entirely formal, in terms of its sufficiency. The topic is interesting, since here there is a relationship between the exponent and the base, as well as the “rate” of growth of functions on the right and left sides of the equation.

The author applies proof by contradiction, based on the fact that the function (a^b+b^a) has polynomial growth, and c^a has exponential growth.

It seems to me that this is not a formal (sufficient) proof, for example, for very large natural numbers. Although in essence these equations are a special case of Fermat’s Last Theorem - if you think about it...

 

[RobertSmart]

The Proteus Theorem, or Heteroscaling Theorem, states that for integers a, b, and c, where c>1, a>1, and b>1, the equations a^b + b^a = c^a and a^b + b^a = c^b do not have complete solutions.

This theorem is a special case of Diophantine equations and hints at the relationship between the exponent and the base, as well as the growth rates of the functions involved. The proof provided by Ib Petrov is based on a contradiction, arguing that a^b + b^a grows polynomially, whereas c^a grows exponentially.

However, the proof is considered informal and may lack sufficiency for very large natural numbers. This informal nature is because it relies on an intuitive understanding of growth rates rather than a rigorous formal argument. Additionally, the theorem's special case relation to Fermat's Last Theorem suggests a deeper underlying complexity.


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