4 ms·
I think a more interesting problem to consider after solving n^m = m^n is solving n^m = m^n + 1 in the natural numbers.
by nmmnthrowaway 3y ago
I think a more interesting problem to consider after solving n^m = m^n is solving n^m = m^n + 1 in the natural numbers.
- mahalex 3y agoIf we exclude 0, the only solutions are 2^1 = 1^2 + 1 and 3^2 = 2^3 + 1
- nmmnthrowaway 3y agoThat is correct. Why are there no others?
- mahalex 3y agoBecause Catalan’s conjecture.
- yarg 3y ago> Catalan's conjecture was proven by Preda Mihăilescu in April 2002. Cool, so the conjecture's a theorem. https://uni-göttingen.academia.edu/PredaMihailescu https://xn--uni-gttingen-8ib.academia.edu/PredaMihailescu
- hgsgm 3y agoWhich has an extremely complicated proof :-(
- mahalex 3y agoI wouldn’t call it extremely complicated; it is much simpler than the proof of FLT.
- nmmnthrowaway 3y agoYes, but there is a much simpler proof.
- judofyr 3y agoSomething related to prime factorization maybe?
- AnimalMuppet 3y agoDo tell.