Prove that the language is not regular without using the Pumping Lemma

Question

Question.

Prove that the language is not regular without using the Pumping Lemma:

$L=\{a^{n}b^{m}c^{k}\,\,|\,\,n^{2}+m^{2}=k^{2}\}$

Attempt.

I assume that L is regular, and I am trying to use closures properties on L and get $\{a^{i}b^{i}\,\,|\,\,i\geq0\}$

which is not regular (or some other non regular language), but it doesn't seem to work :(

Would appreciate any help

Answer 1

Direct from the definition of regular in terms of finite automata: Assume there is a finite automaton for $L$. Then there are distinct $n_1$ and $n_2$ such that the automaton is in the same state after reading $a^{n_1}$ or $a^{n_2}$. In fact, we find these duplicates not only in $\Bbb N$, but in every infinite subset thereof - for example in the infinite set of odd numbers. Then with $k=\frac{n_1+1}2$ and $m=\frac{n_1-1}2$, the automaton should accept $a^{n_1}b^mc^k$ but reject $a^{n_2}b^mc^k$, which is absurd.