Proving a language is non-regular using the Pumping Lemma for non-binary strings [duplicate]

This question already has an answer here: I am unsure of how to prove this language is non-regular. I do not even know where to start to develop a string that would prove the language is non-regular by contradiction. Any help would be appreciated.

I understand that the string has to have a length greater than the pumping length of L. What does that even mean in this context of such a vast alphabet?

I imagine I have to choose a string that fits the properties:

must be in the form of the expression a+b=c no leading zeros in a,b,c the expression on the left must equal the value on the right But I do not know what string would fit all these properties but also provides a contradiction in the proof.

My Attempt: marked as duplicate by David Richerby, Thinh D. Nguyen, Evil, Yuval Filmus, JuhoNov 8 '18 at 8:50

You have demonstrated a pretty good understanding of the pumping lemma for regular languages. Your proof is clean and correct. (That mino typo, $$xy^{iz}$$ where $$z$$ should not be in the superscript, could be ignored by a human reader.)
Here is a minor suggestion. You can just use a specific string such as $$1^p+2^p=3^p$$, instead of a more general one $$a^p+b^p=c^p$$. Your proof will become easier to write and easier to understand.