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We proved Fermat's Little Theorem in class through induction.

That is, we proved that for $a \in \mathbb{N},$ p is prime, and $p\not | a$, then $a^{p-1} \equiv 1$(mod p).

We proved this using mathematical induction with respect to $a$ that $a^p \equiv a$(mod p).

I have been working on the homework, and one of our prompts is the following:

"Suppose that the base case is left out of the proof. Why would this render the proof invalid? However, the remaining lines would prove another proposition. What would that proposition be?"

I have spent two days on this problem, and I feel I am making no progress--what new proposition would result if we left out the base case?

Maria Edge
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  • The proposition wouldn't change, you just would no longer have a valid proof. There's an excellent intuitive explanation here as to how induction works, and why the base case is necessary. – user744868 Mar 20 '20 at 01:46

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Without the base case, you just have the inductive step, without actually setting the thing rolling.

In this case, the inductive step would be that whenever $a^p\cong a\pmod p$, we then have $(a+1)^p\cong a+1\pmod p$.

This, indeed, is different than Fermats little theorem.

(Incidentally, the reason the inductive step works is that, we get the freshman's binomial, mod p.)