Question

Please anyone help me to prove Fermat's Last Theorem.

Answer 1

Answer:

Start by listing the first p-1 positive multiples of a:

a, 2a, 3a, ... (p -1)a

Suppose that ra and sa are the same modulo p, then we have r = s (mod p), so the p-1 multiples of a above are distinct and nonzero; that is, they must be congruent to 1, 2, 3, ..., p-1 in some order. Multiply all these congruences together and we find

a (2a) (3a) ... ((p-1)a) ≡ 1.2.3.....(p-1) (mod p)

which is, a(p-1)(p-1)! ≡ (p-1)! (mod p). Divide both side by (p-1)! to complete the proof.∎

Sometimes Fermat's Little Theorem is presented in the following form:

Corollary.

Let p be a prime and a any integer, then ap ≡ a (mod p).

Proof.

The result is trival (both sides are zero) if p divides a. If p does not divide a, then we need only multiply the congruence in Fermat's Little Theorem by a to complete the proof.∎

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Answer 2

Step-by-step explanation:

That would mean there is at least one non-zero solution (a, b, c, n) (with all numbers rational and n > 2 and prime) to an + bn = cn. ... If the assumption is wrong, that means no such numbers exist, which proves Fermat's Last Theorem is correct. 3. Suppose that Fermat's Last Theorem is incorrect.