*) The KEY Lemma.

All calculations are done with numbers in base n, a prime number greater than 2.

The notations that are used in the proofs:

A', A'', A''', A_(k) – the first, the second, the third, the k-th digit from the end of the number A;

A_[k] is the k-digit ending of the number A (i.e. A_[k] = A mod n^k).

The KEY Lemma about the endings compound-power numbers of the type A^{n^k}.

The equation a^{n^{k+1}}_[k+2]=[(xn^^{k+1}+a^{n^{k+1}}_[k+2] )(yn^^{k+1}+1)]_[k+2],

where k>0, 1<a<n, x and y are digits,

has a unique solution: x=y=0.

Since the proof for any k and 1<a<n is exactly the same, we will present the proof of the Lemma only for k=1 (this is the starting case in the proof of FLT):

1°) The equation a^{n^2}_[3]=[(xn^²+a^{n^2}_[3] )(yn^²+1)]_[3] has a unique solution: x=y=0.

Proof.

2°) Opening the brackets in 1°, we find: (x+ay)'n²=0, hence x=(-ay)'.

Now let's increase of the equality 1° (where x and y are DIGITS) into (n-1)-th degree.

3°) The factor (yn²+1)^n-1_[3] =[(n-1)yn²+1]_[3] and has the third digit [(n-1)y]'.

The three-digit ending of the number xn²+a^{n^2} in (n-1)-th degree is equal to

4°) [(n-1)xn²(a^{n^2})^n-2+1]_[3],

where a^{n^2}'=a (see Little Theorem), and the third digit of the 4° will be equal to

(n-1)xa^n-2, or

5°) (n-1)xa^n-1/a, where a^n-1'=1.

And now the third digit of the product (xn²+a^{n^2}_[3] )^n-1*(yn²+1)^n-1 will be equal to:

6°) (n-1)x/a+a(n-1)y [=0], from which it follows

7°) x=-a²y, where a≠1 (!). And in 2° x=-ay. From here we find that

8°) a²y=ay, or ya(a-1)=0, where a>1. And hence y=0 and x=0.

The theorem is proved.

Victor Sorokine. Mezos. 2 March 2017