Fermat’s little theorem states that if p is a prime number, then for

any integer a, the number a^{p}-a is an integer multiple of p.

a |

Assume that a , b , c naturals and p prime and 0<a≤b<c and a<p

a^{p}≡a(mod p)
b c |

Let’s assume that a^{p}+b^{p}=c^{p}

We can say

a + px + b + py = c + pz with x,y,z positive integers
a + b – c = p(z – x – y) If a + b – c ≤ 0 we have a + b ≤ c we have (a+b) a |

In conclusion the original assumption that a^{p}+b^{p}=c^{p} is false .

If a + b – c>0 because 0<a≤b<c implies b – c < 0 ,0 < a + b – c < a < p

then a + b – c ≥ 1 and because a + b – c < a implies a ≥ 2 and because a < p implies p > 2. *

If a+b -c<p and if a + b – c = p(z – x – y)

implies 0≤z – x – y<1 ,

because x,y,z are positive integers z – x – y=0

which implies

(1) a + b – c = 0 and a + b = c , raising to power p ,

a^{p}+b^{p}<(a+b)^{p}=c^{p}

Again the assumption that a^{p}+b^{p}=c^{p}is false .

(if z – x – y < 0 , a + b – c < 0 and we are in previous case

and if z – x – y ≥1 , a + b – c <a < p < p(z – x – y)

and then a + b – c ≠ p(z – x – y) and a^{p}+b^{p}=c^{p} is false )

We can’t find naturals 0<a≤b<c with a<p with p odd

prime to satisfy a^{p}+b^{p}=c^{p}.

We can extend this to a , b , c rational numbers ,

0<a≤b<c and a<p .

If a^{p}≠ c^{p}– b^{p} is true for a , b , c ,naturals a < p , is true for a rational , a < p and b , c naturals because c^{p}– b^{p} is natural.

We can divide a^{p}≠ c^{p}– b^{p } with k^{p} , k rational k > 1 and note (a/k) = q ,

q^{p}≠ (c/k)^{p}– (b/k)^{p} with q rational q < p.

Let’s pick d positive integer , p ≤ d , d≤b < c and

assume that d^{p}+b^{p}=c^{p }.

We can find k rational number such d/k < p and we have

(d/k)^{p} + (b/k)^{p} = (c/k)^{p } which is

false of course because d/k < p .

If p is not prime we can write p as a product of prime factors .

Assume that q composite , q = rp with p prime we can write

a^{q} = a^{rp} = (a^{r})^{p},

which implies that we can’t find positive integers to satisfy

a^{p}+b^{p}=c^{p} with p > 2.