Fermat Last Theorem for prime numbers

Question

I would like to prove Fermat Last Theorem for prime numbers i.e. that there is no such natural $n>1$ (in original FLastT $n>2$) for any $r,q,p$ primes (given here in increasing order) that $r^n+q^n=p^n$.

It seems much easier than general case for natural numbers as bases because we know that on the left side $r=2$ must be present - additionally $q,p$ are odd numbers.
So we can rewrite equation as

$2^n=p^n-q^n$.

The right side $p^n-q^n=(p-q)(p^{n-1}+p^{n-2}q+p^{n-3}q^2+\dots+q^{n-1})$ so $(p-q)=2^m$ and $(p^{n-1}+p^{n-2}q+p^{n-3}q^2+\dots+q^{n-1})= 2^{n-m}$ for some $m$ .

For $n$ odd the second expression has in bracket odd number of odd summands so the whole sum is also odd. This excludes odd $n$. For them there is no solution.

• However what if $n$ is even ?

For even $n$ $\ \ 2^n$ is divisible not only by $p-q$, but also by $p+q$.
So we could write $p-q=2^m$ and $p+q=2^k$ for some natural $m<k$.

From this follows $2p=2^m+2^k$ , $p=2^{m-1}+2^{k-1}$ and necessary $m=1$ as $p$ is odd.

So it seems that $p,q$ should have been twin primes i.e $p=q+2$.
It simplifies equation to the $2^n=p^n-(p-2)^n$.

• And how to prove further the case taking into account that $n$ is even ?

One of my ideas was to replace $p$ by $a+1$. Now $a$ is even.
Then we receive more symmetrical expression $2^n=(a+1)^n-(a-1)^n$, .. it would lead to the conclusion if in the factorization of $(a+1)^n-(a-1)^n$ for even $n$ odd factors occurred..

Indeed for example:

$(a+1)^4-(a-1)^4=8 a (a^2 + 1)$,
$(a+1)^6-(a-1)^6= 4 (3 a^2 + 1) a (a^2 + 3)$
$(a+1)^8-(a-1)^8 = 16 a (a^2 + 1) (a^4 + 6 a^2 + 1) $
$ (a+1)^{10}-(a-1)^{10} = 4 (5 a^4 + 10 a^2 + 1) a (a^4 + 10 a^2 + 5) $ $\dots$

but how to be sure of the case for greater $n$?

Answer 1

You can exclude the case where $n$ has an odd divisor in the same way as you did for $n$ odd: if $d$ is an odd divisor, let $P=p^{n/d},Q=q^{n/d}$, write $(2^{n/d})^d=P^d-Q^d$ and factor $P^d-Q^d$ to get an odd divisor, as before.

This reduces it to Fermat for $n=4$. This works for any odd $p,q$.

Alternatively, if you don't want to use Fermat for $n=4$: you found $n$ even and $p=q+2$. I thought of using the Lifting the exponent lemma, but in fact we don't need it. I kept the LTE solution below. Without it, write $$2^n=(q+2)^n-q^n \geq 2q^{n-1}$$ so $q \leq 2$, contradiction. (In fact, If $n,k\in\mathbb N$, solve $3^k-1=x^n$. is what shows up here, with $(p,q)=(3,1)$.)

Using Lifting the Exponent, compare $2$-adic valuations on both sides to have $n=v_2(q+1)+1+v_2(n)$, so $q\geq \frac{2^{n-1}}n-1$ yet $$2^n=LHS=RHS =(q+2)^n-q^n>q^{n-1} \geq (2^{n-2}/n)^{n-1}$$ which is a contradiction for $n$ large enough ($2^{n^2}\gg n^n$, in fact $n\geq$ something $10$ish already), and a computer can check the small cases. (Note how $n$ bounds $q$.) This is a bit ridiculous because of the much quicker argument above, but I don't think there's harm in presenting other ideas.

All these proofs work for any odd $p,q$.

Answer 2

Once you have $p-q=2^m$ for some $m\ge1$, you have

$$p^n=(q+2^m)^n=q^n+\cdots+2^{mn}\gt q^n+2^n$$

if $n\gt1$. The strict inequality comes from dropping the "$\cdots$", which at the very least includes a $nq^{n-1}2^m$. In other words, the equation $2^n=p^n-q^n$ has no integer solutions, in primes or not, if $n\gt1$.