# Possible proof of Fermat's Last Theorem for prime exponents greater than 2

I would appreciate if someone could check my attempt in proving the Fermat's Last Theorem for prime exponents greater than $2$.

Firstly, let's prove a couple of lemmas which state that sum or difference of two non-equal odd $p$-th powers cannot be a $p$-th power for prime $p>2$.

Lemma 1:
The difference of two $p$-th powers of non-equal odd positive integers is not a $p$-th power of an integer for prime $p>2$.

Take prime $p>2$, and odd $z,x \in\Bbb Z^+, \; z>x$. Let $a=z-x$. Obviously $a$ is even. $$z=a+x$$ $$z^{\ p}=\sum_{i=0}^p \binom{p}{i}a^{\ p-i}x^{\ i}=x^{\ p}+a\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i}$$ $$z^{\ p}-x^{\ p}=a\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i}$$ Assume that the right side is a $p$-th power of an integer.

**** And here comes my erroneous logic: ****
:(

Then, since $x$ is not divisible by $a$ (because $x$ is odd, and $a$ is even), it must be: $$\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a^{\ p-1})$$

$$\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1}+\sum_{i=0}^{p-2} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1}+a\sum_{i=0}^{p-2} \binom{p}{i}a^{\ p-2-i}x^{\ i} \equiv 0\;(mod\;a)$$ $$px^{\ p-1} \equiv 0\;(mod\;a)$$ However, since both $p>2$ and $x$ are odd, then $px^{\ p-1}$ is also odd, so it cannot be divisible by $a$ which is even. We reached a contradiction so Lemma 1 is proven.

Lemma 2:
The sum of two $p$-th powers of non-equal odd positive integers is not a $p$-th power of an integer for prime $p>2$.

Proof for the sum of two odd $p$-th powers not being a $p$-th power is almost the same. Start with odd $x,y \in\Bbb Z^+$, set $b=x+y$, and raise $x=b-y$ to the p-th power. If the sum $x^{\ p}+y^{\ p}$ were the p-th power, then $py^{\ p-1}$ would have to be divisible by $b$, which cannot be because $py^{\ p-1}$ is odd, while $b$ is even.

Fermat's Last Theorem: (from Wikipedia)

No three positive integers $x, y$, and $z$ can satisfy the equation $x^n+y^n=z^n$ for any integer value of $n$ greater than two.

Proof of the Fermat's Last Theorem for prime exponents greater than 2:

To prove the Fermat's Last Theorem for prime exponents greater than $2$, we assume that there are such $x,y,z$ and then rewrite all $3$ possible types of sums of odd and even $p$-th powers in such a way to get to a contradiction with the one of the above Lemmas.

Assume that for prime $p>2$ there exist $x,y,z \in\Bbb Z^+$, all three different, for which $$x^{\ p}+y^{\ p}=z^{\ p}$$

A) Assume that $x$ is even and $y$ is odd, both positive integers.

First note that then $z$ must be odd too.

We can rewrite it as $$z^{\ p}-y^{\ p}=x^{\ p}$$ however, since both $z$ and $y$ are odd then this is in contradiction with Lemma 1.

Therefore, combinations where one of $x,y$ is even and the other is odd are not possible.

B) Assume that both $x$ and $y$ are odd positive integers, then $x^{\ p}+y^{\ p}=z^{\ p}$ is in direct contradiction with Lemma 2.

Therefore, combinations where both $x$ and $y$ are odd are not possible.

C) Assume both $x$ and $y$ are even, then we can repeatedly divide them with $2$, until one, or both of them gets odd, which reduce this case to A) or B).

Therefore, combinations where both $x,y$ are even are also not possible.

Since neither of A), B), or C) hold, then there are no such $x,y,z \in Z^+$ to satisfy the equation $x^{\ p}+y^{\ p}=z^{\ p}$ for prime $p>2$ and $x,y,z$ all different.

• Why would $z^p-x^p\equiv0\bmod a^{p}$ even if it were a $p$-th power? Commented Jul 28, 2015 at 15:03
• Even if $z^p-x^p$ is a $p$th power and divisible by $a=z-x$, it does not follow that it would be divisible by $a^p$. For example $41^2-9^2=40^2$ is divisible by $41-9=32$, but it is not divisible by $32^2$. You only used $p>2$ later - I'm using $p=2$ here to show that the logic breaks down here. By Wiles' theorem counterexamples with $p>2$ don't exist so this is just for explaining my point :-) Commented Jul 28, 2015 at 15:04
• @AdLibitum I think he/she assumed that $a$ is prime. Commented Jul 28, 2015 at 15:05
• Thanks guys! I appreciate it. :) Commented Jul 28, 2015 at 15:45
• what about Kummer's Lemma? Commented Jan 2, 2022 at 23:12

Then, since $x$ is not divisible by $a$ (because $x$ is odd, and $a$ is even), it must be: $$\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} \equiv 0\;(mod\;a^{\ p-1})$$
If you write $a\sum_{i=0}^{p-1} \binom{p}{i}a^{\ p-1-i}x^{\ i} = y^p$, as you do, then we have that $a | y^p$. This does not imply that $a^p | y^p$, as $a$ is not necessarily prime. For instance $12$ divides $36 = 6^2$ but $12$ does not divide $6$.