Partial proof of first case of Fermat's Last Theorem

Question

Found the following theorem related to the first case of Fermat's Last Theorem is it correct?

Theorem:

Let $p$ be an odd prime and:

$$p \nmid xyz$$

$$\gcd(x,y,z)=1$$

$$x^{p} = y^{p} + z^{p}$$

Then:

$$\sum_{k = 1}^{p - 1}\frac{(p - 1)!u^{k}}{k!(p - k)!}\equiv 0 \pmod{p}$$

for $u \equiv y/z \pmod{p}$

Proof:

Consider the equation:

$$x^{n} - y^{n} \equiv (x - y)nx^{n - 1} \pmod{(x - y)^2}$$

which can be easily proved using induction on $n$.

Now put $n = p$ and note:

$$\gcd((x^{p} - y^{p})/(x - y),x - y)$$

$$= \gcd(x - y,px^{p - 1})$$

$$= \gcd(x - y,p) = 1$$

So we may conclude that:

$$x - y = r^{p}$$

$$(x^{p} - y^{p})/(x - y) = s^{p} \equiv 1 \pmod{p^2}$$

$$z = rs,\gcd(r,s) = 1$$

for some $r,s$.

By means of the binomial expansion of $x^{p} = ((x - y) + y)^{p}$ we get:

$$x^{p} - y^{p} = ((x - y) + y)^{p} - y^{p} = z^{p} = (rs)^{p}$$

$$\implies (rs)^{p} = \binom{p}{0}(x - y)^{p} + \binom{p}{1}y(x - y)^{p - 1} + ... + \binom{p}{p - 1}y^{p - 1}(x - y)$$

Divide by $x - y = r^{p}$:

$$s^{p} = \binom{p}{0}(x - y)^{p - 1} + \binom{p}{1}y(x - y)^{p - 2} + ... + \binom{p}{p - 1}y^{p - 1}$$

Remembering: $s^{p} \equiv 1 \pmod{p^2}$:

$$s^{p} \equiv \binom{p}{0}(x - y)^{p - 1} + \binom{p}{1}y(x - y)^{p - 2} ... + \binom{p}{p - 1}y^{p - 1}\equiv 1 \pmod{p^2}$$

Now note $x - y = r^{p} \implies \binom{p}{0}(x - y)^{p - 1} \equiv r^{p(p - 1)} \equiv 1 \pmod{p^2}$ so we get:

$$\binom{p}{1}y(x - y)^{p - 2} + \binom{p}{2}y^{2}(x - y)^{p - 3} + ... + \binom{p}{p - 1}y^{p - 1}\equiv 0 \pmod{p^2}$$

Divide by $p$:

$$\frac{(p - 1)!y(x - y)^{p - 2}}{1!(p - 1)!} + \frac{(p - 1)!y^{2}(x - y)^{p - 3}}{2!(p - 2)!} + ... + \frac{(p - 1)!y^{p - 1}}{(p - 1)!1!}\equiv 0 \pmod{p}$$

Substitute $u \equiv y/(x - y) \pmod{p}$ to get:

$$\sum_{k=1}^{p - 1} \frac{(p - 1)!u^k}{k!(p - k)!} \equiv 0 \pmod{p}$$

We conclude our theorem is correct.

Idea:

Let:

$$f(u) = \sum_{k=1}^{p - 1} \frac{(p - 1)!u^k}{k!(p - k)!} \equiv 0 \pmod{p}$$

Any $u$ such that $f(u) \equiv 0 \pmod{p}$ gives us a solution to:

$$(y + z)^p \equiv y^p + z^p \pmod{p^2}$$

To see this note:

$$pf(u) = \sum_{k=1}^{p - 1} \binom{p}{k}u^k = (u + 1)^p - (u^p + 1) \equiv 0 \pmod{p^2}$$

$$\implies (y/z + 1)^p \equiv (y/z)^p + 1 \implies (y + z)^p \equiv y^p + z^p \pmod{p^2}$$

Example:

We take $p = 7$, so we have:

$$f(u) \equiv u(u + 1)(u + 3)^{2}(u + 5)^{2} \equiv 0 \pmod{7}$$

Suppose now $u \equiv y/z \equiv -3 \equiv 4 \implies y \equiv 4z \implies x \equiv 5z \pmod{7}$

$$\implies x^{7} = (5z)^{7} \equiv y^{7} + z^{7} \equiv (4z)^{7} + z^{7} \pmod{7^2}$$

$$\implies (5z)^{7} \equiv (4z)^{7} + z^p \pmod{7^2}$$

Why have you deleted and recreated your user id? You have chosen to try to improve your maths by attacking hard problems. That is your choice, but I think you owe it to the people helping you and those trying to learn from the critique supplied that you keep your question history intact. — James Arathoon, Dec 15 '17 at 20:14
Better get to the point, if the above holds it can give proof for lot's of primes — , Dec 16 '17 at 01:20
Primes like: $3,5,11,17,23,29,41,47,53,71,89,101,107,113,131,137,149,..$ — , Dec 16 '17 at 03:29
Now posted also on MO: Question about theorem related to FLT, first case. — Martin Sleziak, Dec 18 '17 at 15:14

Dietrich Burde · Answer 1 · 2017-12-15T20:11:06.913

4

I don't see that $f(u)$ is irreducible modulo $17$. On the contrary, we have $$ f(u)=(u^3 + 2u^2 + 16u + 16)(u^3 + u^2 + 15u + 16)(u^2 + 15u + 15)(u^2 + 4u + 1)(u^2 + u + 8) $$ over $\mathbb{F}_{17}$.

Edit: The text is changed now.

edited Dec 15 '17 at 20:11

answered Dec 15 '17 at 20:05

Dietrich Burde

130,978

Thanks. Isn't it true that there are no solutions for $u$? – Dec 15 '17 at 20:10
Yes it was my misconception about irreducibility I guess. – Dec 15 '17 at 20:11
Could you write out the whole polynomial after $u=y/z$? – Dietrich Burde Dec 15 '17 at 20:12
$u+8u^{2}+40u^{3}+140u^{4}+364u^{5}+728u^{6}+1144u^{7}+1430u^{8}+1430u^{9}+1144u^{10}+728u^{11}+364u^{12}+140u^{13}+40u^{14}+8u^{15}+u^{16}$ – Dec 15 '17 at 21:03

score 0 · Answer 2 · answered Dec 29 '17 at 10:09

For a shorter proof note:

$$ x^p = y^p +z^p$$

$$ \implies (y + z)^p \equiv y^p + z^p \pmod{p^2}$$

Divide by $z^p$ (we have $p \nmid xyz$):

$$ (y/z + 1)^p \equiv (y/z)^p + 1 \pmod{p^2}$$

Substitute $u \equiv y/z \pmod{p}$

$$ (u + 1)^p \equiv u^p + 1 \pmod{p^2}$$

$$ \implies \sum_{k = 1}^{p - 1}\binom{p}{k}u^k \equiv 0 \pmod{p^2}$$

And dividing by $p$ gives gives $f(u)$.

As an example we prove the first case for $p = 5$, we have:

$$ f(u) \equiv u + 2u^2 + 2u^3 + u^4 \equiv (u + 1)(u^2 + u + 1) \equiv 0 \pmod{5}$$

Suppose now $u \equiv -1 \implies y + z \equiv 0 \pmod{5}$ which is impossible.

But then we have $u^2 + u + 1 \equiv 0 \pmod{5}$, which is also impossible because of $-3$ not being a square modulo $5$.

Partial proof of first case of Fermat's Last Theorem

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