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July 11th, 2017, 10:52 AM   #111
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We are now no longer close to cracking it. It is already solved. Coming up in a few hours. Also it indeed is beautiful and simpler than all those trials we've tackled.

This has been possible through trusting instincts.
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July 13th, 2017, 09:42 PM   #112
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Proof of Fermat's last theorem.

lemma
If $a=b$, then $a^n+b^n\neq c^n$ for positive integers $a$, $b$, $c$ and $n>2$.
proof
If $a=b$,
\begin{align*}
a^n+a^n&=c^n\\
2a^n&=c^n\\
2&=\frac{c^n}{a^n}\\
2&=(\frac{c}{a})^n
\end{align*}
If $\frac{c}{a}$ is a fraction then $(\frac{c}{a})^n$ will be a fraction since $(\frac{c}{a})^n=\prod \limits^n\frac{c}{a}$, implying $c^n$ would not divide $a^n$.
Thus, $2=(\frac{c}{a})^n$ will hold iff $\frac{c}{a}$ is an integer. The only value of $\frac{c}{a}$ that satisfies the equation such that $n$ is a positive integer is $\frac{c}{a}=2$ so that $n=1$.
Thus, $a^n+b^n\neq c^n$ for positive integers $a$, $b$ $c$ and $n$ such that $n>2$ and $a=b$.

We therefore assume $a\neq b$.
We multiply both sides of the equation $c^n=a^n+b^n$ by $a^n-b^n$
\begin{align*}
c^n(a^n-b^n)&=(a^n+b^n)(a^n-b^n)\\
c^n(a^n-b^n)&=(a^n)^2-(b^n)^2\\
c^n(a^n-b^n)&=(a^2)^n-(b^2)^n\tag{1}
\end{align*}
When factoring,
\begin{align*}
a^n-b^n&=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})\\
a^n-b^n&=(a-b)\sum\limits_{i=1}^na^{n-i}b^{i-1}
\end{align*}
Therefore
\begin{align*}
(a^2)^n-(b^2)^n&=(a^2-b^2)\{(a^2)^{n-1}+(a^2)^{n-2}b^2+\cdots +a^2(b^2)^{n-2}+(b^2)^{n-1}\}\\
(a^2)^n-(b^2)^n&=(a^2-b^2)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}
\end{align*}
We substitute this in 1.
$$c^n(a-b)\sum\limits_{i=1}^na^{n-i}b^{i-1}=(a^2-b^2)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}$$
We divide both sides of the equation by $(a-b)$
\begin{align*}
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=\frac{(a+b)(a-b)}{(a-b)}\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}\\
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}
\end{align*}
\begin{align*}
c^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}\\
a^n+b^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}
\end{align*}

theorem
$a^n+b^n\neq c^n$ for positive integers $a$, $b$, $c$ and $n>2$.
proof
Let $\frac{\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}$ be $\frac{x}{y}$.
$$a^n+b^n=\frac{(a+b)}{y}x=a\frac{x}{y}+b\frac{x}{ y}$$
The equation above holds iff either $a+b=y$ so that $a^n+b^n=x$ or $\frac{x}{y}=1$ so that $a^n+b^n=a+b$.\\
If $\frac{x}{y}=1$ then $x=y$. Therefore,
$$\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}=\sum\limits_{i=1}^na^{n-i}b^{i-1}$$
$a^n+b^n=a+b$ if $n=1$. Therefore $x=y$ should hold for $n=1$. We substitute $n=1$
\begin{align*}
\sum\limits_{i=1}^1(a^2)^{n-i}(b^2)^{i-1}&=\sum\limits_{i=1}^1a^{n-i}b^{i-1}\\
(a^2)^0(b^2)^0&=a^0b^0\\
1&=1
\end{align*}
If $a+b=y$,
$$a+b=\sum\limits_{i=1}^na^{n-i}b^{i-1}$$
This equation is satisfied if $n=2$ such that
\begin{align*}
a+b&=a^1b^0+a^0b^1\\
a+b&=a+b
\end{align*}
Therefore $a^n+b^n=x$ should hold for $n=2$
\begin{align*}
a^n+b^n&=\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}\\
a^2+b^2&=\sum\limits_{i=1}^2(a^2)^{n-i}(b^2)^{i-1}\\
a^2+b^2&=(a^2)^1(b^2)^0+(b^2)^1(a^2)^0\\
a^2+b^2&=a^2+b^2
\end{align*}
Therefore the equation $$a^n+b^n
=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}$$
holds for $n=1$ and $n=2$ only. Meaning $a^n+b^n=c^n$, such that $a$, $b$ and $c$ are positive integers holds for $n=1$ and $n=2$ only.
Therefore $a^n+b^n\neq c^n$ for positive integers $a$, $b$ and $c$ and $n>2$.
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July 13th, 2017, 10:15 PM   #113
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Originally Posted by Mariga View Post
Proof of Fermat's last theorem.

lemma
If $a=b$, then $a^n+b^n\neq c^n$ for positive integers $a$, $b$, $c$ and $n>2$.
proof
If $a=b$,
\begin{align*}
a^n+a^n&=c^n\\
2a^n&=c^n\\
2&=\frac{c^n}{a^n}\\
2&=(\frac{c}{a})^n
\end{align*}
If $\frac{c}{a}$ is a fraction then $(\frac{c}{a})^n$ will be a fraction since $(\frac{c}{a})^n=\prod \limits^n\frac{c}{a}$, implying $c^n$ would not divide $a^n$.
Thus, $2=(\frac{c}{a})^n$ will hold iff $\frac{c}{a}$ is an integer. The only value of $\frac{c}{a}$ that satisfies the equation such that $n$ is a positive integer is $\frac{c}{a}=2$ so that $n=1$.
Thus, $a^n+b^n\neq c^n$ for positive integers $a$, $b$ $c$ and $n$ such that $n>2$ and $a=b$.

We therefore assume $a\neq b$.
We multiply both sides of the equation $c^n=a^n+b^n$ by $a^n-b^n$
\begin{align*}
c^n(a^n-b^n)&=(a^n+b^n)(a^n-b^n)\\
c^n(a^n-b^n)&=(a^n)^2-(b^n)^2\\
c^n(a^n-b^n)&=(a^2)^n-(b^2)^n\tag{1}
\end{align*}
When factoring,
\begin{align*}
a^n-b^n&=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots +ab^{n-2}+b^{n-1})\\
a^n-b^n&=(a-b)\sum\limits_{i=1}^na^{n-i}b^{i-1}
\end{align*}
Therefore
\begin{align*}
(a^2)^n-(b^2)^n&=(a^2-b^2)\{(a^2)^{n-1}+(a^2)^{n-2}b^2+\cdots +a^2(b^2)^{n-2}+(b^2)^{n-1}\}\\
(a^2)^n-(b^2)^n&=(a^2-b^2)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}
\end{align*}
We substitute this in 1.
$$c^n(a-b)\sum\limits_{i=1}^na^{n-i}b^{i-1}=(a^2-b^2)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}$$
We divide both sides of the equation by $(a-b)$
\begin{align*}
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=\frac{(a+b)(a-b)}{(a-b)}\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}\\
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}
\end{align*}
\begin{align*}
c^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}\\
a^n+b^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}
\end{align*}

theorem
$a^n+b^n\neq c^n$ for positive integers $a$, $b$, $c$ and $n>2$.
proof
Let $\frac{\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}$ be $\frac{x}{y}$.
$$a^n+b^n=\frac{(a+b)}{y}x=a\frac{x}{y}+b\frac{x}{ y}$$
The equation above holds iff either $a+b=y$ so that $a^n+b^n=x$ or $\frac{x}{y}=1$ so that $a^n+b^n=a+b$.\\
If $\frac{x}{y}=1$ then $x=y$. Therefore,
$$\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}=\sum\limits_{i=1}^na^{n-i}b^{i-1}$$
$a^n+b^n=a+b$ if $n=1$. Therefore $x=y$ should hold for $n=1$. We substitute $n=1$
\begin{align*}
\sum\limits_{i=1}^1(a^2)^{n-i}(b^2)^{i-1}&=\sum\limits_{i=1}^1a^{n-i}b^{i-1}\\
(a^2)^0(b^2)^0&=a^0b^0\\
1&=1
\end{align*}
If $a+b=y$,
$$a+b=\sum\limits_{i=1}^na^{n-i}b^{i-1}$$
This equation is satisfied if $n=2$ such that
\begin{align*}
a+b&=a^1b^0+a^0b^1\\
a+b&=a+b
\end{align*}
Therefore $a^n+b^n=x$ should hold for $n=2$
\begin{align*}
a^n+b^n&=\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}\\
a^2+b^2&=\sum\limits_{i=1}^2(a^2)^{n-i}(b^2)^{i-1}\\
a^2+b^2&=(a^2)^1(b^2)^0+(b^2)^1(a^2)^0\\
a^2+b^2&=a^2+b^2
\end{align*}
Therefore the equation $$a^n+b^n
=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}$$
holds for $n=1$ and $n=2$ only. Meaning $a^n+b^n=c^n$, such that $a$, $b$ and $c$ are positive integers holds for $n=1$ and $n=2$ only.
Therefore $a^n+b^n\neq c^n$ for positive integers $a$, $b$ and $c$ and $n>2$.
It is very late at night here; I am sleepy and cannot look at this in detail. But you show that a certain proposition may be valid if n = 1 or if n = 2, and then assert, without any proof that I see in my drowsy state, that the same proposition cannot be valid if n > 2. Don't you need to prove that?
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July 13th, 2017, 10:39 PM   #114
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Quote:
Originally Posted by Mariga View Post
The equation above holds iff either $a+b=y$ so that $a^n+b^n=x$ or $\frac{x}{y}=1$ so that $a^n+b^n=a+b$.\\
Prove it
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July 14th, 2017, 11:04 AM   #115
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Quote:
Originally Posted by JeffM1 View Post
It is very late at night here; I am sleepy and cannot look at this in detail. But you show that a certain proposition may be valid if n = 1 or if n = 2, and then assert, without any proof that I see in my drowsy state, that the same proposition cannot be valid if n > 2. Don't you need to prove that?
Go through it again. I've shown that $a^n+b^=c^n$ for positive integers $a$, $b$, $c$ and $n$ holds iff $n=1$ or $n=2$.

Isn't that a sufficient proof?
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July 14th, 2017, 11:10 AM   #116
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I may address this in a number of posts because I have other demands on my time.

Quote:
Originally Posted by Mariga View Post
Proof of Fermat's last theorem.

lemma
If $a=b$, then $a^n+b^n\neq c^n$ for positive integers $a$, $b$, $c$ and $n>2$.
proof
If $a=b$,
\begin{align*}
a^n+a^n&=c^n\\
2a^n&=c^n\\
2&=\frac{c^n}{a^n}\\
2&=(\frac{c}{a})^n
\end{align*}
If $\frac{c}{a}$ is a fraction then $(\frac{c}{a})^n$ will be a fraction since $(\frac{c}{a})^n=\prod \limits^n\frac{c}{a}$, implying $c^n$ would not divide $a^n$.
Thus, $2=(\frac{c}{a})^n$ will hold iff $\frac{c}{a}$ is an integer. The only value of $\frac{c}{a}$ that satisfies the equation such that $n$ is a positive integer is $\frac{c}{a}=2$ so that $n=1$.
Thus, $a^n+b^n\neq c^n$ for positive integers $a$, $b$ $c$ and $n$ such that $n>2$ and $a=b$.
I do not believe that is a valid proof, but I am pretty sure that a valid proof could be constructed using the Fundamental Theorem of Arithmetic so I am not going to fuss about it.

Quote:
Originally Posted by Mariga View Post
We therefore assume $a\neq b$.... [to get]
\begin{align*}
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=\frac{(a+b)(a-b)}{(a-b)}\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}\\
c^n\sum\limits_{i=1}^na^{n-i}b^{i-1}&=(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}
\end{align*}
\begin{align*}
c^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}\\
a^n+b^n&=\frac{(a+b)\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}
\end{align*}
This looks like algebra that has been checked before several times (in fact I may have got this exact format in a previous post) so I am going to accept it provisionally.


Quote:
Originally Posted by Mariga View Post
Let $\frac{\sum\limits_{i=1}^n(a^2)^{n-i}(b^2)^{i-1}}{\sum\limits_{i=1}^na^{n-i}b^{i-1}}$ be $\frac{x}{y}$.
$$a^n+b^n=\frac{(a+b)}{y}x=a\frac{x}{y}+b\frac{x}{ y}$$
The equation above holds iff either $a+b=y$ so that $a^n+b^n=x$ or $\frac{x}{y}=1$ so that $a^n+b^n=a+b$.
As bruno cogently pointed out, this is merely asserted rather than proved.

Consider the case when n = 1.

$\dfrac{x}{y} = \dfrac{\sum\limits_{j=1}^n(a^2)^{(n-j)}(b^2)^{(j-1)}}{\sum\limits_{i=1}^na^{(n-j)}b^{(j-1)}} \iff$

$\dfrac{x}{y} = \dfrac{(a^2)^{(1-1)}(b^2)^{(1- 1)}}{a^{(1-1)}b^{(1- 1)}} \iff$

$\dfrac{x}{y} = 1 \iff$

$\dfrac{x}{y} * (a + b) = a^1 + b^1.$ Good so far.

How about when n = 2.

$\dfrac{x}{y} = \dfrac{\sum\limits_{j=1}^n(a^2)^{(n-j)}(b^2)^{(j-1)}}{\sum\limits_{i=1}^na^{(n-j)}b^{(j-1)}} \iff$

$\dfrac{x}{y} = \dfrac{(a^2)^{(2-1)}(b^2)^{(1- 1)} + (a^2)^{(2-2)}(b^2)^{(2-1)}}{a^{(2-1)}b^{(1- 1)} + a^{(2-2)}b^{(2-1)}} \iff$

$\dfrac{x}{y} = \dfrac{a^2 + b^2}{a + b} \iff$

$\dfrac{x}{y} * (a + b) = a^2 + b^2.$

Thus, it is true that

$\{n = 1 \text { or } n = 2\} \iff \left \{\dfrac{x}{y} = 1 \text { or } y = a + b \right \}.$

But that says nothing whatsoever if n > 2. Thus the rest of the proof hangs in the air.
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July 15th, 2017, 01:03 AM   #117
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I see your point guys. More work needs to be done on this. I'll see what I'll make of it.
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July 15th, 2017, 05:28 AM   #118
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...and Big Pete Fermat smiles from above...
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July 17th, 2017, 06:09 AM   #119
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Once again: find a specific property of the powers of integers, the difference from n=2 and n>2, and a new tools to play with them... or be sure you cannot say "eureka"...

I believe my Complicate Modulus Algebra is the only chance after Wiles... and I'm still working on...
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July 19th, 2017, 04:05 AM   #120
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Denis, I wouldn't want you to have a heart attack. Brace yourself. The next proof is the real one.

Complicatemodulus you are indeed right. I need to include more tools because I've been just revolving around one concept for quite some time and it isn't sufficient enough on its own.

The proof I am soon going to give might show $a^n+b^n=c^n$ holds iff $n=1$ or $n=2$ and thus I wouldn't need to prove for $n>2$. It will lock on those two. That's the beauty of an algebraic proof.
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