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# Fermat's Lost Theorem

(Please notify if this is already a well known fact, I have no idea where to search) This note will discuss about how Fermat's Equation, $$x^{n}+y^{n}=z^{n}$$, where $$n>2$$, is just a special case of Pythagoras' Equation, and suggest a possible approach to rediscover Fermat's original proof of his last theorem.

We start by proving the connection between the two equation: Fermat's Equation (FE) and Pythagoras' Equation (PE).

Fermat's Equation: $$x^{n}+y^{n}=z^{n}, n>2$$

Pythagoras' Equation: $$x^{2}+y^{2}=z^{2}$$

Proof:

Fermat's Equation can be divided into two cases , when $$n$$ is odd and when $$n$$ is even.

Case 1:$$n$$ is even

If $$n$$ is even, then $$n$$ can be expressed as $$2m$$ for some integer $$m$$ and $$m>1$$.

Substitute it into FE and we get

$$x^{2m}+y^{2m}=z^{2m}$$

$$(x^{m})^2+(y^{m})^2=(z^{m})^2$$

and we're back at our PE.

Case 2: $$n$$ is odd

If $$n$$ is odd, then $$n$$ can be expressed as $$2m-1$$ for some integer m and $$m>1$$.

Same as before, substitute it into FE and we get

$$x^{2m-1}+y^{2m-1}=z^{2m-1}$$

$$\frac{x^{2m}}{x}+\frac{y^{2m}}{y}=\frac{z^{2m}}{z}$$

$$(\frac{x^{m}}{\sqrt{x}})^{2}+(\frac{y^{m}}{\sqrt{y}})^{2}=(\frac{z^{m}}{\sqrt{z}})^{2}$$

Once again, back at PE.

QED

Possible approach:

One thing to note is the way a Pythagorean Triple is constructed.

Now the problem has been reduced to proving no Pythagorean Triple can be written as $$(x^{m},y^{m},z^{m})$$ and $$(\frac{x^{m}}{\sqrt{x}},\frac{y^{m}}{\sqrt{y}},\frac{z^{m}}{\sqrt{z}})$$ for $$m>1$$, either using contradiction or playing with the equations until it no longer holds or both.

Note by Tan Wei Xin
1 year, 10 months ago

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It ain't that simple. What makes you think that $$\frac {x^m}{\sqrt m }$$ is an integer in the first place? Note that there's many invalid proofs attempting to prove Fermat's Last Theorem in the past. · 1 year, 10 months ago

noooo please dont gooooooo · 1 year, 9 months ago

I'm very well aware of many past attempts on the theorem, and I also mentioned that this is a suggestion, not an actual approach.

Also if $$\frac{x^{m}}{\sqrt{m}}$$ is not an integer then it would mean there is no whole number solution for odd $$n$$ and $$m>1$$, and we will be done for odd $$n$$. · 1 year, 10 months ago