(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.

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## Comments

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TopNewestIt 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.

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noooo please dont gooooooo

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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 \(n\).

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