# On Fermat’s last theorem and others...

1. Jul 6, 2015

### PengKuan

On Fermat’s last theorem

This theorem states that for any n except 2, the equation X^n+Y^n=Z^n is not true for any positive integer triplet X, Y and Z. Fermat’s “I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.” has fascinated mathematicians from 1637 but no one has found what his proof was. Let us try to understand this theorem better.

On Fermat’s last theorem
http://pengkuanonmaths.blogspot.com/2015/07/on-fermats-last-theorem.html
or On Fermat’s last theorem

2. Jul 6, 2015

Interesting.

3. Jul 6, 2015

### micromass

Staff Emeritus
As expected, the proof is flawed. I don't get the fascination with a very obscure mathematical result that already has a nice proof. Anyway, since

$$\prod_{j=1}^i\frac{1/n - j +1}{j} X^n$$ is not a natural number in $\{0,1,...,Y^n\}$, it is not the $i$'th digit in the expansion in the basis $Y^n$. So it being repeating or not has no impact whatsoever.

The flaw in the continued fraction thing is clear too since it doesn't satisfy the irrationality criterion.

Furthermore, in neither of your proofs have you really used that $n\neq 2$.

Last edited: Jul 6, 2015
4. Jul 6, 2015

### PengKuan

Thanks.

I know this. This is why I asked for help at the end of the article.

5. Jul 6, 2015

### HallsofIvy

Staff Emeritus
Years after writing this, Fermat published separate proofs for the cases 3 and 4. He would not have done that if he had a proof for "all n". What happened was what happens to everyone- he thought he saw a proof that would work for all n but on later consideration saw that there was an error. Fermat clearly did not have a valid proof.

6. Jul 6, 2015

### PengKuan

It appears all his remarks in margin are correct according the youtube video on Numberphile.

7. Jul 6, 2015

### PengKuan

I have changed my proof for continued fraction. It fits the criterion now.

I have added the mention n>2.

8. Jul 6, 2015

### micromass

Staff Emeritus
Again, where does your proof fail for $n=2$?

9. Jul 6, 2015

### PengKuan

I have computed for X=3 and y=4. The continued fraction converges to 5.

I think for n=2, the Pythagorean triples are exception for the theorem "if a1, a2,… and b1, b2,…are positive integers with ak<bk for all sufficiently large k, then the fraction converges to an irrational limit "

Indeed, all other x and y for n=2 give irrational z.

10. Jul 6, 2015

### micromass

Staff Emeritus
No, that theorem has no exceptions, since it is definitely proven to be true. So either your proof must somehow not work for $n=2$, or it is flawed in general.

11. Jul 6, 2015

### PengKuan

You may be right. But I'm unable to find the flaw.

This theorem is about number. All coefficients are numbers. But in my expression the fractions are functions. The pythagorean triple case may be reducible if they were number.

I have just checked my derivation. The coefficients for even k do not fit. So, this is the cause for n=2 case fail. Maybe for n>2 there can be new theorem that proves the irrationality. But I cannot find.

Last edited: Jul 6, 2015
12. Jul 6, 2015

### HallsofIvy

Staff Emeritus
What? The NumberPhile video says the exact opposite of what you seem to think it does. On the video about Fermat on "Numberphile" at 4:35, the person clearly says "Fermat thought he had a proof but was mistaken."The remark about Fermat's "remarks in the margin", at 3:15, is that "In every case where Fermat said he had a proof he was correct except this one."

That is essentially what I said.

Last edited: Jul 6, 2015
13. Jul 6, 2015

### PengKuan

I have mixed up two video that I have seen. Sorry.

14. Jul 6, 2015

### PengKuan

I have just checked my derivation. The coefficients for even k do not fit. So, this is the cause for n=2 case fail. Maybe for n>2 there can be new theorem that proves the irrationality. But I cannot find.