i mean https://www.techrxiv.org/users/717330/articles/702287-on-fer......

so how much do you just stare at grass and yell at the sky

So you're arguing those are all good things, I don't understand tbh

I would make food shelters and start a school, a dream of mine.

On p.4 you argue that for integers a, b, c and n:

  (a + b − c)^n = (c − a)(c − b)g_1(n)

  => a + b − c = [(c − a)(c − b)g_1(n)]^(1/n)

  => g_1(n) | a + b - c

This doesn't follow as it stands. For example, if a=b=3 and c=n=2, then g_1(n)=16 whereas a + b - c = 4.

This is only true for a,b,c that satisfies a^2+b^2=c^2.

So a=b=3 and c=n=2 is not part of the solution set.

But you're not using a^n+b^n=c^n in your argument at the bottom of p.4. You just say "therefore g_1(n) divides a+b-c". My example shows that yhe implication doesn't follow in general. And it's not clear why it follows specifically if a,b,c are a solution to FLT.

i believe you skipped pages 1-3. g_1(2)=2 for all a,b,c with n=2. g(n) carries with it the assumption of a^n+b^n=c^n as i showed in pages 1-3.

Pages 1-3 simply show that (c-a) and (b-a) divide (a+b-c)^n (for even n), assuming they are a solution to FLT.

You then define g to be (a+b-c)^n/(c-a)(b-a), an integer.

I follow you this far. I do not see why g divides a+b-c, and I don't think the argument on p.4 proves it.

I won't reply further to this question about g, i do think i've been clear. and at this point you can be on your merry way still thinking it's wrong. but you simply misunderstood.

I've read your paper and followed the arguments and this is where I believe it falls down. Either that or it needs much better explanation.

it's a proof by contradiction. g would divide a+b-c IF a+b-c are integers.

for n=2, g(2)=(c-a)(c-b)g_1(2) and g_1(2)=2.

So only when n=2 is it true that g divides a+b-c.

Otherwise we get a contradiction that it divides. since then, g_1(n) for n>2 is not a factor of a+b-c, we can safely assume at least one of them was not an integer.

I honestly don't follow your last sentence. Why does g not being a factor of a+b-c mean they're not integers?

it follows specifically from the form on pages 1-3. i would recommend reading it with fresh eyes after a good night's rest.

It doesn't follow from anything on p1-3. Certainly not directly. If you were being genuine about this I think you would appreciate an opportunity to improve the proof rather than resort to insults!

Why post this? This appears to be the writings of a crank.

which part

On p.4 you argue that for integers a, b, c and n:

  (a + b − c)^n = (c − a)(c − b)g_1(n)

  => a + b − c = [(c − a)(c − b)g_1(n)]^(1/n)

  => g_1(n) | a + b - c

This doesn't follow as it stands. For example, if a=b=3 and c=n=2, then g_1(n)=16 whereas a + b - c = 4.

The part in which FLT is derived from the binomial theorem, lol.

Nah not derived, set equal to from the outset with essentially no explanatory test throughout but with enough effort to insert some arbitrary graphs and label them with 'hey neato look at the vibes'

My favorite part is "The derivative resembles the rhythm of a heartbeat."

This would definitely benefit from a bit more explanatory text as I'm struggling to understand what you've shown. The crux seems to be that if a^n+b^n=c^n then (c-a)(c-b) divides (a+b-c)^n. I haven't been through all the details of this, but I also don't see how that implies FLT.

If I'm not mistaken Fermat's last theroem isn't even featured in the proof. Like nowhere did I see a^n+b^n=c^n referenced in the proof,save for the end of page 1 and 3, but it's never featured in an equality. Just 'this implies this trust me bro'.

I've actually had another quick look and I now have a vague idea of the outline. It's an attempted proof by contradiction, where a solution to FLT is applied to the binomial theorem and some arguments about integrality are made to form a contradiction.

My issue at the moment is with a line at the bottom of p.4, which effectively says that if k^n = xy for integers k, n, x, y, then k must be a multiple of y. Unless I'm missing something this is clearly false, for example 2^4 = 4 x 4.

Exactly. I have my degree as well, so I was doing some research of my own. It’s an ongoing thing.

techrxiv.org/doi/full/10.36227/techrxiv.170629872.29614231/v1

Seems like some kind of novel approach. Looks right to me. But my math is okay, so I need better eyes.