Last Fermat Theorem, sumation, a couple of updates on u-function

By LFT, z^n-y^n=x^n, for x,y,z \in N, ie. x^n=\Sigma_{i=0}^{i=n}{y^i z^{n-i}}. When S(n)=\Sigma_{i=0}^{i=n}{y^i z^{n-i}}, then S(n)-xyS(n-2)=x^n+y^n. From there, we have x^n=(z-y)S_{n-1} and S_n=z^n+y^n+zyS(n-2), which leads to S_n=(y+{{a}\over{S_{n-1}}})^n+y^n+y(y+{{a}\over{S_{n-1}}})S_{n-2} for a fixed a \in N, where a is n-th square of a chosen integer, thus u(a)>7. From there, we have S(n)-xyS(n-2)=z^n. Haven’t gone further with this yet, as more in-depth analysis of the very sumations is required.

Another interesting thing, for n=3 of the LFT, is that from z^3=(z-y)(z^2+zy+y^2) and a={{p}\over{q}} for p,q \in N, where it is trivial to show that a<1, with substitutions z=y+ax and z^2+zy+y^2=x^2a^{-1}, we have determinant where {{3p^3}\over{q^3}} can’t must never be integer for contradiction to hold, assuming no leaks in analysis. As for the substitutions z-y={{x}\over{k}} and z^2+zy+y^2=kx^2 for some k\in N, we must make sure so that generality is not affected. Wanted to use this transformation as some sort (long-vision) of equivalence between the elementary NT and ring homomorphism (and its numerical criterion) from Wiles .

When it comes to u-function, it’s trivial to show that for a,b \in N, u(b)>u(a) either u(a) nor u(b) are not the borderlines for u(a+b), so the questions that arise are: is u(a+b)  limited by either the amount of primes equal to the argument, or – for instance – the smallest prime appearing in the sum equal to the very argument. I will update you on this shortly as soon as I have more information.

Yet another update on learning.


About misha

Imagine a story that one can't believe. Hi. Life changes here. Small things only.
This entry was posted in Mathematics. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s