Millenium Problems

ITT I will try to solve (and in fact solve) all the Millennium Prize Problems one by one. I will do so by a new proof technique that has been proved to be quite powerful. It combined homothopy theory with algebraic geometry. Having said that, the proof technique itself is elementary though. So, let's go ahead.

1. [math] \displaystyle P=NP [/math]

By definition, polynomila algorithms admit decomposition in chains of smaller polynomial algorithms. Consequently, polynomial time algorithms do not solve problems where blocks, whoose order is the same as the underlying problem, require simultaneous resolution. Thus, in fact [math] \displaystyle P \neq NP [/math]

2. Hodge conjecture

Assuming that if a compact Kähler mainfold is complex-analytically rigid, the area-minimizing subvarieties approach complex analytic subvarieties. The set of singularities of an area-minimizng flux is zero in measure. The rest it left to the reader as an easy routine excersize.

3. Riemann hypothesis

This is a simple experimental fact. [math] \displaystyle 10^{13} [/math] roots of the Riemann hypothesis have been already tested and it suffices for all practical applications. In fact, one state a suitable statistical hypothesis and check it on the sample of, say, [math] \displaystyle 10^5 [/math] roots.

4. Yang–Mills existence and mass gap

Well, discrete infinite bosonic energy-mass spectrum of gauge bosons under Gelfand nuclear triples admits non-perturbative quantization of Yang-Mills fields whence the gauge-invariant quantum spectrum is bounded below. A particular consequence is the existence of the mass gap.

5. Navier–Stokes existence and smoothness

(To be continued)

(Cont.)

I haven't worked this one in such detail, but observing that

[math] \displaystyle \| L (u, v) \| ^ 2 = \sum_{n \ge 25} u ^ 2_ {2n} v ^ 2_ {2n +1} / n ^ 2 \le C\|(u_n/\sqrt n)\|_4^2 \|(v_n/\sqrt n)\|_4^2 \le C\|(u_n/\sqrt n)\|_2^2 \|(v_n/\sqrt n)\|_2^2 = C \left (\sum u ^ 2_ {n} / n \right) \left (\sum v ^ 2_ {n} / n \right) [/math]

one can easily find at leat one closed-form solution applying the bubble integral. In the equation, [math] \displaystyle L [/math] is a bilinear operator.

6. Birch and Swinnerton-Dyer conjecture

The problem with former attempts has been in the way elliptic curves have been dealt with. But this really admits a proof with a computer by checking the (finitely many) categories of curves.

I also have a simpler than Perelman's proof of the Poincare conjecture, but it's not worth the prize anymore

P=NP

P/P=NP/P
N=1

What if P=0?

Veeky Forums Pass user since June 2016.

>By definition, polynomila algorithms admit decomposition in chains of smaller polynomial algorithms. Consequently, polynomial time algorithms do not solve problems where blocks, whoose order is the same as the underlying problem, require simultaneous resolution. Thus, in fact P≠NP
I don't think you know what you're talking about

0 equals 1 times 0

Yeah but every other value of N works too

Veeky Forums Pass user since June 2016.

>Veeky Forums pass user since June 2016
what did he mean by this?

I got a Veeky Forums pass on my October, is there any way for it to not show that shitty text?

I don't buy it to brag, I bought it to help the site.

nope, we can all see it

no good deed goes unpunished

Veeky Forums Pass user since June 2016.

I knew you were a shitposter when you said you prefered Paprika

I know math badly. IS there any merit to OP's attempts?

no

this thread is embarassing

>Veeky Forums pass user
>posting on Veeky Forums enough to even consider getting a pass

why

>By definition, polynomila algorithms admit decomposition in chains of smaller polynomial algorithms.
No.
>Consequently, polynomial time algorithms do not solve problems where blocks, whoose order is the same as the underlying problem, require simultaneous resolution.
No.

This triggers me. A co-author once wanted to replace "polynomial complexity" with "exponential complexity" and insisted on the change when I pointed out he is wrong and that they are not the same. He managed by connections, there is absolutely no effing way he would have made it anywhere on merit.

>people will reply to this old ass pasta

>pasta
Why do you think so?

Any comments?