Question

Help i need help with yang mills existence and gap , can someone help me solve it

Answer 1

i can

Answer 2

@satellite73

Answer 3

ty crazy

Answer 4

ok

Answer 5

I can @matlee

Answer 6

Ok ty chosen

Answer 7

this is what i got so far 3^{3n}+[2(3^{n})]^{3}=3^{3n+2}; n \ge1
(a^{n}-1)^{2n}+(a^{n}-1)^{2n+1}=[a(a^{n}-1)^{2}]^{n}; a \ge2, n \ge3
and

[a(a^n+b^n)]^n+[b(a^n+b^n)]^n=(a^n+b^n)^{n+1}; a \ge1, b \ge1, n \ge3
Furthermore, for each solution (with or without coprime bases), there are infinitely many solutions with the same set of exponents and an increasing set of coprime bases. That is, for solution

A_1^{x}+B_1^{y}=C_1^{z}
we additionally have

A_{n}^{x}+B_{n}^{y}=C_{n}^{z}; n \ge2
where

A_{n}= (A_{n-1}^{yz+1}) (B_{n-1}^{yz }) (C_{n-1}^{yz })
B_{n}= (A_{n-1}^{xz }) (B_{n-1}^{xz+1}) (C_{n-1}^{xz })
C_{n}= (A_{n-1}^{xy }) (B_{n-1}^{xy }) (C_{n-1}^{xy+1})
Any solutions to the Beal conjecture will necessarily involve three terms all of which are 3-powerful numbers, i.e. numbers where the exponent of every prime factor is at least three. It is known that there are an infinite number of such sums involving coprime 3-powerful numbers;[8] however, such sums are rare. The smallest two examples are:

\begin{align}
271^3 + 2^3 3^5 73^3 = 919^3 &= 776,151,559 \\
3^4 29^3 89^3 + 7^3 11^3 167^3 = 2^7 5^4 353^3 &= 3,518,958,160,000 \\
\end{align}

Answer 8

what is the question?

Answer 9

ok.......795415385937

Answer 10

Wow thanks!

Answer 11

no problem :)