July 28th, 2014, 10:58 AM  #1 
Senior Member Joined: Sep 2010 Posts: 219 Thanks: 20  Fermat's Last again
Since the proof of FLT posted by MrAwojobi is based on algebraic approach to the problem I think itâ€™s reasonable to outline main steps of another way to prove it by means of algebra. The following has been deduced. 1. To satisfy equation $\displaystyle a^n+b^n=c^n$ it is required $\displaystyle a=uwv +v^n; b=uwv+w^n;c=uwv+v^n+w^n$ In the case of $\displaystyle n$ dividing $\displaystyle b$ these expressions are modified. 2. The $\displaystyle a^n+b^n$ is a product of $\displaystyle a+b=(u_p)^n$ and $\displaystyle a^{n1}a^{n2}b+â€¦+b^{n1}=(c_p)^n.$ Then $\displaystyle c=(u_p)(c_p); u=(u_p)(u_s)$. 3. $\displaystyle a^n+b^n =2(uwv)^n+n(uwv)^{n1}(v^n+w^n)+â€¦+nuwv[v^{n(n1)}+w^{n(n1)}]+v^{n*n}+w^{n*n}$ is a sum of polynomials divided by c $\displaystyle (uvw)^n+n(uwv)^{n1}v^n+â€¦+n(uwv)v^{n(n1)}$; (i) $\displaystyle (uwv)^n+n(uwv)^{n1}w^n+â€¦+n(uwv)w^{n(n1)}$; (ii) $\displaystyle v^{n*n}+w^{n*n}$. (iii) 4. The long division of (iii) by either (i) or (ii) results in remainder not divisible by $\displaystyle u$ i.e by factor $\displaystyle u_p$ of $\displaystyle c$. This contradicts previous conclusion and proves the FLT. 

Tags 
fermat 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fermat's Last Theorem  victorsorokin  Number Theory  84  May 23rd, 2012 03:23 PM 
Fermat's Last Theorem  mathbalarka  Number Theory  2  April 3rd, 2012 11:03 AM 
More Fermat's Last Theorem.  theomoaner  Number Theory  29  November 26th, 2011 11:23 PM 
Fermat  fakesmile  Calculus  0  June 2nd, 2009 05:36 AM 
Fermat  circum  Number Theory  1  June 15th, 2008 05:20 PM 