 mersenneforum.org > Math Help me check the proof, please.
 Register FAQ Search Today's Posts Mark Forums Read 2019-02-13, 04:16 #1 IHeartMath   Feb 2019 Taiwan 1 Posts Help me check the proof, please. Prove a^p+b^p=c^p⇒gcd(c,a+b)=2, where a⊥b⊥c and p is a prime number greater than 2. Proof. Initially, we can write a+b+2c = (c-b)+(c-a)+2(a+b). (1) Assume gcd( c,(a+b) ) = c1. (2) If c1 is equal to 1, then gcd(c^p,(a+b)^p) = c1^p = 1. (3) Since term a+b is the divisor of c^p, we can have gcd(c^p,(a+b)^p) >= gcd(c^p,(a+b)) = a+b > 1, (4) which is against to (3). Hence, we have c1 > 1. (5) We can write c1|(a+b)+2c ⇒ c1|(c-b)+(c-a)+2(a+b) ⇒ c1|(c-b)+(c-a). (6) Since c1|c ⇒ c1|a^p+b^p, (7) we have c1|{ [ (c-b)+(c-a) ][ c^(p-1)+...+b^(p-1) ] }-( a^p+b^p ) ⇒ c1|{ (c-b)[ c^(p-1)+...+b^(p-1) ]-a^p }+(c-a)[ c^(-1)+...+b^(p-1) ]-b^p ⇒ c1|{ 0 }+(c-a)[ c^(-1)+...+b^(p-1) ]-(c-a)[ c^(p-1)+...+a^(p-1) ] ⇒ c1|(c-a){ [ c^(-1)+...+b^(p-1) ]- [ c^(p-1)+...+a^(p-1) ] }. (8) Because term c-a is the divisor of b^p, we have c1|[ c^(-1)+...+b^(p-1) ]- [ c^(p-1)+...+a^(p-1) ] ⇒ c1|b^(p-1)-a^(p-1). (9) Because c1 is the divisor of a+b, we have c1|a^(p-1)+b^(p-1). (10) Considering both (9) and (10), then we have c1|2b^(p-1) ⇒ c1|2. (11) Because c1 is greater than 1, the only reasonable value of c1 is 2. Q.E.D. Discussion. If this lemma is correct, then it will lead to a+b = 2^p, which automatically proves FLT.   2019-02-16, 01:05 #2 dcheuk   Jan 2019 Iowa, US 5×43 Posts What is ⊥ in a⊥b⊥c? And also Fermat's last theorem, which was proven by Andrew Wiles some years ago, states that you can't find any integer n>2 such that a^n+b^n=c^n where a,b,c are integers. Last fiddled with by dcheuk on 2019-02-16 at 01:09   2019-02-16, 03:41   #3
a1call

"Rashid Naimi"
Oct 2015
Out of my Body

110111101102 Posts Quote:
 Originally Posted by dcheuk What is ⊥ in a⊥b⊥c?

https://www.johndcook.com/blog/2010/...atively-prime/   2019-02-16, 17:58   #4
dcheuk

Jan 2019
Iowa, US

5·43 Posts Quote:
 Originally Posted by a1call https://www.johndcook.com/blog/2010/...atively-prime/
I didn't know that, thanks!

I was googling that symbol all over the place.   2019-02-16, 18:54 #5 a1call   "Rashid Naimi" Oct 2015 Out of my Body 110111101102 Posts My pleasure dcheuk, Here is a useful resource for writing formal math statements: https://en.m.wikipedia.org/wiki/List...atical_symbols   2019-02-18, 14:52   #6
Dr Sardonicus

Feb 2017
Nowhere

3×52×41 Posts Here is where I become unable to check your work:

Quote:
 Since c1|c ⇒ c1|a^p+b^p, (7) we have c1|{ [ (c-b)+(c-a) ][ c^(p-1)+...+b^(p-1) ] }-( a^p+b^p )
I don't know what the expression on the right side of the equation after "we have"is supposed to be.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post aaa120 Miscellaneous Math 4 2019-03-09 16:46 opyrt Prime Sierpinski Project 3 2009-01-02 01:50 M0CZY Software 15 2008-10-30 14:20 vtai Math 12 2007-06-28 15:34 jinydu Math 5 2005-05-21 16:52

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