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 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.

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