20190213, 04:16  #1 
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 = (cb)+(ca)+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(cb)+(ca)+2(a+b) ⇒ c1(cb)+(ca). (6) Since c1c ⇒ c1a^p+b^p, (7) we have c1{ [ (cb)+(ca) ][ c^(p1)+...+b^(p1) ] }( a^p+b^p ) ⇒ c1{ (cb)[ c^(p1)+...+b^(p1) ]a^p }+(ca)[ c^(1)+...+b^(p1) ]b^p ⇒ c1{ 0 }+(ca)[ c^(1)+...+b^(p1) ](ca)[ c^(p1)+...+a^(p1) ] ⇒ c1(ca){ [ c^(1)+...+b^(p1) ] [ c^(p1)+...+a^(p1) ] }. (8) Because term ca is the divisor of b^p, we have c1[ c^(1)+...+b^(p1) ] [ c^(p1)+...+a^(p1) ] ⇒ c1b^(p1)a^(p1). (9) Because c1 is the divisor of a+b, we have c1a^(p1)+b^(p1). (10) Considering both (9) and (10), then we have c12b^(p1) ⇒ c12. (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. 
20190216, 01:05  #2 
Jan 2019
Iowa, US
327_{8} 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 20190216 at 01:09 
20190216, 03:41  #3 
"Rashid Naimi"
Oct 2015
Out of my Body
2·3^{4}·11 Posts 

20190216, 17:58  #4  
Jan 2019
Iowa, US
5×43 Posts 
Quote:
I was googling that symbol all over the place. 

20190216, 18:54  #5 
"Rashid Naimi"
Oct 2015
Out of my Body
2×3^{4}×11 Posts 
My pleasure dcheuk,
Here is a useful resource for writing formal math statements: https://en.m.wikipedia.org/wiki/List...atical_symbols 
20190218, 14:52  #6  
Feb 2017
Nowhere
2·5·307 Posts 
Here is where I become unable to check your work:
Quote:


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