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Old 2019-02-13, 04:16   #1
IHeartMath
 
Feb 2019
Taiwan

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Default 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.
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Old 2019-02-16, 01:05   #2
dcheuk
 
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Jan 2019
Iowa, US

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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
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Old 2019-02-16, 03:41   #3
a1call
 
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"Rashid Naimi"
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Quote:
Originally Posted by dcheuk View Post
What is ⊥ in a⊥b⊥c?

https://www.johndcook.com/blog/2010/...atively-prime/
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Old 2019-02-16, 17:58   #4
dcheuk
 
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Quote:
Originally Posted by a1call View Post
I didn't know that, thanks!

I was googling that symbol all over the place.
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Old 2019-02-16, 18:54   #5
a1call
 
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"Rashid Naimi"
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My pleasure dcheuk,

Here is a useful resource for writing formal math statements:

https://en.m.wikipedia.org/wiki/List...atical_symbols
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Old 2019-02-18, 14:52   #6
Dr Sardonicus
 
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Feb 2017
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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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