# Help me check the proof, please.

#### IHeartMaths

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.