If , is a prime, is not a multiple of , then what is:
Assume that , and are positive integers.
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if n is an odd number then
a n + b n = ( a + b ) ( ∑ i = 0 n − 1 a i b n − 1 − i ( − 1 ) i )
so
a p + b p = ( a + b ) ( ∑ i = 0 n − 1 a i b p − 1 − i ( − 1 ) i ) ⟹ a + b a n + b n = ∑ i = 0 n − 1 a i b p − 1 − i ( − 1 ) i
by repeatedly applying Euclidean algorithm, we find the GCD to be
G C D ( ∑ i = 0 n − 1 a i b p − 1 − i ( − 1 ) i , a + b ) = G C D ( p b p − 1 , a + b ) = 1
since ( a + b , p ) = 1 and ( a + b , b t ) = 1 ∀ t ∈ N
Edit:
As requested by Sonu Gupra, I decided to make the proof more accurate, for the case p = 2 .
The GCD would be defined if a + b a 2 + b 2 is a whole number. So, we check the condition, for which is a whole number.
a + b ∣ a 2 + b 2 = ( a + b ) 2 − 2 a b ⟹ a + b ∣ 2 a b
as g c d ( a , b ) = 1 , this is only possible if a + b = 2 . because, if q ∣ 2 a b , for a prime q ∣ a + b , q > 2 , then either q ∣ a or q ∣ b . this is contradiction, since if q ∣ a and g c d ( a , b ) = 1 then q ∤ a + b . Therefore, a + b can only be 2 , meaning a = 1 , b = 1 . Consequently, one realises that the final answer stays the same.