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Let x and y belong to a commutative ring R with characteristic p not equal to 0. The document shows that: 1) (x+y)^p equals x^p + y^p, by using Euler's theorem that (a,p)=1 implies a^φ(p)=1 (mod p) where φ(p) is the totient function. 2) By induction, (x+y)^(p^n) equals (x^(p^n)) + (y^(p^n)) for any positive integer n. 3) An example is given of elements x and y in a ring of characteristic 4, such that (x+y)^4 does not equal (

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- 1. Let x and y belong to a commutative ring R with char(R)= p?0. a. Show that {(x+y)^p}= (x^p)+(y^p). b. Show that, for all positive integers n, (x+y)^(p^n)= (x^(p^n))+ (y^(p^n)). c. Find elements x and y in a ring of characteristic 4 such that (x+y)^4? (x^4)+(y^4). Solution Observe that according to Euler's totient formula, let a be any integer coprime to p, then (a, p) = 1. Then a^(phi(p)) = 1 (mod p). Now since p is prime, all a in Z/PZ is coprime to p and phi(p) = (p-1). Then (x+y)^p = (x+y)(x+y)^(p-1) (mod p) = (x+y)(1) (mod p). Then we have (x+y)^p = (x+y) (mod p), hence (x+y)^p = x (mod p) + y (mod p). Multiply back through by (x+y)^(p-1) gives us (x+y)^p(x+y)^(p-1) = x^p + y^p (mod p) and since (x+y)^(p-1) = 1 (mod p) we have (x+y)^p = x^p + y^p (mod p) as we wished to prove.