Question 4648: Let p be prime. Prove that...
(1 + p)^(p^(n - 1)) = 1(mod p^n) for all n = 1, 2, 3,...
Would using the Binomial Thm help? What about applying the Chinese Remainder Thm?
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! To show :(1 + p)^(p^(n - 1)) = 1 mod p^n ...(**)for all n
You are right, it relates to binomial theorem (but not Chinese Remainder Thm)
Proof: Use Math. Induction and Binomial .
Basic: When n = 1, (1 + p)^(p^(1 - 1)) = (1 + p)^1 = 1+p mod p^1 = 1.
Hence, (**) is true when n=1.
Induction Hypothesis: When n = k, (1 + p)^(p^(k - 1)) = 1 mod p^k,
hence, (1 + p)^(p^(k - 1)) = m * p^k + 1.
Consider (1 + p)^(p^k) = (1 + p)^(p^(k-1)*p) = [(1 + p)^(p^(k-1))]^p
= (m p^k + 1)^p = E C(p,p-i) (m p^k)^(p-i)
[E means summation of i from i =0 to p]
= m^p p^(k+1) + 1 + E C(p,p-i) (m^(p-i)^ (p^(k*(p-i)) [i runs from 1 to p-1]
Since p^(k*(p-i)) >= p^(k+1) for all 1<=i<=p-2
and when i = p-1, p = C(p,p-1) and so p^(k+1) is a divisor of C(p,1)*(m p^k)
Hence, (1 + p)^(p^k) = = m^p p^(k+1) + 1 = 1 mod p^(k+1).
This shows (**) is true when n=k+1.
And, the induction proof is complete.
Plz read carefully about the details.
Kenny
|
|
|
