In Z5 compute (a + b)^5?

Question

okay, I'm not sure how to say it but Z5 is Z(mod 5) I think.
there are four parts to this question and if I know the first two or three I think I might be able to get this so I appreciate the help.

A) In Z5 compute (a + b)^5.
B)In Z3 compute (a + b) ^3.
C) In Z2 compute (a + b) ^2.
D) Based on the results of parts (a) - (c), what do you think (a + b)^7 is equal to in Z7?
Thanks!

Answer 1

I'll do the first one the long way, and then give you a trick that can be used to compute the rest of them much faster.

By the binomial theorem, (a+b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵. But since this is being computed in Z₅, we know that all multiples of 5 are 0, so this is just a⁵ + b⁵.

Now, for the trick: if you know Fermat's little theorem, you know that n^p ≡ n mod p, whenever p is a prime number. So since 5 is prime, you know that (a+b)⁵ ≡ a+b mod 5. This is consistent with the result we got by doing longhand multiplication, because a⁵≡a and b⁵≡b. Similarly, (a+b)³ ≡ a+b mod 3 and (a+b)² ≡ a+b mod 2.

Answer 2

Go to an elementary abstract algebra book (Fraleigh would be fine) and find for Fermat's little theorem:

It tells us that if p is prime, then in Zp, (a+b)^p = a^p+b^p

The proof makes use of binomial theorem which tells you that the coefficient (remember the combinations "n taken k") of the middle terms are multiples of p, therefore they are equivalent to 0(modp) thus we are left with the last and first terms which are a^p+b^p