Modular Arithmetic: Solve (a + b)^5 in Z_5

[Coto]

Homework Statement

Compute:
(a + b)^5
in Z_5 (Z mod 5).

Homework Equations

The end result is apparently:
(a^5 + b^5)

Intuition would tell me to exploit the properties of arithmetic in Z_n, however I don't see how I can reconcile this solution with just a normal expansion of (a+b)^5 (which seems to be allowed due to the def. of arithmetic for these classes.)

The Attempt at a Solution

Well, my best guess would be to show that the middle terms go to zero no matter what. But trying this with for example (a+b)^2 (which has the analagous sol'n) would mean that ab + ba = ab + ab = 0 (mod 5). However, choosing for example a = [2], and b = [2], shows that this would actually be [4]+[4] = [8] = [3], which clearly is not 0... so I'm at a loss.

or.. I could have read this question wrong.. or perhaps my world view of modular arithmetic is one big illusion. ugh.

 

[morphism]

The 5 up there is important. (a+b)^2 doesn't need to be equal to a^2+b^2 mod 5. However, it is equal to that mod 2. In general, (a+b)^p = a^p+b^p mod any prime p.

 

[Coto]

Gotcha. Thanks for the answer, it makes sense now.