Thread: Proving a^b (mod n) ≡ [ a (mod n) ]^b

How would one go about in order to prove



I would try doing something like this



however I'm not sure how to actually interpret . It's periodic function which returns remainders and repeatedly so, what would it's algebraic expression perhaps be?

Let



.

Then you have

Simplify that.

Originally Posted by MathCrusader

How would one go about in order to prove

Your notation is strange. Either you have a relation or you have a function x mod n, which returns the remainder when x is divided by n. In the first case, you can't use mod in both sides of the relation. In the second case, you can't use , or, at least, you need to define it.

Originally Posted by emakarov

Your notation is strange. Either you have a relation or you have a function x mod n, which returns the remainder when x is divided by n. In the first case, you can't use mod in both sides of the relation. In the second case, you can't use , or, at least, you need to define it.

You are correct, let me fix that.

Originally Posted by a tutor

Let



.

Then you have

Simplify that.

Thanks!

Originally Posted by a tutor

Let



.

Hmm
Then you have

Simplify that.

Hm, I would appreciate if you could show me the simplificaton, the multiple mods are confusing me.

Following emakarov's comments perhaps it would be better if we used a % b for the remainder when a is divided by b.

Then rather than



we have





One last step?

if one uses [a] for the remainder of a, upon division by the integer n (where this remainder is between 0 and n-1, inclusive), we can state this as:

[a^b] = [[a]^b]. note that [a] is just an integer.

this can be shown using induction on b:

for b = 1:

[a] = [[a]], since 0 ≤ [a] ≤ n-1.

assume that for b = k,

[a^k] = [[a]^k].

then:

[a^k+1] = [(a)(a^k)] = [[a][a^k]] (see below)

= [[a][[a]^k]] (by our inductive hypothesis)

= [[[a]][[a]^k]] (by our base case, [a] = [[a]])

= [[a][a]^k] (again, see below)

= [[a]^k+1]

(the crucial step is this:

[[x][y]] = [xy].

note that if x = tn + c, y = un + d, where 0 ≤ c,d < n that:

xy = (tn + c)(un + d) = (tun + c + d)n + cd. writing cd = wn + f, for 0 ≤ f < n, xy = (tun + c + d + w)n + f,

so we have:

[[x][y]] = [cd] = f = [xy]. one can view this as "a rule for brackets and multiplying": remove the outer 2 brackets on each side, and the innermost pair, multiply and add outer brackets).