# Proving a^b (mod n) ≡ [ a (mod n) ]^b

• Aug 6th 2012, 02:47 AM
Proving a^b (mod n) = [ a (mod n) ]^b
How would one go about in order to prove

$a^b \mod{n} =\left [ a \mod{n} \right ]^b \, ?$

I would try doing something like this

$a \equiv b \mod{n} \ \Leftrightarrow \ \dfrac{a-b}{n} = k \, , \ \ k \in \mathbb{Z} \, ,$

however I'm not sure how to actually interpret $\left [ a \mod{n} \right ]^b$. It's periodic function which returns remainders and repeatedly so, what would it's algebraic expression perhaps be?
• Aug 6th 2012, 04:57 AM
a tutor
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
Let $a=x+kn$

$0\le x

$x,k,n \in \mathbb{Z}$.

Then you have $(x+kn)^b (\mod n) \equiv ((x+kn) \mod n)^b (\mod n)$

Simplify that.
• Aug 6th 2012, 05:09 AM
emakarov
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
Quote:

How would one go about in order to prove

$a^b \mod{n} \equiv \left [ a \mod{n} \right ]^b \, ?$

Your notation is strange. Either you have a relation $x\equiv y\pmod{n}$ 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 $\equiv$, or, at least, you need to define it.
• Aug 6th 2012, 07:34 AM
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
Quote:

Originally Posted by emakarov
Your notation is strange. Either you have a relation $x\equiv y\pmod{n}$ 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 $\equiv$, or, at least, you need to define it.

You are correct, let me fix that.

Quote:

Originally Posted by a tutor
Let $a=x+kn$

$0\le x

$x,k,n \in \mathbb{Z}$.

Then you have $(x+kn)^b (\mod n) \equiv ((x+kn) \mod n)^b (\mod n)$

Simplify that.

Thanks!
• Aug 6th 2012, 07:40 AM
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
Quote:

Originally Posted by a tutor
Let $a=x+kn$

$0\le x

$x,k,n \in \mathbb{Z}$.

Hmm
Then you have $(x+kn)^b (\mod n) \equiv ((x+kn) \mod n)^b (\mod n)$

Simplify that.

Hm, I would appreciate if you could show me the simplificaton, the multiple mods are confusing me.
• Aug 6th 2012, 08:43 AM
a tutor
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
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

$(x+kn)^b (\mod n) \equiv ((x+kn) \mod n)^b \ (\mod n)$

we have

$(x+kn)^b\ \%\ n \equiv ((x+kn)\% \ n)^b \ (\mod n)$

$(x+kn)^b\ \% \ n \equiv x^b \ (\mod n)$

One last step?
• Aug 6th 2012, 12:35 PM
Deveno
Re: Proving a^b (mod n) ≡ [ a (mod n) ]^b
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:

[ab] = [[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,

[ak] = [[a]k].

then:

[ak+1] = [(a)(ak)] = [[a][ak]] (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).