show that 4^n + 5 is divisible by 3 for all integers n?

Note: This result is true for all non-negative integers n.

Working mod 3:
4^n + 5 = 1^n + 2 = 1 + 2 = 0 (mod 3).

Hence 4^n + 5 is divisible by 3 for any integer n.
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Here's an inductive proof.

Base case (n = 0)
4^0 + 5 = 6, which is divisible by 3.

Inductive Step
Assuming that 3|(4^k + 5):

4^(k+1)+ 5
= 4 * 4^k + 5
= (3 + 1) * 4^k + 5
= 3 * 4^k + (4^k + 5).

Since 3| (3 * 4^k) clearly and 3|(4^k + 5) by inductive hypothesis, it follows that 3|(3 * 4^k + (4^k + 5)).

This completes the inductive step and the proof.

I hope this helps!

and if you are not comfortable with modular arithmetic....true if n = 1 , assume true if n = k....4^k + 5 = 3 w { w an integer }...then 4^k = 3w - 5....for n= [k+1] we see 4^(k+1) + 5 = 4 [ 4^k] + 5 = 4 [ 3w-5] + 5 = 12 w - 15 = 3 [ 4w - 5], multiple of 3

4^n + 5 = 1^n + 2 = 1 + 2 = 3 = 0 (mod 3)

This situation is true when n is a non-negative number only. When n is negative:

4^(-n) + 5 = 1/(4^n) + 5 (<= Since the result is not an integer, so n cannot be negative.)

Note: I think you mean positive integers, that should be specified.
The easiest way (without going to mods) is by induction, with the following 2 steps:
Show-
a) "it's true for n=0"
b) "if it's true for n, it's true for n+1"

a) just plug it in!
4^0 + 5 = 6 = 2*3. Done.

b) Assume 4^n + 5 = 3k for some integer k, and plug in n + 1:

4^(n+1) + 5 = 4*4^n + 5 = 4*4^n + 4*5 - 3*5 = 4(4^n + 5) - 3*5 = 3k - 3*5 = 3(k-5)

Since k - 5 is an integer, the number is a multiple of three. Proof complete.