Prove that 5 | [3^(3n+1)+2^(n+1)] for all integers n >= 0?

Question

A full proof would be needed to solve this problem

Answer 1

I would use induction.
If n=0:
3^(3n + 1) + 2^(n + 1) = 3^1+2^1 = 5

Now, suppose that it is true for n, and prove for n+1

3^[3(n+1)+1] +2^(n + 1 + 1)) = 3^(3n + 3 + 1) + 2*2^(n + 1) =
= 3^3*3^(3n + 1) +2*2^(n + 1)

Now, there must be m1 in N and k in {1,2,3,4} such that 2^(2n + 1) = 5m1 + k, and m2 in N such that 3^(3n+1)=5m2 + 5 -k.
[ Because we suppose that for each n>=0 5 divides the whole expression].

Then 3^3*3^(3n + 1) + 2*2^(n + 1) = 27(5m2 + 5 -k) + 2*(5m1 + k) = 135m2 +135 -27k + 10m1 + 2k =
= 135m2 + 10m1 -25k + 135

QED

Answer 2

Prove that 3^(3n + 1) + 2^(n + 1) is divisible by 5 for all integers n >= 0.

Proof ( by induction ):

Base case: Let n = 0. Then
3^(3n + 1) + 2^(n + 1) = 3^(3(0) + 1) + 2^(0 + 1)
= 3^(1) + 2^1 = 5, which is obviously divisible by 5.

Induction Hypothesis: Assume the formula holds true for n equal to some value k. That is, assume

3^(3k + 1) + 2^(k + 1) is divisible by 5.

{We want to prove that 3^(3(k + 1) + 1) + 2^( (k + 1) + 1) is divisible by 5}

But, what does 3^(3(k + 1) + 1) + 2^( (k + 1) + 1) equal?

3^(3(k + 1) + 1) + 2^( (k + 1) + 1) =

3^(3k + 3 + 1) + 2^(k + 2) =

3^(3k + 4) + 2^(k + 2) =

3^(3k + 1 + 3) + 2^(k + 1 + 1) =

3^(3k + 1) 3^3 + 2^(k + 1) 2^1 =

27 * 3^(3k + 1) + 2 * 2^(k + 1) =

Here, I'm going to do a little trick of adding 0.
I'm going to add and then subtract the value
27 * 2^(k + 1).

27 * 3^(3k + 1) + {27 * 2^(k + 1) - 27 * 2^(k + 1)} + 2*2^(k + 1)

The "adding and subtracting" is done in the { } brackets.

At this point, I'm going to factor the first two terms and the last two terms.

27 [ 3^(3k + 1) + 2^(k + 1) ] - 2^(k + 1) [ 27 - 2 ] =

27 [ 3^(3k + 1) + 2^(k + 1) ] - 2^(k + 1) [ 25 ]

Look at this closely;

27 [ 3^(3k + 1) + 2^(k + 1) ] is divisible by 5 because our induction hypothesis lies in the square brackets.

2^(k + 1) (25) is OBVIOUSLY divisible by 5 (it has 25 as a factor, which in turn has 5 as a factor).

Therefore, we have the sum of two things divisible by 5, which would itself be divisible by 5.

Therefore, the formula holds true for n = k + 1, and we've just proved that if the formula holds for n = k, the formula holds true for n = k + 1. Since we've just proven that, by the principle of mathematical induction,

3^(3n + 1) + 2^(n + 1) is divisible by 5 for all integers n >= 0.

Answer 3

I'm on an apparent one-man crusade to convince people that induction is not the only proof technique when a problem says "for all integers..."

For this particular problem, reduce the number modulo 5 and show that it is zero, independent of n. All double equals signs below mean congruent modulo 5 (i.e. 6==1==11==...)

3^(3n+1)+2^(n+1) = 3*3^(3n)+2*2^n =
3*27^n+2*2^n == -2*2^n + 2*2^n == 0.

Answer 4

3^(3n + 1) + 2^(n + 1)
= 3*27^n + 2*2^n
=3(27^n - 2^n) +(3*2^n + 2*2 ^n)
first term is divisible by 27 - 2 = 25 for any natural number n, so it is divisible by 5, second term does not need any proof that it is divisible by 5. Hence the result

Answer 5

wow!! what the heck is that i have never seen this!!!

Answer 6

NO