Let N be a positive integer. prove that \($2^{2^n} + 2^{2^{n - 1}} + 1$ \)\(\)can be writen as as the product of at least n prime factors, not necessarily distinct.

I've already figured out that the number of prime factors of \(2^{2^n} + 2^{2^{n - 1}} + 1 \) will always be equal to n, but how do I prove it?

tzeperson Oct 20, 2023

#1**0 **

We prove this statement by induction on n.

Base case: When n=1, we have 22n+22(n−1)+1=22+21+1=5. 5 is a prime number, so the statement holds for the base case.

Inductive step: Assume that the statement holds for some positive integer k, i.e., 22k+22(k−1)+1 can be written as the product of at least k prime factors, not necessarily distinct. We want to show that the statement also holds for k+1, i.e., 22k+1+22(k+1)−1+1 can be written as the product of at least k+1 prime factors, not necessarily distinct.

We have:

2^{2^{k + 1}} + 2^{2^{(k + 1) - 1}} + 1 = 2^{2^k \cdot 2} + 2^{2^k} + 1

By the distributive property of multiplication, we can rewrite this as:

(2^{2^k})^2 + 2 \cdot 2^{2^k} + 1 = (2^{2^k} + 1)^2

We know that 22k+1 can be written as the product of at least k prime factors, not necessarily distinct. Let these prime factors be p1,p2,...,pk. Then, we can write:

(2^{2^k} + 1)^2 = (p_1 \cdot p_2 \cdot ... \cdot p_k)^2 = p_1^2 \cdot p_2^2 \cdot ... \cdot p_k^2

This shows that 22k+1+22(k+1)−1+1 can be written as the product of at least 2k=k+1 prime factors, not necessarily distinct. Therefore, the statement holds for k+1.

Conclusion: By the principle of mathematical induction, we can conclude that the statement holds for all positive integers n. In other words, 22n+22(n−1)+1 can be written as the product of at least n prime factors, not necessarily distinct.

bingboy Oct 27, 2023