How to Find All Natural Numbers That Satisfy 2n m^2b, where 0 ≤ b ≤ 2m

Understanding the relationship between natural numbers and their representations is a fundamental concept in mathematics. This article delves into a specific problem where we aim to find all natural numbers n such that 2n can be written in the form m^2b, where 0 ≤ b ≤ 2m. We will explore the conditions and steps required to find these numbers.

Overview of the Problem

The problem at hand is to find all natural numbers n that can be expressed as:

2n m^2b where 0 ≤ b ≤ 2m

To solve this, we need to find the values of m and b that satisfy the given conditions and ensure that the expression is an integer.

Step-by-Step Solution

Let's start by rewriting the given expression:

2n m^2b

We can manipulate this expression further to find the values of m and b that will satisfy the given conditions.

Case 1: 2n is a Perfect Square (i.e., 2n 4m^2)

If 2n is a perfect square, we can write:

2n 4m^2

In this case, we have:

floor(sqrt{2n}) 2m

Since 2m is coprime with 4m^2, we can deduce that:

S {n (m^2 * m) / 2 : for some m in N}

Case 2: 2n is Not a Perfect Square

When 2n is not a perfect square, we can express 2n as:

2n m^2b

where 0 ≤ b ≤ 2m

From the expression above, we know that:

floor(sqrt{2n}) m

Now, we need to determine the condition for b such that:

(m^2b) / (1m) is an integer. This implies:

m^2b 0 (mod m1)

Thus, we have:

m^2 1 (mod m1) rarr; b 0 (mod m1)

This leads us to:

b m

Substituting b m back into the expression, we get:

2n m^2m

Therefore, the condition for the integers to satisfy the given condition is:

2n m^2m

Final Conclusion

The natural numbers n that satisfy the condition can be expressed as: