Positive Integers That Satisfy Specific Divisibility Conditions

This article delves into the mathematical problem of finding all positive integers n leq 10^{100} that satisfy three specific divisibility conditions. We will explore the logical steps required to solve this problem, enhancing the understanding and application of concepts such as divisibility and powers of 2.

Understanding the Problem

The problem involves finding all positive integers n leq 10^{100} such that:

n mid 2^n (n - 1) mid (2^n - 1) (n - 2) mid (2^n - 2)

Condition Analysis

Condition 1: n mid 2^n

The first condition, n mid 2^n, implies that n must be a power of 2. This is because 2^n can only be fully divisible by powers of 2. Therefore, we can express n 2^k for some integer k.

Condition 2: (n - 1) mid (2^n - 1)

The second condition, (n - 1) mid (2^n - 1), requires further analysis using the previous conclusion. For n 2^k, we have:

n - 1 2^k - 1

Checking (2^k - 1) mid (2^{2^k} - 1) using the properties of exponents, we see:

2^{2^k} - 1 2 - 12^{2^k - 1} 2^{2^k - 2} cdots 1

By the Lifting The Exponent Lemma (LTE), we find:

v_2(2^n - 1) v_2(2 - 1) v_2(n) 1 k

Since 2^k - 1 divides 2^{2^k} - 1, this condition holds for k geq 1.

Condition 3: (n - 2) mid (2^n - 2)

The third condition, (n - 2) mid (2^n - 2), can be approached similarly. For n 2^k, we have:

n - 2 2^k - 2

The condition can be checked by factoring:

2^{2^k} - 2 22^{2^k - 1} - 1

Using the LTE lemma again, we have:

v_2(2^n - 2) v_2(2) v_2(n) 1 k

The condition holds if 2^k - 2 divides 22^{2^k - 1} - 1, which holds for k geq 1.

Conclusion

From our analysis, we conclude that the positive integers n leq 10^{100} that satisfy all three conditions are 2^k where k geq 1. The only powers of 2 less than 10^{100} are:

2^1 2 2^2 4 2^3 8 2^4 16 2^5 32 2^6 64 2^7 128 2^8 256 2^9 512 2^{10} 1024 ... 2^{332}

The complete set of solutions is therefore:

{ 2^k mid k 1, 2, ..., 332 }

Additional Insights

Another perspective was provided, stating that the first condition directly indicates that n must be a power of 2, i.e., n 2^k. The second condition implies 2^k - 1 mid 2^{2^k} - 1, which holds if the order of 2 modulo 2^k - 1 divides 2^k. The third condition similarly leads to the conclusion that the order of 2 modulo 2^{k-1} - 1 divides 2^k - 1, which is essentially the same as the second condition.

Combining all these insights, the numbers satisfying the conditions are 2^{2^{2^l}} where l geq 0. The only such numbers less than 10^{100} are:

4 16 65536 2^{256}approx 1.2 times 10^{77}

This comprehensive analysis and the additional insights provide a clear understanding of the problem and its solution within the given constraints.