Exploring Pairs of Positive Integers in Diophantine Equations

In the study of number theory, Diophantine equations play a crucial role, often requiring the finding of integer solutions to polynomial equations. In this article, we delve into the problem of finding all pairs of positive integers (n, k) that satisfy the equation:

Problem Statement and Initial Analysis

The given Diophantine equation is:

$$ n! 2^k - 8 $$

This can be rewritten as:

$$ n! 8 cdot 2^{k-3} - 1 $$

From this form, we can deduce certain constraints on n and k. For the quantity on the right side to be positive, k must be greater than or equal to 4. Consequently, n must also be greater than or equal to 4, as the factorial function n! is defined only for positive integers.

Further Analysis and Constraints

Next, we factor out the 8 from the right side of the equation:

$$ n! 8 cdot 2^{k-3} - 1 $$

Since $2^{k-3} - 1$ is an odd integer for k ≥ 4, it follows that $2^3$ is the highest power of 2 that divides n!

Knowing this, we further deduce that for all n ≥ 6, the equation n! 8 · 2^(k-3) - 1 would imply that $2^4$ divides n!, which is not possible as n! already includes the factor $2^3$ and no higher power of 2 until n reaches higher values. Therefore:

$$ n leq 5 $$

This forces n to be either 4 or 5, as they are the only integers that satisfy the given constraints.

Final Solutions

When n 4

If n 4, we substitute n into the original equation:

$$ 4! 2^k - 8 $$

$$ 24 2^k - 8 $$

$$ 32 2^k $$

$$ k 5 $$

Hence, one solution is (n, k) (4, 5).

When n 5

If n 5, we substitute n into the original equation:

$$ 5! 2^k - 8 $$

$$ 120 2^k - 8 $$

$$ 128 2^k $$

$$ k 7 $$

Hence, another solution is (n, k) (5, 7).

Conclusion

The only pairs of positive integers (n, k) that satisfy the given Diophantine equation are:

$$ (n, k) (4, 5) text{ and } (5, 7) $$

These are the final solutions to the problem.

Further Exploration and Verification

To ensure accuracy, we can verify these solutions by plugging them back into the original equation:

When (n, k) (4, 5):

$$ 4! 24 $$

$$ 2^5 - 8 32 - 8 24 $$

The equation holds true.

When (n, k) (5, 7):

$$ 5! 120 $$

$$ 2^7 - 8 128 - 8 120 $$

The equation holds true.

These results confirm the correctness of the solutions (4, 5) and (5, 7).