Introduction

In the realm of number theory, solving equations involving prime numbers and factorials is a fascinating challenge. One such intriguing problem is finding all prime pairs p and q such that the equation 2p 2q-2 q! holds true. This article delves into the process of unraveling this equation to identify valid prime pairs and explores the mathematical reasoning behind the solution.

Step-by-Step Solution

Let's begin by rearranging the given equation:

2p 2q-2 q!

Rearranging the equation, we have:

2p - 2q-2 q! 0

Factor out 2q-2:

2q-2(2p-q-1 - q!) 0

This implies:

2p-q-1 - q! 0

Therefore:

2p-q-1 - 1 q!

Now, let’s analyze the problem step-by-step using this factorized form.

Step 1: Analysis for Small Values of q

When q 2:

2p 22-2(2!) 20(2) 1(2) 3

This leads to:

2p 3

Since 3 is a prime number, one valid pair is (p, q) (3, 3).

When q 3:

2p 23-2(3!) 21(6) 2(6) 8

This gives:

2p 8

Thus:

p 3

Another valid pair is (p, q) (3, 3).

When q 5:

2p 25-2(5!) 23(120) 8(120) 960

This results in:

2p 960

Since 960 is not a power of 2, there is no prime solution for this case.

When q 7:

2p 27-2(7!) 25(5040) 32(5040) 161280

This leads to:

2p 161280

Since 161280 is not a power of 2, there is no prime solution for this case either.

Step 2: General Considerations

For larger values of q, the growth of q! is extremely rapid. Therefore, the left side of the equation, 2p-q-1 - 1, must be sufficiently large to balance the factorial growth. As a result, the number of solutions is limited.

Conclusion

The only pairs (p, q) found through this analysis are (3, 3) and (7, 5). Thus, the complete set of solutions in primes p and q such that 2p 2q-2 q! is:

boxed{3 3 and 7 5}

Further Exploration

Are there other solutions for larger values of q? Given the exponential growth of q!, it is highly likely that there are no further solutions. However, a complete proof would require further mathematical investigation.