Problems & Puzzles: Conjectures

Conjecture 59. n!+p(n)=m^k

Alexander R. Povolotsky sent (around August 2008) the following conjecture:

No "n" exist, such that ( n! + prime(n) ) = m^k, where k>1

Q. Prove it or find a counterexample.

On April 14, 2023, Alexander R. Povolotsky found, one day before, a solution to his own Conjecture (posted around August 2008) using the ChatGPT AI tool, October 23 version. But he declares emphatically that "I am not even sure that the proof provided by ChatGPT is correct."

Anyhow I (CR) decided to share this result with all of you under my own responsibility.

Alexander's Question:

"Prove that  n! + prime(n) != m^k  for any integer values of k, m and n supplied by ChatGPT October 23 version"

"To prove that the expression "n! + prime(n) ≠ m^k" is true for any integer values of k, m, and n, we can use proof by contradiction.

Assume the opposite, i.e., suppose there exists some integer values of k, m, and n for which the expression "n! + prime(n) = m^k" is true.

Since n! represents the factorial of n, it is always a product of consecutive positive integers from 1 to n. Therefore, n! is always divisible by all the prime numbers less than or equal to n.

Now, let's consider the term "prime(n)". It represents the nth prime number. Since n! is always divisible by all the prime numbers less than or equal to n, adding prime(n) to n! would result in a number that is greater than n!.

Mathematically, we can express this as:

n! + prime(n) > n! ---(1)

On the other hand, "m^k" represents m raised to the power of k, which is also an integer value.

Now, let's consider the case where n is greater than or equal to 2. In this case, n! would always be divisible by 2 and hence an even number. Adding prime(n) to an even number would result in an odd number.

However, m^k represents an integer raised to an integer power, which could result in either an even or an odd number depending on the values of m and k.

Therefore, we can conclude that the left-hand side of the equation "n! + prime(n) = m^k" would always be an odd number, while the right-hand side "m^k" could be either an even or an odd number. This is a contradiction, as an equation cannot hold true when the left-hand side and the right-hand side have different parities (i.e., one is odd and the other is even).

Hence, our assumption that "n! + prime(n) = m^k" is true for some integer values of k, m, and n is false, and we have proved that "n! + prime(n) ≠ m^k" for any integer values of k, m, and n."

***

J. R. Howell wrote:

The proof given is not correct. It is OK through the part that says n! + prime(n) must be odd.

From this, we can conclude  m^k must be odd, since it is equal to that sum. So m itself must be odd. In other words, if some values of k, m, and n exist which satisfy the original equation, then m must be odd.  The fact that m^k is even for some values of k and m (namely m even) only means that those values cannot be part of any solution.

In order for the proof by contradiction to work, it must rule out ALL possible values of k, m, and n. The given "proof" only rules out those cases where m is even.

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