Problems & Puzzles:
Conjecture 59. n!+p(n)=m^k
Alexander R. Povolotsky sent (around
August 2008) the
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.
"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
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.