Conjecture 98. Using AI to solve some Conjectures.

Alexander R. Povolotsky sent on April 14, 2023, the following answer to his own Conjecture 59 from 2008, obtained one day before -he wrote- using the ChatGPT October 23, version. But he declares emphatically that "I am not even sure that the proof provided by ChatGPT is correct."

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"

 

ChatGPT's Answer:

"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."

At the very first day of receiving his email, I was totally skeptical and told my self that I was receiving a joke from Alexander.

Some days after I was now skeptical from my own skepticism and decided to make the same question to my AI tool I'm using very recently, chat.openai.com

Here is the answer I got by my side:

My Question (CR):

"Prove that n! + prime(n) != m^k for any integer values of k, m and n."
 

chat.openai.com Answer:

"To prove that n! + prime(n) != m^k for any integer values of k, m, and n, we will use proof by contradiction.

 

Assume that there exist integer values of k, m, and n such that n! + prime(n) = m^k.

 

First, note that n! is always divisible by all prime numbers less than or equal to n. Since prime(n) is the nth prime number, it is greater than all prime numbers less than or equal to n. Therefore, n! and prime(n) are relatively prime.
 

Now, consider the equation n! + prime(n) = m^k. Since n! and prime(n) are relatively prime, it follows that m must divide one of them. If m divides n!, then m^k is divisible by all prime numbers less than or equal to n, which means that m^k is greater than or equal to n! + prime(n), a contradiction.

 

On the other hand, if m divides prime(n), then m^k is a power of a prime number, which means that it is less than or equal to prime(n). Therefore, n! + prime(n) > m^k, another contradiction.

 

Thus, we have shown that there are no integer values of k, m, and n such that n! + prime(n) = m^k, completing the proof by contradiction."

Because I'm not a mathematician I can not judge if both answers are the same and are both correct.

As a matter of fact, Alexander wrote about the same issue: "Looks the same. I am not sure though that the proof given ChatGPT is correct."

Q1. What is your opinion about both proofs for the Conjecture 59 gotten by AI?

Q2. If you find out that the current tools of AI are working well with the Conjecture 59, can you try it with some other unsolved conjectures?

 

From 22 to 28 April, 2023 contributions came from Alessandro Casini, J. R. Howell, Fred Schneider, Adam Stinchcombe, Emmanuel Vantieghem, Oscar Volpatti

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Alejandro wrote:

Lets talk about the proof attempt of the two AI to the conjecture n° 59. Both proofs are garbage, due to a faulty logical step. The ChatGPT got confused on the parity’s analysis. Where a human would have logically claimed “LHS is odd => RHS must be odd => m must be odd”, ChatGPT instead stupidly assert “LHS is odd, but RHS has no particular parity restriction => the whole Universe must be wrong”. The second AI instead is right until the relatively primality of n! and prime(n), even if it doesn’t argue very well why prime(n) > n (though obviously true). After that, it’s wrong about saying that m must divide one of them (9+16=25, but 5 does not divide neither 9 nor 16), and from that moment a cascade of unfounded deductions begins to come out. Therefore, the two proofs say nothing, although it was a very interesting experiment. If I had to choose between the two, I would choose the second one, given that ChatGPT missed a fundamental pure logical step, while the second AI was wrong on a number theory’s reasoning. Moreover, the second is more similar to a true proof and has the potential to deceive a larger audience than the first one, if we consider the latter as an evaluation parameter. As I said, an interesting experiment. However, I don’t know whether to investigate further or not.

 

Then I wrote to him: "Now I ask you: is there any way to improve and complete the AI answers? For example, to correct the omissions and wrong steps given by AI?...I am asking if a human being can improve the AI answer to fill the gaps and get a complete proof."

 

Alejandro additional commments: " I don’t know, the gaps left by the AI are huge, therefore we’re still waiting a solid proof. Having said that, I verify that there are no solutions for n<200000. Still an open problem."

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Howell wrote:

The second "proof" given here is also incorrect.  It is OK until the line "Now consider the equation n! + prime(n) = m^k."  The rest is basically nonsense.

Since n! and prime(n) are relatively prime, the proof says that m must divide one of them. No. Either m divides both or m divides neither.  Since they are relatively prime, m cannot divide both, therefore it divides neither.  And the rest of the "proof" does not apply.
 

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Fred wrote:

Q1) Both proofs are incorrect and just nonsensical for different reasons
 

The proofs both have a glaring omission.  They didn't check the case for k = 1. 

This conjecture is always wrong for every n >= 1.
 

f(n) = n! + prime(n)
Let m = f(n) and k= 1
 

Now, if you have a restriction where k is an integer > 1 that's a more challenging problem.  Maybe that was omitted from the conjecture?

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Adam wrote:

The problem posed to the AI was poorly formulated.  The AI successfully proved "there exist numbers that do not solve the equation" which is true but easier to prove, just list a non-solution {n=5,m=2,k=25000}.  The equation is not true for any n,m,k (i.e., it is not an identity).  However, the properly formulated conjecture is either (in the affirmative) "there exist n,m,k that solve the equation" and (in the negative) "for all n,m,k the equation is not true."

 

It is impressive that AI can mash those words and concepts together.  It raises a concern for me that AI could generate untruths that the AI supports with such convoluted logic that a human can not reasonably decipher it.  That would be viewed as a success by the AI, thus providing positive pressure (economic and professional rewards for the operators, pervasiveness for the AI on the web---i.e., AI proliferation for that specific brand) for the AI to continue to generate incomprehensible but seemingly seamless logic.

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Emmanuel wrote:

The first "proof" is false.

If the left member of an equation has only even values and the right member can have even or odd values does not lead to a contradiction.

Example : the equation  2c + 1 = y^2.

 

The second proof is false too !

If  p  divides the sum of two coprime numbers a  and  b, it must not divide one of the numbers  a  or  b.

Example : 7 + 20 = 27.

(it does hold for the product of a  and  b).

 

So, I think we should not worry about AI !

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Oscar wrote:

About Conjecture 59:
ChatGPT's "proof" is incomplete;
chat.openai.com's "proof" is incorrect.

ChatGPT attempts a parity-check non-existence proof: the given equation cannot hold true when the left-hand side and the right-hand side have different parities.
The left-hand side n!+prime(n) is odd for n>=2, and it is odd for n=1 too.
The right hand side m^k has the same parity of m itself, which can be either an even or an odd number.
Hence, parity check alone allows to discard triplets (n,k,m) with m even, but not triplets (n,k,m) with m odd.
At best, I judge this "proof" incomplete.

chat.openai.com chooses a different approach: it uses three "divisibility properties" to deduce explicit inequalities between the sides of the given equation.
Unfortunately, none of the claimed "divisibility properties" is actually true.
As an example, chat.openai.com correctly notices that n! and prime(n) are relatively prime, then it states that "m must divide one of them".
But this claim is false: if two addends n! and prime(n) are relatively prime, then their sum m^k is coprime with both of them, and its proper factor m is coprime with both of them too.
 

At best, I judge such "proof" incorrect.
 

 

Of course I'm not a mathematician and I may be wrong.

But notice that both AI "proofs" should also work for the following slightly modified conjecture: 

no n exists such that  n! + prime(n)^2 = m^k,  where k>1.

In this case, we do have a counterexample:

7! + prime(7)^2 = 5040+17^2 = 5329 = 73^2.

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