Problems & Puzzles:
Conjectures
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 lefthand side of the equation "n! + prime(n) = m^k" would
always be an odd number, while the righthand side "m^k" could
be either an even or an odd number. This is a contradiction, as
an equation cannot hold true when the lefthand side and the
righthand 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
***
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."
***
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.
***
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?
***
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 nonsolution {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 webi.e., AI proliferation for that specific brand) for the
AI to continue to generate incomprehensible but seemingly seamless
logic.
***
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 !
***
Oscar wrote:
About Conjecture 59:
ChatGPT's "proof" is incomplete;
chat.openai.com's "proof" is incorrect.
ChatGPT attempts a paritycheck nonexistence proof: the given
equation cannot hold true when the lefthand side and the righthand
side have different parities.
The lefthand 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.
***
