Infinitely many even primes??

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[chrisptak]

The first mistake I can see is using Fermat's little theorem when the exponent isn't necessarily a prime. Not to mention that what it says in that step isn't true even when the exponent is a prime, 3^2-1 is congruent to 2 mod 3 not 1

[mrphlip]

The mistake is trying to use chatgpt as some sort of knowledge engine, rather than a source of text slurry that does a reasonable job of _resembling_ information.

But more specifically...
• it claims that 3^n - 1 is congruent to 1 (mod 3), when it is in fact congruent to 2
• to support this it cites FLT which isn't relevant
• it claims this implies any prime factor of this must also be congruent to 1, which doesn't follow
• it then proceeds to misquote part of the standard infintely-many-primes proof, scrambled somewhat by shoehorning in some references to 3^n that don't make any sense
• it pulls out of nowhere the idea that M must have a prime factor of the form 3^n-1
• probably even more mistakes that I didn't see in a skim through

[fartoxedm]

- N is not congruent to 1 mod 3(Fermat's flipping somewhere rn)
- M is not always of the form 3^n (only if k is even and several other assumptions)
- 3^n must have only 3^i as factors not 3^n-1....
I am trying to beat the machine, yeah...

And also p^n - 1(n > 1) = (p-1) * smth so since p != 2 it's kinda impossible to be prime(well unless some messy cases but it's 2's thing don't take it from it)

[compiling]

Find "the" mistake? Nearly every line is wrong in some way.

[siquod]

I easily spotted the first mistake because I happen to be an export on congruence modulo 3 since childhood. Stupid OCD. But what comes after this error is sheer nonsense, and what comes before is "correct" but probably leads nowhere. Apparently the AI tried to complete the pattern for "prove there are infinitely many primes", but adapted to a situation involving powers of three and subtraction of one.

[thomasolson]

Don't do it. I warn all of you to stay as far away from primes as possible. You will only waste your time. The composite functions are interesting though.

[itzakehrenberg]

N is always congruent to -1 (not +1) modulo 3 since 3^(p_1*p_2*...*p_k) is obviously divisible by 3. In fact, it is easy to show that 3^m-1 is even-composite for all m>1. Poor ChatGPT had better take a refresher on Fermat's Little Theorem.

[jonathanr]

ChatGPT is hopeless when it comes to numbers. It keeps repeating the same mistakes even after being corrected. I think it stores some values of the conversation and can't seem to overwrite it, thus jumps back to the wrong assumption. It also seem to mess up when you want to reason with it. Tell the GPT a story, and somewhere along the line it will completely mess things up.

[goldowolabi]

For starters powers of three are odd and minusing it from one would make it even. Since all even integers greater than 2 is not a prime. It is only true for n = 1

[TranquilSeaOfMath]

The mistake is trusting computers too much.

[jorgesaxon]

Its not surprising that chatgpt gets it wrong however what is quite amazing is that gpt-4 gets it right!
"
The statement that there are infinitely many primes of the form 3^n - 1 for some integer n is not necessarily true. In fact, if you examine the sequence of numbers of the form 3^n - 1 for n ≥ 2, you'll find that they're not prime.

Let's see why that is.

Notice that:

3^2 - 1 = 8, which is not prime.
3^3 - 1 = 26, which is not prime.

And in general, we can prove that all numbers of the form 3^n - 1 where n ≥ 2 are composite (i.e., not prime). This is because:

3^n - 1 = (3^n - 1^2) = (3 - 1)(3^(n-1) + 3^(n-2) + ... + 3 + 1)

Here, 3 - 1 = 2, and (3^(n-1) + 3^(n-2) + ... + 3 + 1) is the sum of a geometric series, which is an integer greater than 1 for n ≥ 2. Thus, 3^n - 1 is a product of two integers greater than 1, meaning it's not prime.

So, unfortunately, the claim that there are infinitely many primes of the form 3^n - 1 isn't true."