I came to a conclusion about GPT (which is very good with English on most topics) when I asked it how many prime numbers, when divided by 35, leave a remainder of 6. It quickly and confidently said there were none. It hadn’t even tried. The correct answer (there’s a proof) is: an infinite number.
Two months later it answered that there were 3. Closer … but no cigar
I’m too lazy to check it now, but here’s GPT-4o mini (via DDG). Maybe I’ll give Gemini a chance later, too.
Elaborate the proof.
Here is a high‐level outline of the proof that there are infinitely many primes in the arithmetic progression
p ≡ a (mod m)
whenever gcd(a, m) = 1. In your case a = 6 and m = 35, so gcd(6, 35) = 1, and the theorem then guarantees infinitely many primes ≡ 6 mod 35.
Dirichlet characters
• A Dirichlet character modulo m is a completely multiplicative arithmetic function
χ : ℤ → ℂ such that
– χ(n+m) = χ(n) for all n,
– χ(n) = 0 whenever gcd(n, m) > 1,
– χ(n) ≠ 0 (and in fact |χ(n)| = 1) whenever gcd(n, m)=1.
• There are exactly φ(m) such characters (including the “principal character” χ₀, which is 1 on all n coprime to m, and 0 otherwise).
Dirichlet L-series
For each character χ mod m one forms the Dirichlet L-series
L(s, χ) = ∑_{n=1}^∞ χ(n) n^(−s) , Re(s) > 1.
This generalizes the Riemann zeta function (which is the special case χ = χ₀ with m=1).
Euler product factorization
Because χ is completely multiplicative, one shows for Re(s)>1:
L(s, χ) = ∏_{p prime} (1 − χ(p) p(−s))(−1).
This factorization links L(s, χ) to the distribution of primes in each residue class mod m.
Nonvanishing at s = 1
A key analytic fact (Dirichlet’s theorem) is that for every nonprincipal character χ (i.e. χ≠χ₀), the L-series L(s, χ) extends to a holomorphic, nonzero function at s=1. Meanwhile L(s, χ₀) has a simple pole at s=1, just like the Riemann zeta function ζ(s).
Summing over characters
Use the orthogonality relations of characters: for any integer n with gcd(n,m)=1,
(1/φ(m)) ∑{χ mod m} χ(n) over all φ(m) characters = 1,
and it’s 0 if n is not coprime to m.
This lets you isolate those primes in a particular residue class a mod m by forming
∑{χ mod m} χ(a)̄ · L(s,χ)
(where χ(a)̄ is the complex conjugate of χ(a)). In the Euler product this picks out exactly the primes p with p≡a mod m.
Logarithmic differentiation and Tauberian theorems
One takes logarithms of the Euler products and looks at
–(L′/L)(s,χ) = ∑{n=1}^∞ Λ(n) χ(n) n^(−s),
where Λ is the von Mangoldt function. Summing these expressions over characters and using orthogonality isolates
∑{p ≡ a (mod m)} p^(−s) ,
plus lower‐order terms. One then applies complex‐analytic Tauberian theorems (or elementary partial summation arguments) to deduce that as x→∞
#{ p ≤ x : p ≡ a (mod m) } ∼ (1/φ(m)) x / log x ,
in particular infinitely many such primes exist.
Conclusion
Dirichlet’s theorem on arithmetic progressions (proven by exactly the steps sketched above) implies that for any m and any a with gcd(a,m) = 1, there are infinitely many primes p ≡ a mod m. Setting m = 35 and a = 6 shows there are infinitely many primes that leave remainder 6 upon division by 35.
One less formal way of looking at it is this: there are an infinite number of multiples of 35. starting with 35, 70, 105. Add 6 to each of the odd multiples. 35+6=41 (prime); 105+6 = 111 (prime). With an infinite # of candidates, you’ve gotta get to an infinite number of solutions (for some value of infinity!)
As for that Dirichlet stuff, it’s way beyond any of the useful stuff I learned too.
Yes, but it obviously wouldn’t work for +5 or +7. I don’t think you can just assume that the number of primes wouldn’t converge to some finite number just because the number of candidates goes to infinity.
Dirichlet’s theorem proves that. I have since looked it up, and that’s correct. I didn’t realize at the time that I was asking it to elaborate the proof for Dirichlet’s theorem. Whether the elaboration is correct is something I will pass.
No, I wouldn’t assume non-convergence either … NOR would I assume that that AI didn’t just grab that ‘high-level’ ‘Elaboration’ from some site … without a citation.
(Very human … Lots of people use quotes to sound smart, hoping they’ll get away with it. LAWYERS! Ministers! Presidents, even! )
Have you ever tried to do math with an AI? And I don’t mean summing up three numbers or something similarly trivial.
Actually, math problems will probably show best that the current LLMs are just parrots with a large dictionary.
I came to a conclusion about GPT (which is very good with English on most topics) when I asked it how many prime numbers, when divided by 35, leave a remainder of 6. It quickly and confidently said there were none. It hadn’t even tried. The correct answer (there’s a proof) is: an infinite number.
Two months later it answered that there were 3. Closer … but no cigar
I’m too lazy to check it now, but here’s GPT-4o mini (via DDG). Maybe I’ll give Gemini a chance later, too.
Elaborate the proof.
Here is a high‐level outline of the proof that there are infinitely many primes in the arithmetic progression
p ≡ a (mod m)
whenever gcd(a, m) = 1. In your case a = 6 and m = 35, so gcd(6, 35) = 1, and the theorem then guarantees infinitely many primes ≡ 6 mod 35.
Dirichlet characters
• A Dirichlet character modulo m is a completely multiplicative arithmetic function
χ : ℤ → ℂ such that
– χ(n+m) = χ(n) for all n,
– χ(n) = 0 whenever gcd(n, m) > 1,
– χ(n) ≠ 0 (and in fact |χ(n)| = 1) whenever gcd(n, m)=1.
• There are exactly φ(m) such characters (including the “principal character” χ₀, which is 1 on all n coprime to m, and 0 otherwise).
Dirichlet L-series
For each character χ mod m one forms the Dirichlet L-series
L(s, χ) = ∑_{n=1}^∞ χ(n) n^(−s) , Re(s) > 1.
This generalizes the Riemann zeta function (which is the special case χ = χ₀ with m=1).
Euler product factorization
Because χ is completely multiplicative, one shows for Re(s)>1:
L(s, χ) = ∏_{p prime} (1 − χ(p) p(−s))(−1).
This factorization links L(s, χ) to the distribution of primes in each residue class mod m.
Nonvanishing at s = 1
A key analytic fact (Dirichlet’s theorem) is that for every nonprincipal character χ (i.e. χ≠χ₀), the L-series L(s, χ) extends to a holomorphic, nonzero function at s=1. Meanwhile L(s, χ₀) has a simple pole at s=1, just like the Riemann zeta function ζ(s).
Summing over characters
Use the orthogonality relations of characters: for any integer n with gcd(n,m)=1,
(1/φ(m)) ∑{χ mod m} χ(n) over all φ(m) characters = 1,
and it’s 0 if n is not coprime to m.
This lets you isolate those primes in a particular residue class a mod m by forming
∑{χ mod m} χ(a)̄ · L(s,χ)
(where χ(a)̄ is the complex conjugate of χ(a)). In the Euler product this picks out exactly the primes p with p≡a mod m.
Logarithmic differentiation and Tauberian theorems
One takes logarithms of the Euler products and looks at
–(L′/L)(s,χ) = ∑{n=1}^∞ Λ(n) χ(n) n^(−s),
where Λ is the von Mangoldt function. Summing these expressions over characters and using orthogonality isolates
∑{p ≡ a (mod m)} p^(−s) ,
plus lower‐order terms. One then applies complex‐analytic Tauberian theorems (or elementary partial summation arguments) to deduce that as x→∞
#{ p ≤ x : p ≡ a (mod m) } ∼ (1/φ(m)) x / log x ,
in particular infinitely many such primes exist.
Conclusion
Dirichlet’s theorem on arithmetic progressions (proven by exactly the steps sketched above) implies that for any m and any a with gcd(a,m) = 1, there are infinitely many primes p ≡ a mod m. Setting m = 35 and a = 6 shows there are infinitely many primes that leave remainder 6 upon division by 35.
Very nice. Makes my generic LLM look like a lumpkin. I’d heard that mathers were impressed by some of them.
Hmm. Interesting question. I haven’t done real math in quite a long time and am absolutely stumped.
One less formal way of looking at it is this: there are an infinite number of multiples of 35. starting with 35, 70, 105. Add 6 to each of the odd multiples. 35+6=41 (prime); 105+6 = 111 (prime). With an infinite # of candidates, you’ve gotta get to an infinite number of solutions (for some value of infinity!)
As for that Dirichlet stuff, it’s way beyond any of the useful stuff I learned too.
Yes, but it obviously wouldn’t work for +5 or +7. I don’t think you can just assume that the number of primes wouldn’t converge to some finite number just because the number of candidates goes to infinity.
Dirichlet’s theorem proves that. I have since looked it up, and that’s correct. I didn’t realize at the time that I was asking it to elaborate the proof for Dirichlet’s theorem. Whether the elaboration is correct is something I will pass.
No, I wouldn’t assume non-convergence either … NOR would I assume that that AI didn’t just grab that ‘high-level’ ‘Elaboration’ from some site … without a citation.
(Very human … Lots of people use quotes to sound smart, hoping they’ll get away with it. LAWYERS! Ministers! Presidents, even! )
The OP article doesn’t say it explicitly, but those mathematicians are getting paid.
The chatbots at duckduckgo don’t have search.