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The IMO are a bit more involved... e.g. IMO 2023: > Problem 1. Determine all composite integers n > 1 that satisfy the following property: if d1, d2, . . . ,

by edf13 3y ago

The IMO are a bit more involved... e.g. IMO 2023: > Problem 1. Determine all composite integers n > 1 that satisfy the following property: if d1, d2, . . . , dk are all the positive divisors of n with 1 = d1 < d2 < · · · < dk = n, then di divides di+1 + di+2 for every 1 ⩽ i ⩽ k − 2 Although ChatGPT 4 makes a pretty decent start on this already: > To determine all composite integers \( n > 1 \) that satisfy the given property, we need to closely examine the condition set forth: for a composite number \( n \) with its divisors \( d_1, d_2, \ldots, d_k \) where \( 1 = d_1 < d_2 < \cdots < d_k = n \), the property \( d_i \) divides \( d_{i+1} + d_{i+2} \) must hold for every \( 1 \leq i \leq k - 2 \). Let's break down the problem: 1. *Composite Number Requirement*: We are looking for composite numbers, which are integers greater than 1 that are not prime. These numbers have more than two distinct positive divisors. 2. *Divisor Property*: The property \( d_i | (d_{i+1} + d_{i+2}) \) means that \( d_{i+1} + d_{i+2} \) is a multiple of \( d_i \). 3. *Systematic Examination*: We will analyze this property systematically for composite numbers starting from the smallest composite number (4) and going upwards. Let's start the analysis: - For \( n = 4 \), the divisors are \( 1, 2, 4 \). Here, \( 1 | (2 + 4) \), so 4 satisfies the condition. - For \( n = 6 \), the divisors are \( 1, 2, 3, 6 \). Here, \( 1 | (2 + 3) \) and \( 2 | (3 + 6) \), so 6 satisfies the condition. - For \( n = 8 \), the divisors are \( 1, 2, 4, 8 \). Here, \( 1 | (2 + 4) \), but \( 2 \nmid (4 + 8) \), so 8 does not satisfy the condition. Let's implement a systematic method to identify such composite numbers up to a reasonable limit (let's say up to 100 for a start) and check if they satisfy the given condition. The composite integers greater than 1 and up to 100 that satisfy the given property are: 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81. Each of these numbers has the characteristic that for all their positive divisors \( d_1, d_2, \ldots, d_k \) (where \( 1 = d_1 < d_2 < \cdots < d_k = n \)), the condition \( d_i \) divides \( d_{i+1} + d_{i+2} \) is true for every \( 1 \leq i \leq k - 2 \).

singularity2001 3y ago
Tangential is there a list of Unicode characters allowed on hacker news? ﷽!
- edf13 3y ago
  Yes - apologies for the poor copy-paste here!
  - Recursing 3y ago
    You can just ask GPT-4 to rewrite it using unicode characters: Rewrite this post using unicode characters: > The IMO are a bit more involved... Answer: > The IMO are a bit more involved... e.g. IMO 2023: > Problem 1. Determine all composite integers n > 1 that satisfy the following property: if d₁, d₂, ..., dₖ are all the positive divisors of n with 1 = d₁ < d₂ < ... < dₖ = n, then dᵢ divides dᵢ₊₁ + dᵢ₊₂ for every 1 ≤ i ≤ k − 2 Although ChatGPT 4 makes a pretty decent start on this already: > To determine all composite integers 𝑛 > 1 that satisfy the given property, we need to closely examine the condition set forth: for a composite number 𝑛 with its divisors d₁, d₂, ..., dₖ where 1 = d₁ < d₂ < ... < dₖ = 𝑛, the property dᵢ divides dᵢ₊₁ + dᵢ₊₂ must hold for every 1 ≤ i ≤ k − 2. Let's break down the problem: 1. Composite Number Requirement: We are looking for composite numbers, which are integers greater than 1 that are not prime. These numbers have more than two distinct positive divisors. 2. Divisor Property: The property dᵢ | (dᵢ₊₁ + dᵢ₊₂) means that dᵢ₊₁ + dᵢ₊₂ is a multiple of dᵢ. 3. Systematic Examination: We will analyze this property systematically for composite numbers starting from the smallest composite number (4) and going upwards. Let's start the analysis: - For 𝑛 = 4, the divisors are 1, 2, 4. Here, 1 | (2 + 4), so 4 satisfies the condition. - For 𝑛 = 6, the divisors are 1, 2, 3, 6. Here, 1 | (2 + 3) and 2 | (3 + 6), so 6 satisfies the condition. - For 𝑛 = 8, the divisors are 1, 2, 4, 8. Here, 1 | (2 + 4), but 2 ∤ (4 + 8), so 8 does not satisfy the condition. Let's implement a systematic method to identify such composite numbers up to a reasonable limit (let's say up to 100 for a start) and check if they satisfy the given condition. The composite integers greater than 1 and up to 100 that satisfy the given property are: 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81. Each of these numbers has the characteristic that for all their positive divisors d₁, d₂, ..., dₖ (where 1 = d₁ < d₂ < ... < dₖ = 𝑛), the condition dᵢ divides dᵢ₊₁ + dᵢ₊₂ is true for every 1 ≤ i ≤ k − 2.
    - CaptainFever 3y ago
      Great idea! I never even thought of using GPT-4 to do this.
magnio 3y ago
That's 300 words for saying "in order for n to satisfy the requirements, it must satisfy the requirements." I tried an easier problem, extending Rolle's theorem to the multidimensional case with Poe.com: Prompt: Let U be an open subset of R^n with compact closure K. Suppose f: K → R is continuous on K, differentiable on U, and satisfies f(x) = 0, for all x in K \ U. Show that there exists a in U with grad f (a) = 0. Poe: To prove that there exists a point a in U such that grad f(a) = 0, we will utilize the concept of connectedness and the Mean Value Theorem. Since U is an open subset of R^n with compact closure K, we know that K is a closed and bounded subset of R^n. Furthermore, since K is a compact set, it is also a connected set. Now, let's define a new function g: K → R^n such that g(x) = (f(x), f(x), ..., f(x)) for all x in K. In other words, each component of g is equal to f(x) for all x in K. Since f is continuous on K and differentiable on U (which is an open subset containing K), it follows that g is continuous on K and differentiable on U. Consider the function h: K → R defined as h(x) = ||g(x)||^2 = f(x)^2 + f(x)^2 + ... + f(x)^2 = nf(x)^2. Note that h is continuous on K since g is continuous on K and ||.||^2 is a continuous function. Now, let's consider the set S = {x in K : f(x) = 0}. Since f(x) = 0 for all x in K \ U, we have K \ U ⊆ S. Therefore, the set S is closed. ... Stochastic parrot indeed.
- dj_mc_merlin 3y ago
  Yeah, this is actually a really good example of its weaknesses.. it's the equivalent of a bullshit essay in math form. Completely tautological, with just a bit of the correct definitions thrown it to seem like it understands.
- Davidzheng 3y ago
  https://chat.openai.com/share/f710b53c-1d87-4228-8940-f48ec69f7af0 https://chat.openai.com/share/f710b53c-1d87-4228-8940-f48ec6... GPT-4 for me solves it.
  - mahogany 3y ago
    That is impressive! It does look like a textbook question so I wouldn’t be surprised if a formulation almost identical to this is in the training data. But the ability to parse the “mathiness” is still impressive. Higher level math becomes pretty verbose so enters the domain of language more so than pure symbolic computation. However there is still complex reasoning under the hood. Gpt-4 in ChatGPT flounders when I ask it questions from my thesis, so I’m not sure how it will do with problems it hasn’t seen, where it needs to apply “new” reasoning. I’d love to know if in your example, GPT is reciting something it’s seem verbatim, or if it is taking multiple sources and combing them.
spi 3y ago
A decent start? It says absolutely nothing about how to solve it, except repeating the question. The part where it tries out the few first numbers is entirely wrong, given that 6 does _not_ satisfy the condition (2 does not divide 3+6=9) and 8 _does_ (2 does divide 4+8=12). Amusingly, the list of the integers <= 100 satisfying the property is correct, and it contradicts itself from the previous paragraph. Maybe if GPT wasn't a one-directional autoregressive model but allowed itself to go back and edit the past, it would have caught up that discrepancy and fixed it - but no such architecture currently exists that would run in decent amount of time. Given that GPT4 is not a model but a full product, behind the scenes it probably coded up and ran a small python code that translated the problem into code, executed it and got its solution for the first few integers. Which would be a good thing to do to start solving a problem like this, except you're not allowed to do that at the IMO, obviously. Looking at the output of this program, it suggests that powers of primes could be a class of solution (or maybe even the only solutions? I guess that's all the problem was _really_ asking to prove, but having never qualified for the IMO myself, I can't be sure). In fact, for n = p^k, the divisors are [1, p, p^2, ..., p^{k-1}, p^k], and clearly always p^i divides p^{i+1} + p^{i+2} = p^i (p + p^2). I guess this small remark would have gained me a point at the IMO, only 41 to go ;-) But the other side of the coin is that having those numbers written down in front of you and not even making a conjecture about powers of prime being the answer would really denote poor mathematical reasoning by GPT4. It's only really proving that it can understand what it's being asked, which, I admit, places it in a better position than maybe 90% of the human population, but unfortunately for GPT4 mathematics is the least democratic science of them all - it's always only the top-1 result that matters in the end. P.S. Being a former mathematician currently working on deep learning, having (or building!) a model that can solve mathematical questions has always been my dream. I'm not even talking about something that can _prove_ things, even just that can understand and rephrase them in different settings (which in mathematics is very, very hard, even for a human). Or spot weaknesses in already stated down proofs. As a graduate student, having something I could chat about to ask silly question while studying a paper would have been a real game changer. Even for best-of-world professionals it would be useful: when the wrong proof about the ABC conjecture came out, it took months of work from the best minds of our world to read through it and disprove it. If Mochizuki had had some tool for automatically checking his proof (and a smaller ego, I guess) he could have caught that early on, saved everybody a lot of work and the whole world some useless drama. And while we're closer than ever to reaching that, I think GPT4 is still quite a far way from it. But with the pace we've seen recently in AI evolution, who knows...
Davidzheng 3y ago
GPT-4+code is surprisingly good. https://chat.openai.com/share/ad42d09a-b366-4b21-b701-782fc8c6b686 https://chat.openai.com/share/ad42d09a-b366-4b21-b701-782fc8... "This observation leads to a hypothesis: only the powers of prime numbers satisfy the given condition. To verify this hypothesis, we need to consider why powers of primes might uniquely satisfy the divisibility condition and then prove or disprove it."
devit 3y ago
In my first attempt at the problem it made this mistake: "The condition fails for i=1 since 1 does not divide p+q unless p+q is a multiple of n, which is not generally true" (1 divides everything) In the second it gives up: "However, this conclusion is based on heuristic reasoning and examples" Third attempt it makes this mistake: "If the immediate next divisor, di+1 , is not a multiple of p (for instance, it could be q or a product involving q), then p does not divide di+1. Hence, p will not divide the sum di+1+di+2 in such a case, violating the condition." (the fact that p does not divide d_i+1 does not imply that it does not divide d_i+1, d_i+2) Then it gives up again: "However, this conclusion is based on heuristic reasoning and examples" Then it makes this basic logic mistake: "However, since p and q are distinct primes, p does not divide q, and it's not guaranteed that p divides q+d_i+2 , especially if d_i+2 is not a multiple of p. Therefore, for such n, the condition fails." (the fact that "it's not guaranteed" doesn't imply that it's false) Overall it proves that prime powers have the property, conjures that non-prime-powers don't, proves that p*q doesn't have it, but completely fails at coming up with a proof strategy that proves that non-prime-powers don't have the property.

by edf13 3y ago

singularity2001 3y ago

Tangential is there a list of Unicode characters allowed on hacker news? ﷽!

edf13 3y ago
Yes - apologies for the poor copy-paste here!
- Recursing 3y ago
  You can just ask GPT-4 to rewrite it using unicode characters: Rewrite this post using unicode characters: > The IMO are a bit more involved... Answer: > The IMO are a bit more involved... e.g. IMO 2023: > Problem 1. Determine all composite integers n > 1 that satisfy the following property: if d₁, d₂, ..., dₖ are all the positive divisors of n with 1 = d₁ < d₂ < ... < dₖ = n, then dᵢ divides dᵢ₊₁ + dᵢ₊₂ for every 1 ≤ i ≤ k − 2 Although ChatGPT 4 makes a pretty decent start on this already: > To determine all composite integers 𝑛 > 1 that satisfy the given property, we need to closely examine the condition set forth: for a composite number 𝑛 with its divisors d₁, d₂, ..., dₖ where 1 = d₁ < d₂ < ... < dₖ = 𝑛, the property dᵢ divides dᵢ₊₁ + dᵢ₊₂ must hold for every 1 ≤ i ≤ k − 2. Let's break down the problem: 1. Composite Number Requirement: We are looking for composite numbers, which are integers greater than 1 that are not prime. These numbers have more than two distinct positive divisors. 2. Divisor Property: The property dᵢ | (dᵢ₊₁ + dᵢ₊₂) means that dᵢ₊₁ + dᵢ₊₂ is a multiple of dᵢ. 3. Systematic Examination: We will analyze this property systematically for composite numbers starting from the smallest composite number (4) and going upwards. Let's start the analysis: - For 𝑛 = 4, the divisors are 1, 2, 4. Here, 1 | (2 + 4), so 4 satisfies the condition. - For 𝑛 = 6, the divisors are 1, 2, 3, 6. Here, 1 | (2 + 3) and 2 | (3 + 6), so 6 satisfies the condition. - For 𝑛 = 8, the divisors are 1, 2, 4, 8. Here, 1 | (2 + 4), but 2 ∤ (4 + 8), so 8 does not satisfy the condition. Let's implement a systematic method to identify such composite numbers up to a reasonable limit (let's say up to 100 for a start) and check if they satisfy the given condition. The composite integers greater than 1 and up to 100 that satisfy the given property are: 4, 8, 9, 16, 25, 27, 32, 49, 64, and 81. Each of these numbers has the characteristic that for all their positive divisors d₁, d₂, ..., dₖ (where 1 = d₁ < d₂ < ... < dₖ = 𝑛), the condition dᵢ divides dᵢ₊₁ + dᵢ₊₂ is true for every 1 ≤ i ≤ k − 2.
  - CaptainFever 3y ago
    Great idea! I never even thought of using GPT-4 to do this.

magnio 3y ago

That's 300 words for saying "in order for n to satisfy the requirements, it must satisfy the requirements." I tried an easier problem, extending Rolle's theorem to the multidimensional case with Poe.com: Prompt: Let U be an open subset of R^n with compact closure K. Suppose f: K → R is continuous on K, differentiable on U, and satisfies f(x) = 0, for all x in K \ U. Show that there exists a in U with grad f (a) = 0. Poe: To prove that there exists a point a in U such that grad f(a) = 0, we will utilize the concept of connectedness and the Mean Value Theorem. Since U is an open subset of R^n with compact closure K, we know that K is a closed and bounded subset of R^n. Furthermore, since K is a compact set, it is also a connected set. Now, let's define a new function g: K → R^n such that g(x) = (f(x), f(x), ..., f(x)) for all x in K. In other words, each component of g is equal to f(x) for all x in K. Since f is continuous on K and differentiable on U (which is an open subset containing K), it follows that g is continuous on K and differentiable on U. Consider the function h: K → R defined as h(x) = ||g(x)||^2 = f(x)^2 + f(x)^2 + ... + f(x)^2 = nf(x)^2. Note that h is continuous on K since g is continuous on K and ||.||^2 is a continuous function. Now, let's consider the set S = {x in K : f(x) = 0}. Since f(x) = 0 for all x in K \ U, we have K \ U ⊆ S. Therefore, the set S is closed. ... Stochastic parrot indeed.

dj_mc_merlin 3y ago
Yeah, this is actually a really good example of its weaknesses.. it's the equivalent of a bullshit essay in math form. Completely tautological, with just a bit of the correct definitions thrown it to seem like it understands.
Davidzheng 3y ago
https://chat.openai.com/share/f710b53c-1d87-4228-8940-f48ec69f7af0 https://chat.openai.com/share/f710b53c-1d87-4228-8940-f48ec6... GPT-4 for me solves it.
- mahogany 3y ago
  That is impressive! It does look like a textbook question so I wouldn’t be surprised if a formulation almost identical to this is in the training data. But the ability to parse the “mathiness” is still impressive. Higher level math becomes pretty verbose so enters the domain of language more so than pure symbolic computation. However there is still complex reasoning under the hood. Gpt-4 in ChatGPT flounders when I ask it questions from my thesis, so I’m not sure how it will do with problems it hasn’t seen, where it needs to apply “new” reasoning. I’d love to know if in your example, GPT is reciting something it’s seem verbatim, or if it is taking multiple sources and combing them.

spi 3y ago

A decent start? It says absolutely nothing about how to solve it, except repeating the question. The part where it tries out the few first numbers is entirely wrong, given that 6 does _not_ satisfy the condition (2 does not divide 3+6=9) and 8 _does_ (2 does divide 4+8=12). Amusingly, the list of the integers <= 100 satisfying the property is correct, and it contradicts itself from the previous paragraph. Maybe if GPT wasn't a one-directional autoregressive model but allowed itself to go back and edit the past, it would have caught up that discrepancy and fixed it - but no such architecture currently exists that would run in decent amount of time. Given that GPT4 is not a model but a full product, behind the scenes it probably coded up and ran a small python code that translated the problem into code, executed it and got its solution for the first few integers. Which would be a good thing to do to start solving a problem like this, except you're not allowed to do that at the IMO, obviously. Looking at the output of this program, it suggests that powers of primes could be a class of solution (or maybe even the only solutions? I guess that's all the problem was _really_ asking to prove, but having never qualified for the IMO myself, I can't be sure). In fact, for n = p^k, the divisors are [1, p, p^2, ..., p^{k-1}, p^k], and clearly always p^i divides p^{i+1} + p^{i+2} = p^i (p + p^2). I guess this small remark would have gained me a point at the IMO, only 41 to go ;-) But the other side of the coin is that having those numbers written down in front of you and not even making a conjecture about powers of prime being the answer would really denote poor mathematical reasoning by GPT4. It's only really proving that it can understand what it's being asked, which, I admit, places it in a better position than maybe 90% of the human population, but unfortunately for GPT4 mathematics is the least democratic science of them all - it's always only the top-1 result that matters in the end. P.S. Being a former mathematician currently working on deep learning, having (or building!) a model that can solve mathematical questions has always been my dream. I'm not even talking about something that can _prove_ things, even just that can understand and rephrase them in different settings (which in mathematics is very, very hard, even for a human). Or spot weaknesses in already stated down proofs. As a graduate student, having something I could chat about to ask silly question while studying a paper would have been a real game changer. Even for best-of-world professionals it would be useful: when the wrong proof about the ABC conjecture came out, it took months of work from the best minds of our world to read through it and disprove it. If Mochizuki had had some tool for automatically checking his proof (and a smaller ego, I guess) he could have caught that early on, saved everybody a lot of work and the whole world some useless drama. And while we're closer than ever to reaching that, I think GPT4 is still quite a far way from it. But with the pace we've seen recently in AI evolution, who knows...

Davidzheng 3y ago

GPT-4+code is surprisingly good. https://chat.openai.com/share/ad42d09a-b366-4b21-b701-782fc8c6b686 https://chat.openai.com/share/ad42d09a-b366-4b21-b701-782fc8... "This observation leads to a hypothesis: only the powers of prime numbers satisfy the given condition. To verify this hypothesis, we need to consider why powers of primes might uniquely satisfy the divisibility condition and then prove or disprove it."

devit 3y ago

In my first attempt at the problem it made this mistake: "The condition fails for i=1 since 1 does not divide p+q unless p+q is a multiple of n, which is not generally true" (1 divides everything) In the second it gives up: "However, this conclusion is based on heuristic reasoning and examples" Third attempt it makes this mistake: "If the immediate next divisor, di+1 , is not a multiple of p (for instance, it could be q or a product involving q), then p does not divide di+1. Hence, p will not divide the sum di+1+di+2 in such a case, violating the condition." (the fact that p does not divide d_i+1 does not imply that it does not divide d_i+1, d_i+2) Then it gives up again: "However, this conclusion is based on heuristic reasoning and examples" Then it makes this basic logic mistake: "However, since p and q are distinct primes, p does not divide q, and it's not guaranteed that p divides q+d_i+2 , especially if d_i+2 is not a multiple of p. Therefore, for such n, the condition fails." (the fact that "it's not guaranteed" doesn't imply that it's false) Overall it proves that prime powers have the property, conjures that non-prime-powers don't, proves that p*q doesn't have it, but completely fails at coming up with a proof strategy that proves that non-prime-powers don't have the property.