#diophantine — Public Fediverse posts
Live and recent posts from across the Fediverse tagged #diophantine, aggregated by home.social.
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🚀 Ah, the age-old question: how many API calls can you squeeze into an hour using quadratic equations and pure ✨wizardry✨? Because obviously, when facing a straightforward 10 requests per hour limit, the only logical step is to consult your dusty #Diophantine tome and channel your inner #Pythagoras 🤓🔢. Forget practical solutions; let's dive into the mathematical abyss for fun! 😂📉
https://vivekn.dev/blog/rate-limit-diophantine #APIcalls #QuadraticEquations #MathHumor #HackerNews #ngated -
Right, so, if we then gave all the cows typewriters, how long would it take one of them to reproduce all the works of Shakespeare?
https://www.youtube.com/watch?v=V2S39AgjFpQ
#Archimedes #Eratosthenes #maths #Diophantus #Diophantine #NumberTheory #mathematics #math #cattle #cows #history -
It's well known you can find a solution to the linear Diophantine equation
\[ a_1x_1 + a_2x_2 + \ldots + a_nx_n = c \]
in polynomial time using the Extended Euclidean Algorithm. But I haven't been able to find a clear answer for the complexity of finding a non-negative integer solution, that is, \( x_i \geq 0 \).
The unbounded knapsack problem is NP-complete, and this is the case where the weights and costs are equal, but I'm not sure that the NP-hardness reduction applies given that additional constraint.
I have also found a statement that the "multidimensional knapsack problem" is NP-hard even with a single row, which seems to match this, but I lack a copy of Papadimitriou and Steiglitz to verify that statement, and can't figure out the reduction.
If this problem is NP-hard, then I'm also interested in showing that a related problem is also NP-hard: if we have one non-negative solution, can we find a second one? I'm trying to answer a question about the special case \( c = \sum_{i=1}^{n} a_i \).
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If I'm really committed to the fact that there is no such thing as actual infinity - and I am: https://www.wisdomneverdies.com/blog/no-infinity -
then there are no actual irrational numbers. By Kronecker's Theorem then, all apparently chaotic behavior is ultimately periodic. An interesting corollary 🙂#infinity #math #philosophy #Cavendish #Cantor #Locke #Aristotle #Knuth #chaos #Kronecker #Diophantine
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Over on Quora, I answered a question about integer solutions to \( a^2 - b^2 = b^2 - c^2 \). If we rewrite it as \( a^2 - 2b^2 = -c^2 \) it becomes clear that it's a general Pell equation. It looks like we get a nontrivial family of solutions whenever \( c \) is a prime that is +1 or -1 mod 8, but I'm not seeing why that should be.
For c = 1 this is exactly the negative Pell equation, so we have solutions like \( 7^2 - 5^2 = 5^2 - 1^2 \). c = 17 gives us a pair of infinite families starting with \( 7^2 - 13^2 = 13^2 - 17^2 \) and \( 31^2 - 25^2 = 25^2 - 17^2 \).
c = 5783 has primitive solutions a = 4369, b = 5125 and a = 7393, b = 6637.
Can we show that there is always a nontrivial solution for primes of this form?
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@geekbrit pulling a #Diophantine on unsuspecting public, I see 🤣
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@jbeardsleymath I think Pythagorean triples are just super cool. When I checked wikipedia to make sure that a² + b² = c² is a #Diophantine equation (it is), there was a potentially-interesting list of others?
About Pythagorean triples, I was astounded to find out there's a *formula* for them. It's not even that complicated! One method of proof is to find rational points on the circle, and it boils down to the quadratic formula (really). I found that to be amazing!
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Last couple of nights I've been falling asleep on another little mental calculation.
Oh yeah, it's time for another episode of #CountingSheep, bay-bee!
This time we're doing linear #Diophantine equations, i.e. the Euclidean algorithm.