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An equation that takes Pythagoras to a new level

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21–24 of 24 posts

Re: An equation that takes Pythagoras to a new level

#21

The caption "The equation 1⁰² + 1¹² + 1²² = 1³² + 1⁴², whose answer is that both sides equal 365, was immortalized in a different form in this 1895 painting: “Mental Arithmetic. In the Public School of S. Rachinsky.” (NIKOLAY BOGDANOV-BELSKY)" has funky exponents. Medium requires that I make an account to say that, so I am saying it here.

FUCK MEDIUM, assholes

Re: An equation that takes Pythagoras to a new level

#22
post #18

I thought this was going to help me with Project Euler 261 [0], but it's not exactly the same, so no luck so far. Cool article anyway; linking geometry and algebra is always satisfactory. [0] https://projecteuler.net/problem=261

How much CPU time and number theory do problems in that range generally require?

Brute force search of 10^10 (with some automatic monitoring of how wide a range of 'm' to scan at each candidate) is a lot but not a lot of a lot, if you code in a CPU efficient language and maybe use parallel computation and are willing to wait a while.

Re: An equation that takes Pythagoras to a new level

#24
post #18

I thought this was going to help me with Project Euler 261 [0], but it's not exactly the same, so no luck so far. Cool article anyway; linking geometry and algebra is always satisfactory. [0] https://projecteuler.net/problem=261

How much CPU time and number theory do problems in that range generally require? Brute force search of 10^10 (with some automatic monitoring of how wide a range of 'm' to scan at each candidate) is a lot but not a lot of a lot, if you code in a CPU efficient language and maybe use parallel computation and are willing to wait a while.

The difficulty and the required run time vary a lot starting from problem 150 or so. This one seems particularly difficult (it's rated at 85% difficulty, which is also the highest I've managed to solve). The problem here is that the search space is not 10^10, but higher, since if you iterate over k you don't know which values of m or n need to be tested. For m=1 it's clear that you can just use Pythagorean triples, but for bigger values I don't know if there is a reliable procedure to generate solutions.
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