3 ms·
A ridiculous problem I remember seeing was approximately this one: a/(b+c) + b/(a+c) + c/(a+b)=896. a,b,c positive integers I think it probably used rando
by dan-robertson 6y ago
A ridiculous problem I remember seeing was approximately this one:
a/(b+c) + b/(a+c) + c/(a+b)=896.
a,b,c positive integers
I think it probably used random symbols instead of letters and didn’t have the integer requirement. Indeed if you allow real numbers or 0 then it is easy to find a solution. I think it didn’t go particularly viral because it was too hard. I think there’s a slightly easier version if you replace 896 by 16.
- contravariant 6y agoThe version I first saw used '4' instead of '896'. Which frankly made the answer large enough that no reasonable amount of brute force could find it. A good write up can be found here: https://www.quora.com/How-do-you-find-the-positive-integer-solutions-to-frac-x-y%2Bz-%2B-frac-y-z%2Bx-%2B-frac-z-x%2By-4/answer/Alon-Amit https://www.quora.com/How-do-you-find-the-positive-integer-s...
- kens 6y agoThat is an amazing, detailed explanation. (Don't be fooled by the quora location.) Spoilers: the solution to the equation is 80-digit numbers, which can be obtained by using elliptic curves, which the link explains in detail.
- IanCal 6y agoThat is fascinating thank you, and a great lead into some more advanced topics.