A Beautiful Cyclic System of Equations | Polish Mathematical Olympiad 2012

[@samwalko]

I wonder if there are some complex solutions too. I think what would change is that at the a^3 = -b^3, instead of just a=-b we also get potential solutions of a = (1+i√3)/2 b and a = (1-i√3)/2 b. (This corresponds to e^iπ, e^iπ/3, and e^5iπ/3 = e^-iπ/3; and is related to the third roots of unity). From here I'm not exactly sure where to go though. But I suspect the answers would have the same magnitude.

[@russelltownsend6105]

Love your logic and clarity thanks heaps!

[@nexusmas9]

Bro this is insane , I don't get how you could solve letters . Mashallah ❤

[@sanjaysurya6840]

🤣😊

[@shankhadeepghosh8574]

Unique thought process👍❤️

[@necro5379]

Great🔥

[@xactxx]

In 6:50, you had earlier proved that both pairs (a,c) and (b,d) have the same sign. That means that ac and bd must be positive. So both a^2+ac+b^2+1 and b^2+bd+d^2+1 must be both positive immediately.

[@زينالعابدينماجدمحمد]

THANK YOU so much ❤

[@crustyoldfart]

At the risk of appearing like an untutored peasant, my reaction to this kind of puzzle is that it can barely be considered any more than a storm in teacup. Clearly, a,b,c,d are homogeneous and any solution for one of them can be equally asserted for each of the others. Accordingly we would expect there to be four interchangeable solution sets. FOUR ? Yes, if 0 is considered non-trivial. I don't have the inclination to rummage round finding specific complex solutions, but it is reasonable to suppose there might be at least one such, as others have pointed out. For instance solving of any one of the four individual equations there are three solutions for each - of the form A, A/2 +/- sqrt(3)/2*i. For example, solving for d : A= ( b - a)^(1/3) -> d = { A, A/2*( -1+sqrt(3)/2*i), A/2*(-1-sqrt(3)/2*i) }

[@kanishkaasde]

Supercalifragilisticexpialidocious

[@alboris8203]

Cool

[@bait6652]

Is it sufficient o note the longer term is pos...so The sign(linear terms) equal...thus sgn(d-b)=sgn(b-d) which should = 0 ..thus b=d , a=c, sub in sum a =-b

[@តន្ដ្រីស្រុកខ្ញុំ]

I like hard questions

[@prometheus3899]

a³+b³+c³+d³ is 0 and subtraction 2 from 1 a³-b³=-b-d c³-d³=-d-b from eqn 3 and 4 Thus a³-b³ = c³-d³ ; a³+d³=b³+c³ =0 so a=-d b=-c a³=-2b -b³= 2a Solve this to get reqd answers

[@isaiaherb6116]

I don’t think the subtraction step works. The c’s and a’s are constructive.