An Unusual Trick That Quickly Solves These Equations | Romanian National Maths Olympiad 2002

[@crazy4hitman755]

I can’t believe how beautiful this solution is

[@nit_man]

Exactly

[@Szynkaa]

my first step was to put a=2cosA and so on, but i didn't know what to do next. Idea with cos5x is very nice but it didn't come to my mind at all.

[@goodhuman4189]

i dont understand at 6:08 why each cos(5theta) - 1 = 0 how did he get it from sum cos(5theta) - 1 = 0 just because the sum is equal to 0 doesnt mean each individual term is = 0 can someone explain please I'm confused

[@pyrodynamic4144]

every cos(5theta)-1 is less than or equal to 0. If even one of them is negative then the result would be negative, not 0. Therefore they're all 0.

[@goodhuman4189]

@@pyrodynamic4144 thanks a lot i was confused

[@redaabakhti768]

this is astoundingly brilliant

[@themathsgeek8528]

Wow this was absolutely amazing!

[@lossenidiaby9677]

Please put the domain of solving your equation,when you post new video, because I'm trying to solve these equation without knowing if it is in Integers or Reals perhaps in complex numbers,so that I have to watch the solution without solving the problem.Thanks you

[@isrogaganyaan6923]

I would say that's a brilliant approach thanks I learnt a new technique 😀

[@mayoor357]

Brilliant idea for substitution, you made the problem easy

[@lalitmarwaha3537]

I did not expect trigonometry to show up

[@خلدوناللشوفي]

You are very smart person

[@sarthjoshi6325]

Elegant solution 👌

[@himu1901]

What a mind boggling technique

[@ankit..901]

Determine whether or not there are any positive integral solutions of the simultaneous equations (X1) ^2+(X2) ^2+(X3) ^2+......(X1985)^2=y^3 and (X1) ^3+(X2)^3+(X3) ^3+......(X1985)^3=z^2with distinct integers X1, X2, X3.... X1985. (USAMO 1985)

[@jitendramohan7500]

What's the logic behind a = 2cos A as how can one sure that a lies bw -2 to +2

[@yasg384]

For any A we have -1

[@jakobr_]

It’s given at the start of the problem. I agree, without that it would make no sense to bring in the cosine (unless you want to go complex but that’ll get messy)

[@akirakato1293]

@@yasg384 he was asking why the substitution was legitimate

[@anggalol]

Because there are many trigonometric identity which we can use to solve the problem

[@isrogaganyaan6923]

BTW @lets think critically is the total no of solns 15 ??

[@socerdemon8]

should be 5 choices for 2 then 4c2 for the next, so 5*6=30

[@isrogaganyaan6923]

@@socerdemon8 vro I used grouping theory 5!/2!2!1!2! Which is 15

[@petersievert6830]

@@isrogaganyaan6923 I think you have one 2! too many in the denominator.

[@isrogaganyaan6923]

@@petersievert6830 but because of 2 ,2! We divide by another 2! Due to identical division don't we ??

[@shivamvishwekar3652]

5!/2!2!=30

[@replicaacliper]

How do we know there must be two pairs

[@sebastianw.1217]

The complete sum is an integer so the sqrt(5) must cancel

[@bubbletea-ol4lr]

There's two because when testing one pair, the sum does not equal 0, but then when two pairs are tested it does. I guess he just left it out because it's a pretty trivial step.

[@replicaacliper]

@@bubbletea-ol4lr yeah idk how I didn't see that thanks

[@goodhuman4189]

@@sebastianw.1217 hey man can you help please i dont understand at 6:08 why each cos(5theta) - 1 = 0 how did he get it from sum cos(5theta) - 1 = 0 just because the sum is equal to 0 doesnt mean each individual term is = 0 can someone explain please I'm confused

[@RGP_Maths]

There's two pairs because x^2+x-1 was a repeated factor of the quintic. Therefore each root of that quadratic is a repeated root of the quintic.

[@Monolith-yb6yl]

How did you decide to consider cos 5A=...

[@krzysk5388]

(cos(x) + i*sin(x))^5 = cos(5x)+i*sin(5x) After expanding you can set real parts of left and right side to exual (or imaginary parts to solve for sin(5x)). You can similarly find values of sin(nx) or cos(nx) by using de Moivres formula

[@Monolith-yb6yl]

@@krzysk5388 thanks, but i ask how and why did he has the idea to consider, not how to prove formula ;)

[@sarthjoshi6325]

I think he must have gone for cos 5a as it contains terms of cos^5 , cos^3 and cosa , which makes sense as we have the values of individual summation of this 3 terms

[@Monolith-yb6yl]

@@sarthjoshi6325 its hard to understand before you watch the video

[@oguzhanozdogan4915]

Çok güzel çözmüşsünüz. Türkiyeden selamlar..

[@yoav613]

Noice