Radical/Power/Floor Function --- SMO 2013

#olympiad #math #algebra #jee #trigonometry #geometry #gmat #mathstrick #olympiad2022

PLAYLISTS 🎵 :

[matematicasybarcos]

Very beautiful solution, SQRT. Thanks for sharing it!

[SQRTime]

Hello everyone. x^4 - x^2y^2 + y^4 = (x^2 + y^2)^2 - 3 (xy)^2. In the video, we have 2(xy)^2 which has a typo. The final answer is totally correct. Hope it helps. This is around time 1:21.

[jessiechua]

Here is a more brute force approach to solving the problem. Note that a=(3+root(17))/2 and b=(3-root(17))/2 are both roots of the equation x^2-3x-2=0. By Vieta's formula, a+b=3 and ab=-2. This gives us (a+b)^6=729. By applying the Binomial Theorem, we derive a^6+b^6=2041. But since -1<b<0, b^6<1. Thus, 2040<a^6<2041, implying that floor(a^6)=2040.

[jessiechua]

Another alternative solution using recurrence relations. Define a and b as per my previous comment. Let u_n=a^n+b^n, and note that the characteristic equation of the second-order linear recurrence Au_n=Bu_(n-1)+Cu_(n-2) is given by Ax^2-Bx-C=0. If we interpret the quadratic x^2-3x-2 as the characteristic equation of a second-order linear recurrence, then we obtain A=1, B=3, C=2, i.e. u_n=3u_(n-1)+2u_(n-2). We can easily verify by hand that this is correct. Since u_0=2, u_1=3, we can successively calculate u_2=13, u_3=45, u_4=161, u_5=573, and u_6=2041. The rest is then simple.

[tanalan]

I think that it's a little wrong something, -3(xy)^2，not -2(xy)^2

[MrPacha]

Sorry, I am not mathematician but: calculate sqrt (17) then add 3, divide by 2 and finally power 6 = 2040.78 !