It's "How Many", But Not An Inequality

Algebra Level 3

{ x 3 + y 3 = 1 x 4 + y 4 = 1 \large\begin{cases} \sqrt[3]{x}+\sqrt[3]{y}=1\\ \sqrt[4]{x}+\sqrt[4]{y}=1\end{cases}

How many real pairs of ( x , y ) (x,y) satisfy the system above?

0 1 2 More than 2

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2 solutions

P C
Mar 21, 2016

Set t = x 12 t=\sqrt[12]{x} and u = y 12 u=\sqrt[12]{y} and we'll get the following system { t 4 + u 4 = 1 t 3 + u 3 = 1 \begin{cases} t^4+u^4=1 \\ t^3+u^3=1\end{cases} { t 4 + u 4 = t 3 + u 3 t 3 + u 3 = 1 \Leftrightarrow\begin{cases} t^4+u^4=t^3+u^3 \\ t^3+u^3=1\end{cases} Considering the first equation t 4 + u 4 = t 3 + u 3 t^4+u^4=t^3+u^3 t 3 ( t 1 ) = u 3 ( 1 u ) ( ) t^3(t-1)=u^3(1-u) \ (*) Since t , u 0 t,u\geq 0 and t 3 + u 3 = 1 t^3+u^3=1 , we get t , u [ 0 ; 1 ] t,u\in [0;1] . So t 3 ( t 1 ) 0 t^3(t-1)\leq 0 and u 3 ( 1 u ) 0 u^3(1-u)\geq 0 , therefore ( ) (*) happens when t 3 ( t 1 ) = u 3 ( 1 u ) = 0 t^3(t-1)=u^3(1-u)=0 { t 3 ( t 1 ) = u 3 ( 1 u ) = 0 t 3 + u 3 = 1 \Rightarrow\begin{cases} t^3(t-1)=u^3(1-u)=0 \\ t^3+u^3=1\end{cases} ( t , u ) = ( 1 ; 0 ) a n d ( 0 ; 1 ) \Leftrightarrow (t,u)=(1;0) \ and \ (0;1) So finally ( x , y ) = ( 1 ; 0 ) , ( 0 ; 1 ) (x,y)=(1;0),(0;1) , 2 2 is the correct answer

@Gurīdo Cuong It's necessary to specify that x , y x,y must be real in order to claim that t , u 0 t,u\geq 0

Abdur Rehman Zahid - 5 years, 2 months ago

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Thanks, I'll fix it

P C - 5 years, 2 months ago
Adams Ayoade
Mar 21, 2016

Well,i discovered that for the equations to be true, x and y must be greater than or equal tozero. So by mere observation, (1,0) and (0,1) seem to be the only valid entry.

How can we confirm that those are the only solutions?

Calvin Lin Staff - 5 years, 2 months ago

A mere observation may be useful in this problem

Evan Huynh - 5 years, 2 months ago

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