2
$\begingroup$

Solve following system of equations over real numbers: $$ x-y+z-u=2\\ x^2-y^2+z^2-u^2=6\\ x^3-y^3+z^3-u^3=20\\ x^4-y^4+z^4-u^4=66 $$ This does not seem as hard problem. I have tried what is obvious here, to write $x^2-y^2$ as $(x-y)(x+y)$, $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$ etc. Problem is because if I substitute $x-y=u-z+2$ in second equation, I cannot eliminate $x+y$ or if I substitute $x^2-y^2=u^2-z^2+6$ in fourth equation, I cannot eliminate $x^2+y^2$ etc. When I realized that this is not the best way, I made a second attempt.
My second attempt is to find come constraints, because all of these numbers are real. I have proven that if $m+n=a$ for some $m,n\in\mathbb{R}^+$, then $\frac{a^2}2\le m^2+n^2\le a^2$. I have applied this formula to second and fourth equation (because in these equations variables have even exponents and that means they are positive), but this didn't give me anything.
What should be the easiest way to solve it?

$\endgroup$
3
$\begingroup$

\begin{align} x-y+z-u&=2 \quad (1)\\ x^2-y^2+z^2-u^2&=6 \quad (2)\\ x^3-y^3+z^3-u^3&=20 \quad (3)\\ x^4-y^4+z^4-u^4&=66 \quad (4) \end{align}

The easiest way to solve it is to start of testing the simplest solutions first. The first eqn suggests to check if $x-y=1$, $z-u=1$ fits. In this case (2) becomes \begin{align} x+y+z+u&=6, \quad(3) \\ (1)+(3)\Rightarrow& \\ x+y&=4, \\ z+u&=2 \end{align} and we have a solution $u=0,z=1,x=3,y=2$, which is indeed a valid solution of the system (1)--(4). And due to the symmetry, there are three more valid combinations.

$\endgroup$
0
$\begingroup$

rewrite the equations:

$x+z-(y+u)=2\\ x^2+z^2-(y^2+u^2)=6\\ x^3+z^3-(y^3+u^3)=20\\ x^4+z^4-(y^4+u^4)=66$

$a=x+z,c=xz,b=y+u,d=yu, \\x^2+z^2=(x+z)^2-2xz=a^2-2c,\\x^3+z^3=(x+z)(x^2+z^2-xz)=(x+z)((x+z)^2-3xz)=a(a^2-3c)\\x^4+z^4=(x^2+z^2)^2-2(xz)^2=((x+z)^2-2xz)^2-2(xz)^2=a^4-4a^2c+2c^2$

you can go from here now.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy