System of three equations - Iran First Round 2018, P14

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System of three equations - Iran First Round 2018, P14

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#1 h Mar 7, 2021, 1:02 AM • 1 Y

Y by Mango247

For how many integers $k$ does the following system of equations has a solution other than $a=b=c=0$ in the set of real numbers? $\begin{align*} \begin{cases} a^2+b^2=kc(a+b),\\ b^2+c^2 = ka(b+c),\\ c^2+a^2=kb(c+a).\end{cases}\end{align*}$

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#2 h Mar 7, 2021, 1:46 AM • 1 Y

Y by Amir Hossein

Subtracting pairs of equations, we get
$(c-a)(a+kb+c)=0$ $(b-a)(a+b+kc)=0$ $(b-c)(ka+b+c)=0$
Case 1:
Obviously if any 2 pairs of the variables are equal, then we get the parametrization $(t,t,t)$ . Plugging into any of the equations, we get
$2t^2(1-k)=0$ Since we want $t\not\equiv 0$ for nontrivial solutions, we have $k=1$

Case 2:
Let's say that we have WLOG $a=c$ . Then the first equation is true.
We have
$(b-a)((k+1)a+b)=0$ For a nontrivial solution, we would want $b=(-1-k)a$
So we get the parametrization $(t,(-1-k)t,t)$
Plugging this in into the third of the original equations, we get
$2t^2=(k-k^2)(2t^2)$ $2(k^2-k+1)t^2=0$ Since we want $t\not\equiv 0$ , we require that $k^2-k+1=0$ , which has no integer solutions.

Case 3:
In this last case, we will have none of the variables equal. This leaves
$a+kb+c=0$ $ka+b+c=0$ $a+b+kc=0$ (I'm going to show off my linear algebra here, but this is pretty easy to do without linear algebra)
For this homogenous system to have nontrivial solutions we need the determinant equal to $0$ .
This can be rearranged to the circulatory matrix with polynomial $f(x)=k+x+x^2$ . The determinant is $(k+2)(k+\omega+\omega^2)(k+\omega+\omega^2)=(k+2)(k-1)^2$
The only new integer solution is when $k=-2$

So there are only $\boxed{2}$ integer values of $k$

This post has been edited 1 time. Last edited by natmath, Mar 7, 2021, 1:47 AM

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#3 h Today at 9:56 AM

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Easy!

:-D

We see that if $a=0$ then we get $a=b=c=0$ . So we must have $a, b, c\neq 0$
Note that if $(a, b, c)$ is a solution then $(-a, -b, -c)$ is also a solution. Also note that there is symmetry so we can suppose WLOG that $a\geq b\geq c$
So it is enough to check just two cases

Case 1: $a, b, c> 0$

Then we must have $k>0$
From the hypothesis we have
$a\geq c \Leftrightarrow a^2 + b^2 \geq b^2 + c^2 \Leftrightarrow kc(a+b) \geq ka(b+c) \Leftrightarrow ca+bc\geq ab+ca \Leftrightarrow c\geq a$
But because $a\geq b\geq c$ we must have $a=b=c=t$
Then we have $2a^2=2ka^2 \Leftrightarrow \boxed{k=1}$

Case 2: $a, b>0,\: c<0$

Then we have $0<a^2+b^2 = kc(a+b) \Rightarrow kc > 0 \Rightarrow k<0$
So we have $a^2 + c^2\geq b^2+c^2 \Leftrightarrow kb(c+a)\geq ka(b+c) \Leftrightarrow bc+ab\leq ab+ca \Leftrightarrow bc\leq ca \Leftrightarrow b\geq a$
But from hypothesis we have $a\geq b$ . So we must have $a=b$ .
Then the first equation is equally written as $2a^2=2akc \Leftrightarrow a=kc$ . Replacing to the second equation we get
$a^2+c^2=ka^2+kac \Leftrightarrow k^2c^2 + c^2= k^3c^2+k^2c^2 \Leftrightarrow c^2=k^3c^2 \Leftrightarrow k^3=1 \Leftrightarrow k=1$ which is impossible because we showed that $k<0$
So we have only one possible value of $k$ , the $\boxed{k=1}$ which gives us the solution $(a, b, c)=(t, t, t), \: t\in \mathbb{R^*}$

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