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Source: Polish Math Olympiad 2025 Finals P1

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blug

#1 h Apr 4, 2025, 11:46 AM

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Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
$\begin{aligned} \begin{cases} a+b+c+d=0,\\ a^2+b^2+c^2+d^2=12,\\ abcd=-3.\\ \end{cases} \end{aligned}$

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sqing

#2 h Apr 4, 2025, 12:06 PM

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blug wrote:

Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
$\begin{aligned} \begin{cases} a+b+c+d=0,\\ a^2+b^2+c^2+d^2=12,\\ abcd=-3.\\ \end{cases} \end{aligned}$

$\{a, b, c,d\} = \{-3,1,1,1\}$ and $\{a, b, c,d\} = \{-1,-1,-1,3\}$

This post has been edited 1 time. Last edited by sqing, Apr 4, 2025, 12:10 PM

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#3 h Apr 4, 2025, 1:05 PM

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sqing wrote:

blug wrote:

Find all $(a, b, c, d)\in \mathbb{R}$ satisfying
$\begin{aligned} \begin{cases} a+b+c+d=0,\\ a^2+b^2+c^2+d^2=12,\\ abcd=-3.\\ \end{cases} \end{aligned}$

$\{a, b, c,d\} = \{-3,1,1,1\}$ and $\{a, b, c,d\} = \{-1,-1,-1,3\}$

what about non interger values ?

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blug

#4 h Apr 4, 2025, 2:54 PM

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kjhgyuio wrote:

what about non interger values ?

These are the only solutions.

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#5 h Apr 4, 2025, 4:59 PM • 1 Y

Y by teomihai

Here is a complete solution. Simply observe that if $(a,b,c,d)$ is a solution so is any permutation of it. Further, note that $ab<0$ and $cd<0$ simultaneously is impossible. So, assume in the remainder $n:=ab>0$ .

We have $-(c+d) = a+b\Rightarrow a^2+b^2+2ab-c^2-d^2-2cd=0$ . Adding this to the second, we get $a^2+b^2+ab-cd=6$ . Using $cd=-3$ , we find
$m^2 - n + \frac{3}{n} = 6\Leftrightarrow m^2 n - n^2 + 3 =6n$ where $m=a+b$ and $n=ab>0$ . Using the fact $m^2\ge 4n$ and $n>0$ , we thus obtain $6n = m^2n-n^2+3\ge 3n^2+3$ , i.e., $3(n-1)^2\le 0$ . So, $n=ab=1$ and $m^2 = (a+b)^2 =4$ . The rest is routine casework, and omitted.

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#6 h Apr 5, 2025, 1:26 AM • 1 Y

Y by teomihai

Using elementary symmetric polynomials, the system is $\begin{cases}e_1=0\\e_1^2-2e_2=12\\e_4=-3\end{cases}$ the solution set to this is $(e_1,e_2,e_3,e_4)=(0,-6,r,-3)$ for $r\in\mathbb{R}$ and so $(a,b,c,d)$ consists of the roots of $x^4-6x^2-rx-3=0$ .

Write this as $(x^2-3)^2=rx+12$

At $x=t$ we have the tangent line to LHS as $y=4t(t^2-3)(x-t)+(t^2-3)^2$ and so we want $-4t^2(t^2-3)+(t^2-3)^2=12$ .

$12=-4t^2(t^2-3)+(t^2-3)^2\implies 3t^4-6t^2+3=0\implies 3(t-1)^2(t+1)^2=0\implies t=\pm 1\implies r=4t(t^2-3)=\mp 8$

But note that for the points of inflection of LHS, $4(x^2-3)+8x^2=0\implies 12x^2-12=0\implies x=\pm 1$ .

So the only way to have $(a,b,c,d)\in\mathbb{R}^4$ is with $r=\pm 8$ where three of the variables are equal.

$x^4-6x^2\pm 8 x-3=0\implies(x\pm 3)(x\mp 1)^3=0$ .

So the solution set is all cycles of the tuples $(3,-1,-1,-1)$ and $(-3,1,1,1)$ .

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#7 h Apr 7, 2025, 10:26 AM

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I found this surprisingly tricky for P1.
As above, use Vieta to reduce to roots of $x^4-6x^2-rx-3=0$ , i.e. $x^3-6x-\frac{3}{x}=r$ .
However, the function $x^3-6x-\frac{3}{x}$ is monotonic on each of $x<0$ and $x>0$ , as can be easily checked using the derivative.
Hence for any $r$ , there will be at most two solutions.
Clearly $a=b=c=d$ is impossible, as well as $a=b, c=d$ .
Hence we are left with the case $b=c=d, a=-3d$ from where we easily get $d=\pm 1$ . Done.

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kokcio

#8 h Apr 7, 2025, 11:20 AM • 1 Y

Y by elfi

By multiplying by $-1$ if necessary, we can assume that $3$ of this numbers are positive.
WLOG $a,b,c>0$
From first equation we have $a+b=-(c+d)$ , so $a^2+2ab+b^2=c^2+2cd+d^2$ , but from last equation we know that $cd=\frac{-3}{ab}$ , so $c^2+d^2=a^2+b^2+2ab+\frac{6}{ab}$ . We put this in second equation and divide by $2$ to get $a^2+b^2+ab+\frac{3}{ab}=6$ , but we from AM-GM inequality we know that $ab+\frac{1}{ab}\geq 2$ and $a^2+b^2+\frac{1}{ab}+\frac{1}{ab}\geq 4$ , so $6=a^2+b^2+ab+\frac{3}{ab}\geq 6$ . Therefore, we have to have equalities in our inequalities, so $a=b$ . Similarly $a=c$ . Now we just have $3a+d=0$ , so $12=3a^2+d^2=12a^2$ , so $a=1$ , because $a>1$ , and $d=3$ .
This means that only solutions are $\{a,b,c,d\}=\{1,1,1,-3\}$ and $\{a,b,c,d\}=\{-1,-1,-1,3\}$ .

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Avron

#9 h Apr 7, 2025, 3:41 PM

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First note that if $(a,b,c,d)$ is a solution then so is $(-a,-b,-c,-d)$ and from the third equation we know that either $1$ or $3$ of the numbers are negative so WLOG assume $d<0$ and $a,b,c>0$
Substituting $d=-(a+b+c)$ we get:
$a^2+b^2+c^2+ab+bc+ca=6 \text{ and } abc(a+b+c)=3$ If $abc\geq1$ then by AM-GM $a+b+c\geq 3$ so $3=abc(a+b+c)\geq 3$ so equality holds everywhere and $a=b=c=1, d=-3$ . Now if $abc<1$ then $a+b+c=\frac{3}{abc}>3$ , but from the first equation $6=a^2+b^2+c^2+ab+bc+ca=(a+b+c)^2-(ab+bc+ca)>9-(ab+bc+ca)$ so $ab+bc+ca>3$ and again from the second equation we get $a^2+b^2+c^2=6-ab-bc-ca<3$ so $a^2+b^2+c^2<ab+bc+ca$ which is a well known contradiction, so the only slutions are $(1,1,1,-3), (-1,-1,-1,3)$ and its permutations.

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