Find all quadruplets $(a, b, c, d)$ of real numbers satisfying the system $(a + b)(a^2 + b^2) = (c + d)(c^2 + d^2)$ $(a + c)(a^2 + c^2) = (b + d)(b^2 + d^2)$ $(a + d)(a^2 + d^2) = (b + c)(b^2 + c^2)$ (Slovakia)
Problem
Source: Czech-Polish-Slovak Match 2016,P1,day 2
Tags: algebra
detrin
01.11.2016 11:41
Any idea?
RagvaloD
01.11.2016 13:52
Case 1: $b=c$ $a^3+a^2b+ab^2=d^3+d^2b+db^2$ $(a-d)(a^2+d^2+ad+ab+db+b^2)=0$ $a=d$ or $(a+d)^2+(b+d)^2+(a+b)^2=0 \to a+d=b+d=a+b=0 \to a=b=d=0$ then from third $4a^3=4b^3 \to a=b=c=d$ Case 2: $b\neq c$ $(a+b)(a^2+b^2)-(a+c)(a^2+c^2)=(c+d)(c^2+d^2)-(b+d)(b^2+d^2)$ $a^2b+ab^2+b^3-a^2c-ac^2-c^3=d^2c+dc^2+c^3-d^2b-db^2-b^3$ $(b-c)(a^2+b^2+c^2+ab+ac+bc)=(c-b)(d^2+b^2+c^2+bd+cd+bc)$ $a^2+b^2+c^2+ab+ac+bc+d^2+b^2+c^2+bd+cd+bc=0$ $(a+b)^2+(a+c)^2+(b+c)^2+(b+d)^2+(b+c)^2+(c+d)^2=0$ $a+b=a+c=b+c=a+d=0 \to a=b=c=d=0$ Answer: $a=b=c=d$
MATH1945
28.09.2017 11:01
Obviously $a=b=c=d$ works.
Assume $a \geq b \geq c \geq d$. Define $f(x,y)=(x+y)(x^2+y^2)$.
If $(a,b,c,d)$ is a solution, then so is $(-a,-b,-c,-d)$.
Now, if all of the are positive, we have $f(a,b) \geq f(c,d)$, equality at $a=b=c=d$.
If $a,b,c$ are positive, and $d$ is negative, we divide into 2 cases. If $d \geq -c$, then $a^2+b^2 \geq c^2+d^2$ and $a+b \geq c+d$,hence $f(a,b) \geq f(c,d)$, contradiction. If $d=-c$ then either $a=b=0$ or $a=-b$, but $a,b$ are both positive, hence it's impposible. If $a=b=0$, then $d^3=c^3=-d^3$ so $d=c=a=b=0$. If $d \leq -c$ then $a+b \geq 0 \geq c+d$, imposibble
If $a,b$ are positive and $c,d$ are negtive, then $a+b \geq 0 \geq c+d$, impossible
If only one or zero are positive,the negative of $a,b,c,d$ is a solution, and 3 or 4 of them are positive, impossible by above
Hence, $a=b=c=d$