Find all the possible values of $(a,b,c,d)$.

Find all quadruples of real numbers $(a,b,c,d)$ satisfying the system of equations

$(b+c+d)^{2010}=3a$

$(a+c+d)^{2010}=3b$

$(a+b+d)^{2010}=3c$

$(a+b+c)^{2010}=3d$

I tried to find the solutions using hit and trial and by using some logic also I find two solutions which are $(0,0,0,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$

I think these are the only solutions that exist.

Solution 1:

All solutions must be non-negative because $2010$ is even.

Suppose $d \gt a$. This would lead to a contradiction comparing $(a+b+c)^{2010}$ and $(b+c+d)^{2010}$.

So all four must be equal and non-negative, say $x$ and you have $(3x)^{2010}=3x$, which only has the solutions $3x=0$ and $3x=1$, i.e. $$a=b=c=d=0$$ or $$a=d=c=d=\frac13$$

Solution 2:

If $(a,b,c,d)$ satisfies the equations, then we may well assume that $a\leq b\leq c\leq d$.These are non-negative because an even power of a real number is always non-negative. It follows that $$ b+c+d\geq a+c+d\geq a+b+d\geq a+b+c$$ since $x$ and hence $x^{2010}$ is increasing for $x\geq 0$, we have that $$3a=(b+c+d)^{2010}\geq (a+c+d)^{2010}\geq (a+b+d)^{2010}\geq (a+b+c)^{2010}=3d.$$Thus, we conclude that $a=b=c=d$ and all the equations take the form $(3a)^{2010}=3a$, so $a=0$ or $3a=1$. Finally, all solutions are $(0, 0, 0 ,0)$ and $(\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})$. Hope it helps.