# High School Olympiad - System of Equations [closed]

For real numbers $$a,b,c$$ solve the following system of equations:

$$\begin{split} a(b^2 + c) = c(c+ab) \end{split}$$ $$\begin{split} b(c^2 +a) = a(a+bc) \end{split}$$ $$\begin{split} c(a^2 + b) = b(b+ac) \end{split}$$

There are many possibilities to solve this system of equations, but the recommended solutions include algebraic manipulations which may not come to my head during live solving. I would greatly appreciate your take on this problem.

Side question: how does one get $$abc = 1$$?

• Can you update your post with what you've tried and where you're stuck? You also mention $x,y,z$, but there are only $a,b,c$ – David P Mar 7 '20 at 9:24
• Worthy note: the only solution really is only $a=b=c$, is there any neat way to prove it? – Danjel Mar 7 '20 at 9:52

If $$a=0$$, we get immediately that $$a=b=c=0$$. Now assume that $$abc\neq 0$$. The first equation can be written as:

$$ab(b-c)=c(c-a)$$

Write the other equations similarly and multiply them:

$$a^2b^2c^2(a-b)(b-c)(c-a)=abc(a-b)(b-c)(c-a)$$

or $$abc(abc-1)(a-b)(b-c)(c-a)=0$$. $$abc\neq 0$$ and if $$a=b$$ this leads immediately to $$a=b=c$$. The only case that may lead to different solutions is $$abc=1$$. In this case the system is equivalent with:

$$\begin{cases}b-c=c^2(c-a)\\ c-a=a^2(a-b)\\ a-b=b^2(b-c)\end{cases}$$

Notice that this implies the $$a-b, b-c$$ and $$c-a$$ have the same sign. However since their sum is $$0$$, this is only possible if $$a-b=b-c=c-a=0$$ and thus $$a=b=c=1$$.

In conclusion the only solution is $$a=b=c$$.

Summing both sides

$$ab^2+bc^2+ca^2+ac+bc+ab=a^2+b^2+c^2+3abc$$

$$a^2(c-1)+b^2(a-1)+c^2(b-1)=ab(c-1)+bc(a-1)+ac(b-1)$$

This leads to one of the solutions

$$a=b=c$$

But there is a case where

$$a^2(c-1)\neq ab(c-1)$$

$$b^2(a-1)\neq bc(a-1)$$

$$c^2(b-1)\neq ac(b-1)$$

Now WLOG, Let $$a>b>c>0$$

Then taking the first equation

$$a(b^2+c)=c(c+ab)$$

Now, Comparing LHS terms with RHS terms

$$ab^2>abc ; b>c$$

$$ac>c^2 ; a>c$$

Since the LHS is strictly greater than the RHS, we conclude that

$$a\neq b\neq c$$

produces no solutions.

• I'm sorry, but how does one get $a=b=c$ from the equation? – Danjel Mar 7 '20 at 9:42
• Compare coefficients of $(a-1),(b-1),(c-1)$ – h-squared Mar 7 '20 at 9:42
• @h-squared No, it doesn't work that way. – Calvin Lin Mar 7 '20 at 9:49
• It is one of the solutions – h-squared Mar 7 '20 at 9:50
• @h-squared, you cannot let WLOG $a>b>c$ because the system is cyclic, not symmetric. You also have to discuss $a<b<c$. – LHF Mar 7 '20 at 10:09