RMO 2015!

We shall use casework to determine the answer to this question. We must use the following cases:

1. $xy= 0$ . Then it is clear that $x = y = 0$ and $(x,y) = (0,0)$ is a solution.

2. $xy < 0$ . By the symmetry of the equations of the original question, we can assume that $x > 0 > y$ . Then $(1 + x)(1 + x^2)(1+x^4)>1$ and $1+y^7<1$ . There are no solutions in this case.

3. $xy > 0$ and $x \neq y$ . Once again, by the symmetry of the equations we can assume that $x > y > 0$ . Then

$(1+x)(1+x^2)(1+x^4) > 1 + x^7 > 1 + y^7$

showing that there are no solutions in this case.

4. $x,y < 0$ and $x \neq 0$ . By the symmetry, we can assume that $x < y < 0$ . Multiplying by $1 - x$ and $1 - y$ to the first and second equation, respectively, the system now reads

$1 - x^8 = (1 + y^7)(1 - x) = 1 - x + y^7 - xy^7$

$1 - y^8 = (1 + x^7)(1 - y) = 1 - y + x^7 -x^7y$ .

Subtracting the first equation from the second results in

$x^8 - y^8 = (x - y) + (x^7 - y^7) - xy(x^6 - y^6)$ .

Since $x < y < 0$ , $x^8 - y^8 > 0$ , $x - y < 0$ , $x^7 - y^7 < 0$ , $-xy < 0$ and $x^6 - y^6 > 0$ . Therefore the left hand side of the equation is positive whereas the right hand side is negative. This can not be possible, hence there are no solutions in this case.

5. $x = y$ . Then solving

$1 - x^8 = 1 - x + y^7 - xy^7 = 1 - x + x^7 - x^8$

which leads to $x = -1, 0, 1$ , which implies that $(x,y) = (0,0)$ or $(-1,-1)$ .

Therefore, $(x,y) = (0,0)$ or $(-1,-1)$ are the only solutions. Since there are 2 solutions, the answer is 2.

---

For a function $f:\mathbb{R} \rightarrow \mathbb{R}$ , it is strictly increasing if for any pair of real numbers $a > b$ , we have $f(a) > f(b)$ . If $f$ and $g$ are strictly increasing functions, and $x$ and $y$ are real numbers such that $f(x) = g(y)$ and $g(x)=f(y)$ , we have $x=y$ .

Proof : Suppose not. WLOG $x < y$ . Then we have $f(x) = g(y) > g(x)=f(y)$ , which contradicts $f(x) < f(y)$ . Therefore, $x=y$ .

This property can be applied in this question to reduce the casework. All that's needed to be done is to prove $(1+y^7)$ and $(1+x)(1+x^2)(1+x^4)$ are strictly increasing.

It is easy to observe that when $x=y$ , both the given equations become identical which is:

$(1+x)(1+x^2)(1+x^4) = 1+x^7 \\ \Rightarrow 1+x+x^2+x^3+x^4+x^5+x^6+x^7=1+x^7 \\ \Rightarrow x+x^2+x^3+x^4+x^5+x^6 = 0 \\ \Rightarrow \dfrac{x(x^6-1)}{x-1} = 0 \dots (a) \\ \Rightarrow x(x^6-1)=0$

So we have $x=0$ or $(x^6-1)=0$ .

When $x^6-1=0 \Rightarrow x^6=1 \Rightarrow x = 1 \ or \ -1$ . But when $x=1$ , the equation $(a)$ becomes undefined , thus $x=-1$ is a solution.

So the ordered pairs are : $(0,0) \ , \ (-1,-1)$ which are $\boxed{2}$ in number.