How do you solve this?

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. $x,y > 0$ 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.

(not a complete solution)

Most people would assume that $x = y$ , and then expand it to get:

$x (x+1) ( x^2 - x + 1 ) (x^2 + x + 1) = 0$

Since the values are real, we have $x = 0$ or $x = -1$ , which gives us two solutions only.

So, why must we have $x = y$ ?

Hint: If the value of $x$ is known, is the value of $y$ unique?

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Definition: For a function $f: \mathbb{R} \rightarrow \mathbb{R}$ , it is strictly increasing if for any pair of real number $a < b$ , we have $f(a) < f(b)$ .

Theorem: If $f$ and $g$ are a strictly increasing function, and $x$ and $y$ are real numbers such that $f(x) = g(y)$ and $g(x) = f(y)$ , then we must have $x = y$ .

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

Application: It is obvious that $g(x) = 1 + x^7$ is a strictly increasing function. It takes some work to show that $f(x) = (1+x)(1+x^2)(1+x^4)$ is also a strictly increasing function. (The easiest way is to appeal to the graph. Most of sharky's proof is to establish this fact.). Hence, in the solution set for the problem, we must have $x = y$ .

Note: We have the following chain of implications for (everywhere) differentiable functions:

$f'> 0 \Rightarrow f \text{ is strictly increasing } \Rightarrow f' \geq 0.$

The arrows do not work the other way. I believe (might be wrong) that

$f \text { is strictly increasing } \Leftrightarrow \forall x, f'(x) \text{ or } f'' (x) \text{ is non-zero}$

Right side odd function and left aide increasing function both will cut at 2 diff points

Since the variables in the equations are interchangeable, $x=y$ . Then, the equation to solve is $(1+x)(1+x^2)(1+x^4)=1+x^7$

After expanding, simplifying and factoring, we get $x(x-1)(x^4+x^2+1)=0$ , the real solutions of which are $0$ and $-1$ .

Therefore, the ordered pairs of real numbers are $(0,0)$ and $(-1,-1)$ .

Without loss of generality we assume x>y .But with this assumption we get a contradiction so x=y,is the only case to be considered.By putting x=y we get x=0,-1.

Proof of $x = y$

$\begin{cases} 1+x^{1} + x^{2} + x^{3} + x^{4} + x^{5} + x^{6} + x^{7} & = & 1+y^{7} & (1)\\ 1+y^{1} + y^{2} + y^{3} + y^{4} + y^{5} + y^{6} + y^{7} & = & 1+x^{7} & (2)\\ \end{cases}$

Multiply (1), (2) by $x-1$ and $y-1$

$\begin{cases} x^{8}-1 & = & (y^{7}+1)(x-1) & (1)'\\ y^{8}-1 & = & (x^{7}+1)(y-1) & (2)'\\ \end{cases}$

Expand and factor

$\begin{cases} (x-1)(x^7-y^7) & = & 0 & (1)''\\ (y-1)(y^7-x^7) & = & 0 & (2)''\\ \end{cases}$

From (1)'' and (2)'', we get $(x=1 \text{ or } x=y)$ AND $(y=1 \text{ or } y=x)$

Suppose, WLOG, that $x = 1$ , we get $y=x=1$ , which doesn't satisfy the equation. (that's possible to have fake answers because we multiplied by $x-1$ and $y-1$ .)

Therefore, $x=y$ .

Substitute in (1) and solve

$x^6 + x^5 + x^4 + x^3 + x^2 + x = 0$ .

Which gives $x(x+1)(x^{2}-x+1) (x^{2}+x+1) = 0$ .

That gives $\boxed{2}$ real roots and $4$ complex roots. ~~~

Extra: Answers (including complex) are $\displaystyle x=y= 0,-1,\frac{1\pm \sqrt{3}i}{2},\frac{-1\pm \sqrt{3}i}{2}$ .