Algebra question to find all ordered pairs to a set of equations Find all ordered pairs of real numbers ($x,y$) for which
$$(1+x)(1+x^{2})(1+x^{4})=1+y^{7}$$
$$(1+y)(1+y^{2})(1+y^{4})=1+x^{7}$$
I am not sure on how to solve this. There is a clear symmetry in the problem. Also on expanding the LHS, we will get the GP
$$1+x+x^2+x^3+x^4+x^5+x^6+x^7=1+y^7$$ and similarly for the second equation. I tried manipulating the two equations and factorise them but I was unsuccesful. Any help would be much appreciated. Thanks a lot

Observe that

*

*If $ x = y$, then we have the solutions $ (0, 0), (-1, -1)$.

*We will show that there are no solutions with $ x \neq y$.

*If $ x = 0$, then from the first equation $ y = 0$.

*If $ x > 0, y < 0 $ then $ 1 + x + x^2 + \ldots + x^7 > 1 > 1 + y^7 $, so no solutions. Likewise for $ x < 0, y > 0$.

*If $x, y > 0$ then notice for for real $x$, $ (1+x)(1+x^2)(1+x^4) > 1 + x^7$. Hence, we have $(1+x)(1+x^2)(1+x^4) = 1 + y^7 <  (1+y)(1+y^2)(1+y^4 ) = 1 + x^7 $ which is a contradiction.

*If $ x , y  < 0$, then multiplying the first equation by $ 1 - x \neq 0$ and the second by $ 1 - y \neq 0$, and subtracting the two, we get

$$ y^ 8 - x^8 = (y - x) + (y^ 7 - x^7) + xy (x^6 - y^6)$$
$\qquad$ If $x < y < 0 $, then the LHS is negative, but each term on the RHS is positive. Hence contradiction.   Likewise if $ y < x < 0 $.

Note: A very common misconception when solving symmetric system of equations is thinking that all solutions to $x = f(y), y = f(x)$ are of the form $ y = x$.
In fact, what we truly require is $ f(f(x)) = x, y = f(x)$.
This is most obvious when looking at self-inverse functions like $f(x) = x, f(x) = -x, f(x) = \frac{1}{x}$. However, it's not about the function, but about the "self-inverse" point, namely $f(x) = f^{-1} (x)$.
