Algebra question to find all ordered pairs to a set of equations

Question

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

Answer 1

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 $.

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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 $f(x) = f^{-1} (x)$.

Answer 2

Since we have powerful graphing calculators, let's see what Desmos tells us

OK, we have some clues, don't we? The first curve seems to lie above the line $y=x$ for $x<-1$ and for $x>0$; for $-1<x<0$ it lies below. But the second curve is the symmetric with respect to the line $y=x$.

Now let's prove the statement above rigorously. Let's try and see where $$ \sqrt[7]{x+x^2+x^3+x^5+x^5+x^6+x^7}>x $$ This becomes $$ x(1+x+x^2+x^3+x^4+x^5)>0 $$ and, factoring, $$ x(x+1)(x^4+x^2+1)>0 $$ hence this holds for $x<-1$ or $x>0$.

Thus, using the symmetry, the curves described by the two equations are always in the opposite half-planes defined by $y=x$. Hence they can only intersect on $y=x$.