# Thread: find all real x,y

1. ## find all real x,y

find all real x,y such that

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

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

2. The two equations are inverses of each other, so any points in either equation that have y=x will be solutions. Substituting this into one of the equations and expanding easily nets you 2 solutions.
However, this method will not necessarily get you all solutions so you still have to prove that there are no more. I am not sure how you should do this.

The two equations are inverses of each other, so any points in either equation that have y=x will be solutions. Substituting this into one of the equations and expanding easily nets you 2 solutions.
However, this method will not necessarily get you all solutions so you still have to prove that there are no more. I am not sure how you should do this.
Looking at the graph this should be easy to prove. Just show that, for x > 0 that the graph of $(1 + x)(1 + x^2)(1 + x^4) = 1 + y^7$ is greater than the graph of $(1 + y)(1 + y^2)(1 + y^4) = 1 + x^7$ and that the same holds true for x < -1. Thus they only intersect in the interval [-1, 0], where there are two intersections.

-Dan

4. Originally Posted by topsquark
Looking at the graph this should be easy to prove. Just show that, for x > 0 that the graph of $(1 + x)(1 + x^2)(1 + x^4) = 1 + y^7$ is greater than the graph of $(1 + y)(1 + y^2)(1 + y^4) = 1 + x^7$ and that the same holds true for x < -1. Thus they only intersect in the interval [-1, 0], where there are two intersections.

-Dan
Can we analytically find the solutions to this question? I mean doing some algebra and all....

perash where did you find this problem? some math contest?

5. If you multiply the LHS of the first equation by $(1-x)$, it becomes $(1-x^8)$. Hence

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

Since the equations are symmetrical in $x$ and $y$, we can guess that the solutions will be in the form $x = y$. Therefore

$-x^8\ =\ -x+x^7-x^8$

$\Rightarrow\ x(1-x^6)\ =\ 0$

$\Rightarrow\ x = 0\ \mbox{or }x=-1$

( $x=1$ is rejected; it came into the working because I multiplied both sides of the first equation by $(1-x)$.)

So the solutions are $(0,0)$ and $(-1,-1)$.

6. Originally Posted by Isomorphism
Can we analytically find the solutions to this question? I mean doing some algebra and all....

perash where did you find this problem? some math contest?
When I suggested my method of proving there were no other solutions I hadn't actually solved the problem, just graphed it. But solving the problem really isn't all that hard. badgerigar gave the key to it. Also, as I was posting this JaneBennet gave a perfectly acceptable solution as well, but didn't actually show that the only two real solutions are x = -1, 0.

When subbing in the y = x the problem turns into solving the 6th degree polynomial $x^6 + x^5 + x^4 + x^3 + x^2 + x = 0$. We automatically know that x = 0 is a solution, so we are left solving
$x^5 + x^4 + x^3 + x^2 + x + 1 = 0$

We may employ the rational root theorem to find that x = -1 is a solution, so by division we need to solve
$x^4 + x^2 + 1 = 0$

This is a "biquadratic equation." We may solve it by using the substitution $z = x^2$, whence
$z^2 + z + 1 = 0$

Using the quadratic formula I get that
$z = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$

As z is complex, so is its square root, so the remaining four solutions to the original problem are complex, not real. Thus there are only two real solutions.

-Dan

7. Yes, I had got all that.

But I do not understand the logic behind the solutions being of the form x=y. Should all symmetric equations have solutions only in that form??
But the ease with which everyone is taking x=y as the only solution for granted,i am led to ask:"Am I missing something very basic?"

8. Originally Posted by Isomorphism
Yes, I had got all that.

But I do not understand the logic behind the solutions being of the form x=y. Should all symmetric equations have solutions only in that form??
But the ease with which everyone is taking x=y as the only solution for granted,i am led to ask:"Am I missing something very basic?"
Given a relation $y = f(x)$ the inverse relation $y = f^{-1}(x)$ may be computed by switching the roles of x and y in the original relation, $x = f(y)$ and solving for y. So your two given relations are inverses of each other.

The role switching of x and y is known as a "symmetry transformation." This particular one represents a reflection over the line y = x. So by definition the inverse of a relation $y = f(x)$ is the graph of the relation reflected over the line y = x.

If you think about this for a moment you will see that the only points that a relation and its inverse can intersect at are for the values such that x and y are the same.

-Dan

Edit: By the way, this argument about a "relation and its inverses" is typically mentioned in the context of functions. But of course any relation (at least the continuous ones anyway) may be broken into pieces that are functions. We can apply this reflection argument to the individual pieces of the relation and get the same results.

9. Originally Posted by topsquark
If you think about this for a moment you will see that the only points that a relation and its inverse can intersect at are for the values such that x and y are the same.
By the way, this argument about a "relation and its inverses" is typically mentioned in the context of functions.
Actually, in the standard foundations course, typically a second year course, the topic relations is the introduction to the material on functions.

More importantly, any relation has an inverse: $R^{ - 1} = \left\{ {\left( {a,b} \right):\left( {b,a} \right) \in R} \right\}$. Now any function is a relation with some additional restrictions, one of is that no two pairs may have the same first term. Hence the requirement that a function must be one-to-one in order to have an inverse.

10. If you think about this for a moment you will see that the only points that a relation and its inverse can intersect at are for the values such that x and y are the same.
This isn't actually true. Although all values where x and y are the same will be points of intersection there can be others. One example is the function y=-x.

11. Originally Posted by topsquark
If you think about this for a moment you will see that the only points that a relation and its inverse can intersect at are for the values such that x and y are the same.
This is a surprisingly common misconception among both students and teachers.

This isn't actually true. Although all values where x and y are the same will be points of intersection there can be others. One example is the function y=-x.
You beat me to the punch, badgerigar!

Another simple but more interesting example is the relation

f: R --> (-oo, 1] where f(x) = -x^2 + 1.

Two of the four intersection points of f and its inverse lie on the line y = -x + 1.

It's easy to show that the points of intersection of a relation and its inverse always lie on one or more of the lines y = x and lines of the form y = -x + c, where c is a constant. Those not lying on the line y = x (obviously) always occur in pairs of the form (a, b) and (b, a).

When there are intersection points NOT lying on the line y = x, those lying on the line y = x can be used to find them and hence all the intersection points .... I can email an (edited) electronic copy (word) of a small conference paper I have presented entitled "The intersection of a relation and its inverse" to interested persons.

Another less simple and even more interesting case in point: I think we can all agree (after some thought) that there are more solutions to y^x = x^y than just the points lying on the line y = x ....... (2, 4) and (4, 2) for example (which btw are the only integer solutions not lying on the line y = x. The others can be found exactly - in parametric form.)

So the question of uniqueness raised by badgerigar and Isomorphism is I think a non-trivial one and is still open .....