# Do these polynomials exist?

1. Aug 5, 2008

### jonas.hall

Are there any ploynomials p(x) such that

p(x)^2 -1 = p(x^2+1) for all x?

To cut it short: With CAS software I have verified that there are no solutions except p(x) = 1.618... and p(x) = -0.618... (constants) up to order 53 or so, but I have to prove this (or find the other solutions).

For example: If I assume p(x) is fourth order p(x) = c4x^4 + c3x^3 + ... + c0 and put this into the above relation I get two new polynomials that must be equal for all x, so I identify terms and put equal terms equal to get an equation system like the following:

8: c4^2 = c4
7: 2c4c3 = 0
6: c3^2 + 2c4c2 = 4c4 + c3
5: 2c4c1 + 2c3c1 = 0
4: c2^2 + 2c4c0 + 2c3c1 = 6c4 + 3c3 + c2
3: 2c3c0 + 2c2c1 = 0
2: c1^2 + 2c2c0 = 4c4 + 3c3 + 2c2 + c1
1: 2c1c0 = 0
0: c0^2 - 1 = c4 + c3 + c2 + c1 + c0

(Notice how the right hand coefficients are elements in Pascals triangle?)

Now to solve this we start from the top. The only solution is c4=1 och c4=0 but this would not make a 4th degree polynomial and so we disallow this since we presumably already has handled this case. So we set c4 = 0. Then the following 4 equations are such that they allow for the solution of one coefficient at a time. Thus c3 = 0, c2 = 2, c1 = 0 and c0 = 2.

All odd coefficients can be proven to be = 0.

Then we are left with equations 3 down to 0 that also have to be satisfied to produce a valid solution. Surprisingly, everyone but the last sometimes are. If the order of the polynomial is 2^n (2, 4, 8, 16...) then they are. Why? No idea!

The last equation is a special case however. It can be written as

c0^2 - 1 = c0 + S where S is the sum of all coefficients except for c0.

Now all c from c4 and down to c1 are obviously rational. Unfortunately the last equation DO have rational solutions for some (infinitly many) values of S so I can't prove there are no more solutions this way.

I have tried looking at the roots of the polynomial but this doesn't seem to yield much at all. I have various semi-interesting results like
p(0) = c0
p(1) = p(0)^2 - 1 = c0^2 - 1
p(2) = p(1)^2 - 1 = c0^4 - 2c0^2
p(5) = p(2)^2 - 1 = c0^8 - 4c0^6 + 4c0^4 -1
etc (and yes, the 5 is correct)

...but none of this seems to bring me closer to closing the case. I might also add that the numerical discrepancy between the rational value of c0 given by the first equations and the irrational value given by the last equation diminishes steadily for all n-th order polynomials where n = 2^j (j integer) as n increases so it is by no means obvious that a 64th or perhaps a 256th order polynomial will fail.

So I guess I need some help to prove that no solutions exist, alternatively find the rest of them, wherever they may lurk.

Anyone fancy giving some help?

2. Aug 5, 2008

### gel

I expect that there is no solution, but don't immediately see an easy proof.

Set f(x)=x^2+1 and g(x)=x^2-1, so your equation reads g(p(x))=p(f(x)). You can iterate this to get $g^np(x)=pf^n(x)$.

So, the sequence f^n(x) goes to infinity if and only if g^n(p(x)) goes to infinity, and p gives a map from the Julia set for c=1 onto the julia set for c=-1. I'd have to search about on google, but I expect that those two Julia sets have very different properties and can't be related by a d to 1 map, such as a polynomial.

3. Aug 5, 2008

### MrJB

This is a sketch of how I think you can prove that the only possible p(x) are the two constant solutions. There are some important details missing at the end.

Let c be one of the roots of x^2 - x + 1 = 0.
Let's see what happens at x=c in p(x)^2-1 = p(x^2+1).

p(c)^2 -1 = p(c^2+1), but c^2+1=c.
p(c)^2 -1 = p(c)

Now solve for p(c)
p(c) = 1/2 ± Sqrt[5]/2

The value of p(c) is consistent with the constant solutions you've already found. But are those the only solutions? All I've shown is that at point c the value of p(c) must be one of 1/2 ± Sqrt[5]/2.

Let's take the derivative of each side of the original equation.
d/dx (p(x)^2 -1) = d/dx p(x^2+1)
2p(x)p'(x) = 2x p'(x^2+1)

Let's see what happens to p'(x) at x=c.
2p(c) p'(c) = 2c p'(c^2+1)
p(c) p'(c) = c p'(c)
But p(c)!=c, so p'(c)=0.
Which is also consistent with your constant solutions.

In order to prove that the only possible p(x) are the constant solutions, you would have to show that then nth derivative at p(c) is zero for all n>1.
If you know that the first to (n-1)th derivatives are zero at c, I believe that you can show that the nth derivative at c is zero. I'm not sure how to generalize the steps to show this.

4. Aug 7, 2008

### jonas.hall

Thanks for both answers - they gave me clues I can work on but unfortunately MrJB has done some mistakes. The equation x^2 + 1 = x has no real solutions so a large part of the reasoning fails. If one evaluate the derivatives at x=0 however some progress can be made but every even derivative is = 0 only if another one is = 0 at x=1 so for instance D^4(p) = 0 at x=0 only if D^2(p) = 0 at x=1. Annoying. As for the julia sets I'll have to read up on them but thanks for the tips.

Any other suggestions?

5. Aug 7, 2008

### Ben Niehoff

Have you tried writing

$$p(x) = \sum_{k=0}^N a_kx^k$$

and then putting this into your equation? You will have to find a new series that is the square of this series, but that's doable. Then simply compare the coefficients, and you ought to be able to show by induction that $a_k = 0$ for all k > 0.

6. Aug 8, 2008

### jonas.hall

I can try.... but I'm not sure I'm up to showing by induction something I can't show i the special cases. Though maybe you're saying: I have started by assuming the highest order term = 1, instead I should start by the lowest term? I'll think about it anyway.