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?