that's not a very hard problem but how do you now when you are done taking solutions?!

Err...whatdo you mean its not very har?

LOL YAY POLYNOMIAL FUNC EQS

ok let's see all nonzero coefficents have to be of even power-ed stuff

so with polynomials I think the idea is, after normal func eq solving stuff, to be like "lawl infinitely many roots implies 0-ness"

so let's see?

P(Q(x)) = Q(P(x)), P = x^2+1

so like... let's look at roots. Roots are fun. Ok so there is a root when Q(x) = which is... not fun.

OH YEAH ALSO DEGREES OF STUFF:

say you have like P(x) =

so now RHS is



LHS is

bleargh that has constant term

ok so now you know

that could be helpful

oh snap also so due to coeff of

blah now look at coeff of ? but wait I think it is too late so I am getting silly stuff... you can probably continue from here

I also bet you want polynomials with integer coeffs so you maybe shoudl say that

Hmm...I think the original problem had integer coefficients. 3 of the solutions are x, , and x is the only odd solution....

EDIT: I have solved it; define the sequence as and for those are the soluitons. So there are infinitelty many solutions.

i did not understand ONE THING THAT NOT_TRIG SAID but inverse functions. That was the stupidest advice you will ever be given

indeed umm inverse functions suck in this case because polynomials blah

Hmm...my proof was basicly using odd and even functions, proving the only odd solutions was f(x)=x, and induction.

Could someone post a link to this blog now that pythag has made it have no title?