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        riddles >> medium >> helpful polynimial
        
(Message started by: Mickey1 on Oct 18^th, 2012, 1:50pm)

Title: helpful polynimial
Post by Mickey1 on Oct 18^th, 2012, 1:50pm

I wonder (for reasons related to Pell’s equation ) whether for an arbitrary non-constant polynomial with integer coefficients , P1(n), there is (always?) another similar polynomial (non-constant polynomial with integer coefficients) P2(n) , so that P1(P2(n))=(P3(n))^2 for a third similar polynomial P3(n)

For instance P1(n)=n+1 and P2(n)=nn+2n would yield P1(P2(n))=nn+2n+1=(n+1)^2 (also =P1(n)^2). For n=0 we get P1(P2(0))=(P3(0))^2 = C^2 if C is the constant in P3(n).

Title: Re: helpful polynimial
Post by pex on Oct 18^th, 2012, 2:01pm

on 10/18/12 at 13:50:16, Mickey1 wrote:

No. Try P₁(n) = n² + 1.

Title: Re: helpful polynimial
Post by Mickey1 on Oct 20^th, 2012, 11:59am

Your comment is not quite clear to me. I need also your f2(n) and f3(n) to see the light!

My problem:

For D in x*x-Dy*y=1, I am looking at solutions for D around squares, D=n*n. For the number following a square, D= n*n+1, we have

(2n*n+1)^2-( n*n+1)*(2n)^2 =1

For the next higher D, i.e. D=n*n+2 we have

(n*n+1)^2-( n*n+2)*n*n =1

for D=n*n +3 the solution (assuming it exist) is less obvious. We have for n=7 and D= n*n+3=52 , y=90 and x=649

We can write y=90 as 2n*n-8 or as 2n*n-n-1
but Dy*y+1 does not give me a square for these expressions. But if there is one, I speculate that I can find it by a substitution into a first attempt version.

I believe that the corresponding real argument polynomial would yield the derivate

f1'(f2(x))=2*f3(x)*f3'(x) (isn't that so and can this perhaps be used in my investigation?)

There are also some follow up issues i) would such a method always work, i.e. starting with any non-constant polynomial will I always find the square polynomial (P3(n))^2? and ii) if a solution for x was found how would I find the lowest?

Title: Re: helpful polynimial
Post by pex on Oct 20^th, 2012, 2:35pm

on 10/20/12 at 11:59:26, Mickey1 wrote:

Your comment is not quite clear to me. I need also your f2(n) and f3(n) to see the light!

Sorry for not being clearer. My point was that I do not think P₂ and P₃ can be found in this case:
P₁(anything) = a square plus one = usually not a square.

Title: Re: helpful polynimial
Post by Mickey1 on Oct 21^st, 2012, 12:21pm

Thanks

I take this to mean that there is no on the shelf technique. You know that xx-Dyy= not usually a square either. I could still be lucky with nn+3. Then again it could be that the squares occurring allow general solutions to be available only for D= nn +-1 and +- 2 ?

Perhaps xxx-Dyyy=1 has similar properties for nnn+- 1 to 3. I tried to find examples on the internet but found only a reference to Hilbert's 10th problem. That may be a good or bad omen.

In any case

n^3-(n^3-1)*1^3=1