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        riddles >> hard >> An Astute Polynomial Puzzle
        
(Message started by: K Sengupta on Feb 16^th, 2007, 12:09am)

Title: An Astute Polynomial Puzzle
Post by K Sengupta on Feb 16^th, 2007, 12:09am

Determine all real polynomials P(x) such that there exists one real polynomial Q(x) satisfying the conditions Q(0) = 0 and:
x + Q( y + P(x)) = y + Q( x + P(y)) for all real numbers x and y.

Title: Re: An Astute Polynomial Puzzle
Post by Eigenray on Feb 16^th, 2007, 4:54pm

[hideb]First, if P=0, then there's the unique solution Q(x)=x, and if Q=0, there is no solution.

So suppose P(x) has leading term ax^m, and Q(x) has leading term bxⁿ, n>0.

If m=0, i.e., P(x)=a is constant, then x+Q(y+a) has leading term (with respect to x) x, while y+Q(x+a) has leading term bxⁿ, so n=1 and Q(x)=x is the unique solution.

If m>1, then the lead term (with respect to x) of x+Q(y+P(x)) is baⁿx^mn, while the lead term of y+Q(x+P(y)) is bxⁿ, which is impossible since mn>n.

So suppose m=1, say P(x) = c + ax, and x+Q(y+c+ax) = y+Q(x+c+ay). Then we can't have a=1.

If n=1, so Q(x)=bx, then we have x+b(y+c+ax) = y+b(x+c+ay), which holds iff 1+ab=b, i.e., b=1/(1-a). So we have one solution in this case.

Now suppose n>1. By equality of leading terms, b(ax)ⁿ = bxⁿ, so aⁿ=1. Since a != 1, and a is real, we must have a=-1. But in this case, Q(x) = dx+bx² is a solution when

x + d(y+c-x) + b(y+c-x)² = y + d(x+c-y) + b(x+c-y)²,

which holds iff 1 - d - 2bc = d + 2bc, i.e., d=(1-4bc)/2. So there are infinitely many possible Q.

To summarize, there exists a Q for a given P iff P(x)=a is constant, or P(x)=c+ax, with a != 1. The Q will be unique iff P(x)=a or P(x)=c+ax, with a != +/-1.[/hideb]