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Topic: An Astute Polynomial Puzzle (Read 722 times) |
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K Sengupta
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An Astute Polynomial Puzzle
« on: Feb 16th, 2007, 12:09am » |
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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.
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| « Last Edit: Feb 16th, 2007, 12:11am by K Sengupta » |
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Eigenray
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Re: An Astute Polynomial Puzzle
« Reply #1 on: Feb 16th, 2007, 4:54pm » |
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| hidden: | 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 axm, 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 banxmn, 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. |
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