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Real Polynomial Puzzle (Posted on 2015-06-27) Difficulty: 3 of 5
Given that x is a real number, determine all polynomials with real coefficients satisfying this relationship:
P(x2) = P(x)*P(x+2)

No Solution Yet Submitted by K Sengupta    
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Many, if not all. Comment 1 of 1
Among constant functions P(x)=0 and P(x)=1 are clearly the only possibilities.

Among linear functions P(x)=ax+b
a little algebra shows a=1, b=-1 is the only possibility.
So P(x)=x-1.

Among quadratics P(x)=ax²+bx+c
a lot of algebra shows a=1, b=-2, c=1 is the only possibility.
So P(x)=x²-2x+1=(x-1)²

From which appears a nice generalization:
P(x)=(x-1)^n for all non-negative integers n.

Proof:
(x^2 -1)^n = (x-1)^n*(x+1)^n

The bold above are the solutions.

I did not check for other higher order polynomials.  It seems these are the only ways to make the zeros to work out right.



  Posted by Jer on 2015-06-27 16:40:37
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