 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  All possible polynomials puzzle (Posted on 2015-05-25) Determine all possible polynomials P(x) having real coefficients that satisfy the equation:

(x-2010)*P(x+67) = x*P(x) for every integer x.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Rooting it out (spoiler) Comment 1 of 1
let x = 0
Then p(67) = 0

Let x = 67
Then p(134) = 0

Similarly, p(k*67) = 0
where k between 1 and 30

When k = 30, then 30*67 = 2010.

If we let x = 2010,
then no new info is available,
as 0*p(2077)= 0

But, P(x) has at least 30 roots.

So, P(x) is of the form R(x)*(x-67)*(x-134)*...*(x-1943)*(x-2010)

Then
x*P(x) = x*R(x)*(x-67)*(x-134)*...*(x-2010)

Also
(x-2010)p(x+67) = (x-2010)*p(x+67) =
(x-2010)*R(x+67)*x*(x-67)*(x-134)*...*(x-1943)

Setting these two equal gives R(x) = R(x+67) for all x.
Because R(X) is a polynomial, it can only be a constant.

So, P(x) = r*(x-67)*(x-134)*...*(x-1943)*(x-2010)
where r is any real number

 Posted by Steve Herman on 2015-05-25 13:39:51 Please log in:

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