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.
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
Rooting it out (spoiler)