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