Find a polynomial as $2836x^2-12724x+16129$ I found a polynomial function with integer coefficients:$f(x)=2836x^2-12724x+16129$
and $f(0)=127^2,f(1)=79^2,f(2)=45^2,f(3)=59^2,f(4)=103^2,f(5)=153^2.$
My question is:can we find a polynomial function with integer coefficients,called $f(x)$,which has no multiple roots,and $f(0),f(1),f(2),f(3),……,f(k)$ are distinct square numbers?($k>5$ is a given integer)
Thanks all.
PS:I'm sorry,guys.I lost a very important condition:$f(x)$ should be a quadratic function:$f(x)=ax^2+bx+c$.($a,b,c$ are integers and $b^2-4ac≠0$)
So the Lagrange interpolation method does not work.
I wonder is there always such a quadratic polynomial when $k$ is arbitrarily large? 
 A: One such quadratic 
$$p(t)=-4980t^2+32100t+2809$$
$p(0)=53^2,p(1)=173^2,p(2)=217^2,p(3)=233^2,p(4)=227^2,p(5)=197^2,p(6)=127^2$
Source : Polynomials E.J Barbeau
A: An interesting link is the following question:
Least power. Squares again
Quoting the answer given there by Ivan Loh:

In the paper http://www.mast.queensu.ca/~murty/poly2.pdf, it is proven that if a polynomial $P(x_1,x_2,…,x_n)\in \Bbb{Z}[x_1,x_2,…,x_n]$ is such that $P(n)$ is a perfect square for all choices of $x_1,x_2,…,x_n$ with $\lvert x_i \rvert \le c$, where $c$ is sufficiently large, then $P(x)$ must be the square of a polynomial.

See also A quadratic polynomial getting square values in consecutive points
already mentioned in a comment.
A: You can use Lagrange interpolation to find a polynomial with any (finite) set of values you want.
A: Some brute force search gives $289 + 2940 t - 420 t^2$,
which satisfies
$$f(0)=17^2, \quad f(1)=53^2, \quad f(2)=67^2, \quad f(3)=73^2, \quad f(4)=73^2, \quad f(5)=67^2, \quad f(6)=53^2, \quad f(7)=17^2.$$
This is the only solution I've found for $k=7$.  I'll update this post if/when I can best it.
