Quartic polynomial taking infinitely many square rational values? I was wondering whether the value of
$$P(x)=x^4-6x^3+9x^2-3x,$$
is a rational square for infinitely many rational values of $x$. Is there a general method to check this for a polynomial (in one variable)? If not, how would one go about figuring this out?
 A: If your quartic polynomial to be made a square,
$$F(x) = z^2$$
has an initial rational point, then there is a birational transformation that can transform this to an elliptic curve's Weierstrass normal form. However, if you are in a rush to learn a simple method (known back to Fermat), then one way is this: Using any non-zero initial solution $x_0$, do the transformation,
$$x = y+x_0\tag1$$
For your curve, we have $x_0 = 1$ and I get,
$$F(y) = y^4-2y^3-3y^2+y+1$$
Assume it to be equal a square,
$$y^4-2y^3-3y^2+y+1 = (ay^2+by+c)^2\tag2$$
Expand, then collect powers of $y$ to get the form,
$$p_4y^4+p_3y^3+p_2y^2+p_1y+p_0 = 0\tag3$$
where the $p_i$ are polynomials in $a,b,c$. Then solve the system of three equations $p_2 = p_1 = p_0 = 0$ using the three unknowns $a,b,c$ to eliminate the $y^2, y^1, y^0$ terms. One ends up with,
$$105/64y^4+3/8y^3=0$$
Thus, $y =-8/35$ or,
$$x = y+x_0 = -8/35+1 = 27/35$$
and you have a new rational point $x_1 = 27/35$. Use this on $(1)$,
$$x = y+27/35$$
and repeat the procedure. Then,
$$x_2 =  3\times 4286835^2/ 37065988023371$$
Repeating it ad infinitum, one finds an infinite number of rational points $x_i$.
