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?
-
4I believe there is. Another way of asking is whether the curve $y^2 = x^4 - 6x^3 + 9x^2 - 3x$ has infinitely many rational points, which suggests that a connection with elliptic curves is promising. – hardmath Aug 08 '15 at 00:03
1 Answers
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 $x_i$.
- 61,356
-
1@Servaes: Yes, it is possible that $x_0 \to x_0 \to \dots$. Or $x_0 \to x_1 \to x_0 \to \dots$. However, I seem to recall a result that if the procedure yields at least a certain $n$ number of distinct rational points $x_i$, then there is in fact an infinite number of distinct $x_i$. Anyway, the best way is to convert your curve to Weierstrass form and look at its rank. If the rank is zero, then it has a finite number of $x_i$. If it is greater than zero, then there is an infinite number of $x_i$, and you can use the procedure to churn out an infinite subset of them. – Tito Piezas III Aug 08 '15 at 12:46
-
@TitoPiezasIII: How did you go from $y^4-2y^3-3y^2+y+1 = (pq^2+qy+r)^2$ to $105/64y^4+3/8y^3 = 0$? – Sandeep Silwal Aug 08 '15 at 22:43
-
@SandeepSilwal: I edited the answer with more details. It's just a simple system of three equations in three unknowns. – Tito Piezas III Aug 09 '15 at 02:36
-