Prove $x^4-18x^2+36x-27$ can never be a square rational (excluding 0), when x is rational
I have tried to use modulus, but didn't get anywhere, any help would be greatly appreciated.
Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational
diophantine equationselementary-number-theorynumber theory
Related Solutions
Assume on the contrary that $4mn-m-n=k^2$. Then $(4m-1)(4n-1)=(2k)^2+1$. Since $4m-1 \equiv 3 \pmod{4}$ and $4m-1>0$, $4m-1$ has a prime factor $p \equiv 3 \pmod{4}$. Then $p \mid (2k)^2+1$, which gives a contradiction since $(\frac{-1}{p})=(-1)^{\frac{p-1}{2}}=-1$.
First, note that $n^2$ can't be one of the elements as the other element would also need to be a perfect square for the product to be a perfect square. As gnasher729 commented to the answer, this is why $\left(n + 1\right)^2$ is not included.
Assume there are $2$ such elements, $n^2 + a$ and $n^2 + b$, with $a \neq b$, $1 \le a, b \le 2n$ and, WLOG, $a \lt b$. Thus, consider their product to be a perfect square of $n^2 + c$, i.e.,
$$\left(n^2 + a\right)\left(n^2 + b\right) = \left(n^2 + c\right)^2 \tag{1}\label{eq1}$$
for some integer $a \lt c \lt b$. As such, for some positive integers $d$ and $e$, we have that
$$a = c - d \tag{2}\label{eq2}$$ $$b = c + e \tag{3}\label{eq3}$$
Substitute \eqref{eq2} and \eqref{eq3} into \eqref{eq1} to get
$$\left(n^2 + \left(c - d\right)\right)\left(n^2 + \left(c + e\right)\right) = \left(n^2 + c\right)^2 \tag{4}\label{eq4}$$
Expanding both sides gives
$$n^4 + 2cn^2 + \left(e - d\right)n^2 + c^2 + c\left(e - d\right) - ed = n^4 + 2cn^2 + c^2 \tag{5}\label{eq5}$$
Removing the common terms on both sides and moving the remaining terms, apart from the $n^2$ one, to the right gives
$$\left(e - d\right)n^2 = -c\left(e - d\right) + ed \tag{6}\label{eq6}$$
Note that $e \le d$ doesn't work because the LHS becomes non-positive but the RHS becomes positive. Thus, consider $e \gt d$, i.e., let
$$e = d + m, \text{ where } m \ge 1 \tag{7}\label{eq7}$$
Using \eqref{eq3} - \eqref{eq2}, this gives
$$b - a = e + d \lt 2n \Rightarrow 2d + m \lt 2n \Rightarrow d \lt n - \frac{m}{2} \tag{8}\label{eq8}$$
Also,
$$ed = \left(d + m\right)d \lt \left(n + \frac{m}{2}\right)\left(n - \frac{m}{2}\right) = n^2 - \frac{m^2}{4} \lt n^2 \tag{9}\label{eq9}$$
Since $e \gt d$ means that $-c\left(e - d\right) \lt 0$, the RHS of \eqref{eq6} cannot be a positive integral multiple of $n^2$, so it can't be equal to the LHS.
Best Answer
The equation $y^2=x^4-18x^2+36x-27$ is birationally equivalent to the elliptic curve $w^2=z^3-432$, with $x=\frac{w-36}{2(z-12)}$. This curve has rank $0$ and torsion group $\mathbb Z/3\mathbb Z$, hence only two rational points $(z,w)=(12,\pm36)$. If we substitute these points into the formula for $x$, the positive $w$ gives the excluded $x=3$ solution and the negative $w$ incurs a division by zero. Hence $x^4-18x^2+36x-27$ can never be a nonzero rational square for rational $x$.