# Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational

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.

• Square integer? Square rational? Apr 15, 2020 at 19:53
• Sorry, square rational Apr 15, 2020 at 19:55
• I tried bounding it between consecutive squares as well, but that did't work Apr 15, 2020 at 19:56
• Try factoring the expression, that may help!
– mpnm
Apr 15, 2020 at 19:57
• I did factorise it into $\left(x-3\right)\left(x^3+3x^2-9x+9\right)$ Apr 15, 2020 at 19:58

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 $$x$$.

Say $$x^4-18*x^2+36*x-27 = n^2$$

The polynomial factors up

$$(x-3)*(x^3+3*x^2-9*x+9) = n^2$$

It means that if $$x-3$$ is a perfect square, then $$x^3+3*x^2-9*x+9$$ is also a perfect square or both aren't perfect square

Just like $$36=9×4=12×3$$

If $$x-3$$ is not a perfect square, then polynomial $$x^3+3*x^2-9*x+9$$ is not also a perfect, therefore it must factor up further, Just like $$36=12×3=4×3×3$$

But $$x^3+3*x^2-9*x+9$$ does not factor

It doesn't factor as supposed, it has to factor into $$(x-3)*(x-a)^2$$ in my definition, but there's no value of $$a$$ to fix this therefore $$x^3+3*x^2-9*x+9$$ does not factor to $$(x-3)*(x-a)^2$$

Say $$x-3 = k^2$$, $$x = 3+k^2$$

$$(3+k^2)^3+3*(3+k^2)^2-9*(3+k^2)+9 = (n/k)^2$$

$$k^6+12*k^4+36*k^2+36 = (n/k)^2$$

$$k^2*(k^2+6)^2+36 = (n/k)^2$$

$$(k\cdot(k^2+6))^2 + 6^2 = (n/k)^2$$

It's easy to see that no integer $$k$$ exist to that make this a perfect square

Therefore $$x^4-18*x^2+36*x-27$$ is not a perfect square

• The fact that the polynomial $x^3+3x^2-9x+9$ doesn't factor does not imply that its values cannot individually factor. Apr 15, 2020 at 21:46
• I do not follow the argument from the penultimate to the final line, and I also agree that just because $x^3+3x^2-9x+9$ does not factorise doesn't mean that its values cannot factor individually. Apr 15, 2020 at 22:02
• It doesn't factor as supposed, it has to factor into $(x-3)*(x-a)^2$ in my definition, but there's no value of $a$ to fix this therefore $x^3+3*x^2-9*x+9$ does not factor to $(x-3)*(x-a)^2$ Apr 15, 2020 at 22:02
• I dont think the logic is correct, a polynomial does not have to factor to a polynomial^2, in order to have square solutions (ie x^3 +1), which is what I think your proof is implying. Apr 15, 2020 at 22:22
• Am not saying it must factor to a polynomial^2, I'm just saying it factors $x^3+1 = (x+1)*(x^2-x+1)$, knowing that it factors gives me a definition I'll use against it.... try re-reading my proof Apr 15, 2020 at 23:29