<s>[INST] 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.
 [/INST] 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
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