An algebra problem by Alex Harman

$\begin{aligned} \frac {x^2-18x-18}9 & = \frac {144=14x^2}{x^2} \\ x^4-18x^3-18x^2 & = 1296 - 126x^2 \\ x^4 - 18x^3 + 108x^2 & = 1296 \quad \quad \small \color{#3D99F6}{\text{Divide both sides by }x^2.} \\ x^2 - 18x + 108 & = \frac {1296}{x^2} \end{aligned}$

Since $x$ is an positive integer, the LHS is integral, therefore, the RHS must also be integral. This means that $x^2|1296 = x^2|2^43^4$ , implying that $x$ is a multiple of 2 and/or 3, since it is obvious that $x \ne 1$ . And when:

$\begin{aligned} x = 2: \quad 2^2 - 18(2) + 108 & = \frac {1296}{2^2} \\ 76 & \ne 324 \end{aligned}$

$\begin{aligned} x = 3: \quad 63 & \ne 144 \\ x = 6: \quad 36 & = 36 \end{aligned}$

$\implies x = \boxed{6}$

I'm not going to lie, It took me awhile to figure out a way to prove this after making the problem. It's a tricky one.
$\frac { { x }^{ 2 }-18x-18 }{ 9 } =\frac { 144-14x^{ 2 } }{ { x }^{ 2 } } \\ \frac { { x }^{ 2 } }{ 9 } -2x-2=\frac { 144 }{ { x }^{ 2 } } -14\\ \frac { { x }^{ 2 } }{ 9 } -\frac { 144 }{ { x }^{ 2 } } -(2x-12)=0\\ \frac { { x }^{ 4 }-1296 }{ 9{ x }^{ 2 } } -(2x-12)=0\\ \frac { \left( { x }^{ 2 }-36 \right) \left( { x }^{ 2 }+36 \right) }{ 9{ x }^{ 2 } } -2(x-6)=0\\ \frac { \left( x-6 \right) (x+6)\left( { x }^{ 2 }+36 \right) }{ 9{ x }^{ 2 } } -2\left( x-6 \right) =0\\ \frac { \left( { 6 }^{ 4 }-1296 \right) }{ 9\left( { 6 }^{ 2 } \right) } -2(6-6)=0$

Cross multiplying, x^4 - 18x^3 + 108x^2 - 1296 =0. Let x + 6t, getting: 1296t^4 - 3888t^3 + 3888t^2 - 1296 = 0. Dividing by 1296, t^4 - 3t^3 + 3t^2 - 1 =0,which has the obvious root t =1, so x = 6. Further division reveals this is a multiple root. Ed Gray