how would you find the solutions for $x$ where the quadratic $-27x^2+234x+169$ is a perfect square

Question

how would you find the solutions for $x$ where the quadratic $-27x^2+234x+169$ is a perfect square of an integer and $x$ is positive integer

I know that $x$ is from $1$ to $11$ because anything greater than $11$ makes the polynomial negative ,but how would compute the correct $x$ values without testing each of the 11 values?

Answer 1

As the comments point out, it's not really any faster to use algebraic number theory to solve the problem.

On the other hand, you can consider

$$y^2=-27x^2 + 18px + p^2$$

where $p$ is a prime number (your question is the special case $p=13$). Then solving this equation can be reduced to finding $a$, $b$ such that

$$a^2+ab+b^2=p^2\text{.}$$

Finding solutions to this last equation depends on whether the remainder of $p$ when divided by $3$ is $0$, $1$, or $2$.

Answer 2

$$ y = -27x^2 + 234x + 169 $$

This is a parabola with a vertex (in this case, the highest point) at $x = -\frac{b}{2a}=-\frac{234}{2(-27)}=4\frac{1}{3}$ and $y=676$.

The roots are at $\dfrac{13}{9}(3 \pm 2\sqrt 3) \approx \{-0.67, 9.34\}$

So we only need to test $x = 0,1,2,\dots,9$

\begin{array}{r,|,c} x & y=-27 x^2 + 234 x + 169 \\ \hline 0 & \color{red}{169=13^2} \\ 1 & 376 \\ 2 & \color{red}{529 = 23^2} \\ 3 & 628 \\ 4 & 673 \\ 5 & 664 \\ 6 & 601 \\ 7 & \color{red}{484 = 22^2}\\ 8 & 313 \\ 9 & 88 \\ \hline \end{array}