<s>[INST] Integer solution to equation Does $$b =a-23\sqrt 3 \sqrt a+432$$ have infinitely many integer solutions?
$a = 3$ gives one;
$a = 27$ gives one;
$a = 27\times 9$ gives one.
 [/INST] Let $a = 3n^2,\;$ where $n$ is any integer (and hence $\,a = 3n^2$ must also be an integer.)
So
$$
\begin{align} 
b &= a-23\sqrt{3}\sqrt{a}+432 \\ \\
&= a -23\sqrt{3a}+ 432 \\ \\
 &= (3n^2)-23\sqrt{3^2n^2}+432\\&=3n^2-23\cdot3n+432\\ \\
\iff b &= 3(n^2 - 23n + 144)
\end{align}
$$
Hence, for $a = 3n^2,\;b = 3(n^2 - 23n + 144)$ is an also an integer for any given any integer $n$. Since $n$ can be any integer, and since there are infinitely many $n$, there are infinitely many solutions $(a, b)$, one solution per $n$, given by $$a = 3n^2;\;\;b = 3(n^2 - 23n + 144).$$
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