Reversing digits in an equation

Let the positive integer $n = \overline{ab}$ be the 2-digit number, we have that:

$\begin{cases} (\overline { ab } )^{ 2 }=\overline { cde } \\ (\overline { ba } )^{ 2 }=\overline { edc } \end{cases}$

When rewritten algebraically, it is:

$\begin{cases} (10a+b)^{ 2 }=100c + 10d + e \\ (10b+a )^{ 2 }= 100e+10d+c \end{cases}$

$\begin{cases} 100a^2 + 20ab + b^2 = 100c + 10d + e \\ 100b^2 + 20ab +a^2 = 100e+10d+c \end{cases}$

Switch all terms to the left side

$\begin{cases} 100(a^2 - c) + 10(2ab-d) + (b^2-e) = 0 \\ 100(b^2 - e) + 10(2ab - d)+(a^2-c) = 0 \end{cases}$

We can see a correspondence between two equations, that is:

$100(a^2 - c) = 100(b^2 - e)$ so therefore $a^2-c = b^2-e$ $(a-b)(a+b) = c-e$

Now, $a$ and $b$ are integers only if $a+b+a-b = 2a$ is equal to an even number. This is because when we solve for such system of equation:

$\begin{cases} a-b=... \\ a+b = ... \end{cases}$

Adding those two equations lend us $2a = ....$ and for $a$ ( therefore $b$ ) to be an integer, the number must be even.

So we go ahead and find all possible solutions such that for digits $c-e$ is an odd prime, or that when factored out ( exclusive of $1\times itself$ ) are expressible as a product of two even numbers.

There's also another fact: $n^2 \le 999 \Rightarrow n \le 31$ since we need $n^2$ to have 3-digits at max.

We can see that $(a,b) = (2,1)$ is the only solution i.e $\boxed{n = 12 , 21}$

So there are two more numbers that are reversible