while I was solving linear equations for my math high school classes I found an interesting problem to solve.
Here is the equation
$a^2x+4=a\cdot(x+4)\\ a^2x+4 =ax+4a\\ a^2x-ax=4a-4\\a\cdot(a-1)\cdot x=4\cdot(a-1)$
CASE 1: Now we can see that we can divide by $a-1$ and then by $a$ to solve for $x$ and we can see that $a\ne0 \text{ and }\\a-1\ne0, a\ne1$
$x=\frac{4\cdot(a-1)}{a\cdot(a-1)}=\frac{4}{a}$
CASE 2: $a=0$
$0\cdot(0-1)\cdot x=4\cdot(0-1)\\0=-4$
We can see that for $a=0$ the equation has no solution in $\mathbb R$
CASE 3: $a=1$
$1\cdot(1-1)\cdot x = 4\cdot(1-1) \\ 0=0$
We can see that for $a=1$ the solution for the equation is every number in $\mathbb R$.