Having two matrices:
$A = \left(\begin{matrix}36a & -170 - a \\ -2a - 144 & 26 \end{matrix}\right)$
$A^2 = \left(\begin{matrix}1298a^2 + 495a + 24480 & -36a^2 - 6146a - 4420 \\ -72a^2 - 5236a - 3744 & 2a^2 + 484a + 25156 \end{matrix}\right)$
I need to find $a$ so that $A = A^2$. I've tried $A^2 - A = 0$ but it doesn't seem to be right.
I was hoping someone here would be able to point me in the right direction.
HINT: $$A^2=\left[ \begin {array}{cc} 1296\,{a}^{2}+ \left( -170-a \right) \left( -2\,a-144 \right) &36\,a \left( -170-a \right) -4420-26\,a \\ 36\, \left( -2\,a-144 \right) a-52\,a-3744& \left( -170-a \right) \left( -2\,a-144 \right) +676\end {array} \right] $$