Determine all integral solutions of $a^2+b^2+c^2=a^2b^2$ I started like this :
$a^2+c^2=b^2(a^2-1)\\c^2 +1=(a^2-1)(b^2-1)$
but it's leads to nowhere.
can you help please ?

This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is  
$a^2+b^2+c^2=t^2$ ,∀ t=ab
all integer positive solution given by
$a=(l^2+m^2−n^2)/n , b=2l, c=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then
$2l(l^2+m^2−n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2−n^2)(2l−1)=0$
suppose $2l−1=0$ refused
then
$l^2+m^2−n^2=0⇒a=0 $ ,and $ab=2n$
but $ab=0⇒n=0⇒l^2+m^2=0⇒b=0,c=0$
therefore the only solution is (0,0,0)
