Solving $a^5=a^3bc+b^2c$ in integers

Question

Solving $a^5=a^3bc+b^2c$ in integers.

I tried assuming there is a common divisor first of a,b,c, then a,b and 2 other pairs, but not sure how to arrive to a contradiction, trying some things right now

Answer 1

Hint:

$cb^2+a^3cb-a^5=0 \Rightarrow a^6c^2+4a^5c$ is a square number, so $a^2c^2+4ac$ should be square number.