$a^2=b^3+bc^4$ has no solutions in non-zero integers this problem is from number theory book ,
$$a^2=b^3+bc^4$$
has no  solutions in non-zero integers
This book hint ：First show that $b$ must be a perfect square.and how to do?

It is clear that $b\ge 0$. Suppose $b\gt 1$, and let $p$ be a prime that divides $b$. Let $p^k$ be the highest power of $p$ that divides $b$. 
There are $2$ cases. If $p$ does not divide $c$, then since $p^{3k}$ divides $b^3$, it follows that the highest power of $p$ that divides $a^2$ is $p^k$, so $k$ is even.
If $p$ divides $c$, then the highest power of $p$ that divides $bc^4$ is $k+4t$ for some $t$.  If $3k\ne k+4t$, then the highest power of $p$ that divides $a^2$ is $p^u$, where $u=\min(3k, k+4t)$. If $k$ is odd, then $u$ is odd, impossible. 
Finally, suppose $3k=k+4t$. Then $2k=4t$, so $k$ must be even.
