$a^2=b^3+bc^4$ 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?
 A: Obviously, $(1,1,0)$ is a solution.
A: Did the book instead say that there was no solution in naturals?
Note that $a^2 = b(b^2+c^4)$. Then, $b$ must be a perfect square (as the book hints). Let $b = d^2$. We we must also have $d^4+c^4 = a^2$, which is impossible by Fermat's Last Theorem in natural numbers. The proof of this is quite standard and involves simply bashing the formula for Pythagorean Triples and achieving Infinite Descent and can be found here.
If we let $c=0$ we have the solution $(1,1,0)$ and if we let $b=0$ we have the solution $(0,0,x)$ for any $x$.
There are solutions such as $(1,1,0)$ that are integral solutions.
A: 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.
