Wonder whether anybody here can provide me with a hint for this one.
Is $c=1$ the only case in which the expression $(c^2+c-1)(c^2-3(c-1))$
returns a perfect square?
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Yes, $c=0$ is the only such value. For the proof, it is useful to let $c=x+1$. Then our expression becomes $$(x^2+3x+1)(x^2-x+1).$$ Note that $x^2-x+1$ is always odd. Any common divisor of $x^2+3x+1$ and $x^2-x+1$ must divide the difference $4x$. But such a common divisor must be odd, so any common divisor must divide $x$. But then it must divide $1$.
Thus $x^2+3x+1$ and $x^2-x+1$ are relatively prime. Since $x^2-x+1$ is always positive, it follows that if their product is a perfect square, each must be a perfect square.
But that can only happen when $x=0$. To prove this, use the fact that for any integer $u$, there is no perfect square strictly between $u^2$ and $(u+1)^2$. Since you asked for a hint, I will, unless you request otherwise, leave out the rest of the argument. It is short.