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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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@Ben, so, what's the cutoff? and, why? – Gerry Myerson Sep 18 '12 at 6:43
@Ben, my experience with people at zero percent is that they are unaware of the accept mechanism and are grateful when it is pointed out to them. If they are aware of the mechanism and truly don't like any of the answers they've had on any of the questions they have asked, why would they keep asking questions here? – Gerry Myerson Sep 18 '12 at 12:51
@GerryMyerson: well, fair enough, but do you really think that Don Antonio's original comment was helpful in that regard? – Ben Millwood Sep 18 '12 at 13:27
@Ben, I'm not convinced that anything in this discussion has been helpful, especially as it really belongs elsewhere. Maybe you want to begin a thread on meta, or contribute to one of the already-existing meta threads on accepting answers. – Gerry Myerson Sep 18 '12 at 23:10
up vote 7 down vote accepted

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.

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