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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,802

- Thread starter
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- Feb 14, 2012

- 3,802

Now, $g(h(1)),\,g(h(2)),\,g(h(3)),\,g(h(4))$ are the roots of the quadratic polynomial $f(x)$, so $g(h(1))=g(h(4))$ and $g(h(2))=g(h(3))$ , which implies $h(1)+h(4)=h(2)+h(3)$. For $h(x)=Ax^2+Bx+C$, this would force $A=0$, a contradiction.