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# polynomial

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• #24419
Aritra
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#24441
Tarit Goswami
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We proceed using induction on the degree of  $p$. It’s evident that  $p$ must have even degree, and the proposition is trivial for $p$  constant. Since  $p$ has even degree, it has a non-negative global minimum, say $m$ . Then $p-(\sqrt{m})^2$  is tangent to the x-axis and so is divisible by $(x-c)^2$  for some real $c$ . Then $\frac{p-(\sqrt{m})^2}{(x-c)^2}$  is expressible as a sum of squares by assumption, and we simply need to multiply through by  $(x-c)^2$ and add  $\sqrt{m}^2$ to get that $c$  is as well.

This argument can be easily refined to again get that  $k=2$ suffices.

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