Let $P\in\mathbb{R}[x, y]$ be a polynomial such that there are exactly $n$ pairs $(x, y)$ in $\mathbb{R}^2$ such that $P(x, y)=0$. Is it possible for $P$ to have degree less than $2n$?
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8$\begingroup$ Yes. Factor n into rs, and pick interesting polynomials P and Q with r and s real roots. Use P(x)^2 + Q(y)^2 to get your n pairs at cost max (2r , 2s) <= n for composite n. $\endgroup$– The Masked AvengerCommented May 2, 2014 at 22:43
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$\begingroup$ There are also versions for n a large prime that allow you to reduce the degree to slightly below n. I leave that to you. $\endgroup$– The Masked AvengerCommented May 2, 2014 at 22:49
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