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Let $m \leq n$ be two positive integers. For every $m$ and $n$, find two polynomials of degree $m$ and $n$ with $m$ distinct intersection points.

This question seems hard since we need to make a polynomial up with this property. Firstly, we can say $f(x) = a(x-x_1)(x-x_2)\cdots(x-x_n)$ and $g(x) = b(x-x'_1)(x-x_2')\cdots(x-x'_m$). Then for $f-g = c(x-x''_1)(x-x''_2)\cdots(x-x''_n)$ which we need to have $m$ real roots and as a result of the complex root conjugate theorem $n-m$ is a multiple of $2$ if we assume our polynomial to have real coefficients.

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The way to think about this problem is critical points. Picture the graph $f(x)=\sin(x)$ and $g(x)=-\sin(x)$. The way these graphs interact is the key. For $m=n$, just take $p(x)=(x-1)(x-2)\cdots (x-m)$ and $q(x)=-p(x)$. For $m<n$, you want to do something similar.

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  • $\begingroup$ What do I do for $m < n$? $\endgroup$ – user19405892 Mar 18 '16 at 21:08
  • $\begingroup$ @user19405892 Sorry I got your notation wrong. I edited it $\endgroup$ – Stella Biderman Mar 18 '16 at 21:10
  • $\begingroup$ Yes, but what do I do for it? In other words, how should I pick $p(x)$ and $q(x)$? $\endgroup$ – user19405892 Mar 18 '16 at 21:13
  • $\begingroup$ @user19405892 pick $p(x)$ in the same manner, and do something for $q(x)$ that makes it not notably different then the example I gave. $\endgroup$ – Stella Biderman Mar 18 '16 at 23:02
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Start with the $p_m(x)$ with m distinct roots.

If $q = n-m$ is even, consider $p_n(x) = (2 + x^q)p_m(x)$.

If $q$ is odd, let $x_0$ be a root of $p_m$, and consider $p_n(x) = (x - x_0)(2 + x^{q-1})p_m(x)$

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