show that

there are not polynomials $p,q$ such that $$\sqrt{x^2-4}=\dfrac{p(x)}{q(x)}$$

there a book say it is clear,because if such polynomials existed,then each zero of$x^2-4$ should have even multiplicity? I can't understand this ,can you explain detail?


If $\sqrt{x^2-4}=\frac{p(x)}{q(x)}$, squaring we get $x^2-4=\frac{p^2(x)}{q^2(x)}$. Now decomposing LHS you have $x^2-4=(x+2)(x-2)$ i.e. all and only roots are $\pm2$, so they are simple roots (is this clear?). Then $p^2(x)$ is again a polynomial, which must vanish at $\pm2$, but in this case the roots has order $\ge2$ and this is impossible.

  • $\begingroup$ I can't understand last ,why in this case the roots has order $\ge 2$, this order is meaning? $\endgroup$ – user223800 Apr 4 '15 at 9:38
  • 2
    $\begingroup$ @Australia if $p(x)/q(x)$ is to have the same roots as you function, then all roots in $\frac{p(x)^2}{q(x)^2}$ must have the same multiplicity as $x^2-4$. That polynomial has two roots with multiplicity $1$, however a polynomial of the form $p(x)^2$ has even multiplicity on all roots, and so does $q(x)^2$, since the multiplicity of common roots is subtracted when polynomials are divided you are left with $\text{even} - \text{even} = 1$ $\endgroup$ – Alice Ryhl Apr 4 '15 at 9:43
  • $\begingroup$ Might want to include that in the answer, helps a lot in understanding. $\endgroup$ – Arpan Apr 4 '15 at 10:02

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