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Find the minimal polynomial of $\sqrt3 + \sqrt5$ over $\Bbb Q$.

I got that if $\alpha = \sqrt3 + \sqrt5$, then $\alpha^2 = 8+2\sqrt{15}$ and from there $\alpha^2 - 8 = 2\sqrt{15}$. Squaring again to get rid of the radical I get that $$(\alpha^2-8)^2 = 60 \iff \alpha^4 - 16\alpha^2 + 4 = 0$$

but looking at the solution the minimal polynoimal should be $x^4 -10x^3 +1$. How did they come to that conclusion?

Walker
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    No idea, but what you did looks fine. See this for confirmation. – lulu Aug 29 '22 at 18:46
  • Minimal polynomial is unique up to a nonzero constant factor. Thus, at most one of $x^4-16x^2+4$ and $x^4-10x^3+1$ can be right. Because yours is right, theirs must be wrong. –  Aug 29 '22 at 18:49
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    The minimal polynomial is not $x^4-10x^3+1$. That would make $\sqrt{3}+\sqrt{5}$ an algebraic integer unit, but it is not. – Arturo Magidin Aug 29 '22 at 18:50
  • $(\sqrt{3}+\sqrt{5})^4 -10(\sqrt{3}+\sqrt{5})^3 + 1 = 124+32\sqrt{15} - 10(18\sqrt{3}+14\sqrt{5}) + 1 = 125 - 180\sqrt{3}-140\sqrt{5}+32\sqrt{15}\neq 0$. – Arturo Magidin Aug 29 '22 at 18:54
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    "The solution" really can't be right because you can see that the roots of the minimal polynomial will be $\pm \sqrt 3\pm \sqrt 5$, and in particular the negative of any root will be a root. The minimal polynomial must therefore be even (contain only even powers of $x$). – Mark Bennet Aug 29 '22 at 19:18
  • This was example $8$ in "Abstract Algebra: Theory and Applications". Looking at the newest version it seems to be corrected to be the same as what I got. – Walker Aug 29 '22 at 19:42

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