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I recently found a problem saying "Find 2 irrational numbers such that their sum and product is both rational."

After a while I noticed any pair like $(a+\sqrt{b},a-\sqrt{b})$ work .From this I could easily say this statement is true for any even integer.But then I thought whether the same is true for odd integers. I realized proving it for $3$ would imply the statement being true for all odd integers. I tried to work with similar expression as above.I also tried to work with the cubic polynomial.But I couldn't make any significant progress.

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    $\begingroup$ Take any polynomial of odd degree with no rational roots, like $x^3-5x^2+3x+2$. That one has real roots (not sure you wanted to require that). the product of the roots is $-2$ and the sum is $5$. A similar approach works for any degree. $\endgroup$ – lulu May 30 '20 at 19:35
  • $\begingroup$ How to make sure that all its roots are real? $\endgroup$ – Yes it's me May 30 '20 at 19:37
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    $\begingroup$ I just adjusted the constant to get something that works. $\endgroup$ – lulu May 30 '20 at 19:42
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    $\begingroup$ To prove existence in general, it's probably easiest to use some Galois theory. See, e.g., this question. $\endgroup$ – lulu May 30 '20 at 19:44
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You could take $\{j\:\sqrt[2k+1]{2}|1\le j\le 2k\}\cup\{-k(2k+1)\:\sqrt[2k+1]{2}\}$. The sum is $0$; the product is $-(2k+1)!2k$.

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