# Polynomials with Integer Coefficients and irrational roots

Is there a polynomial with integer coefficients which has √2 +√7 ￼ as a root?

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What is that ￼ ? – Robert Israel Mar 11 '14 at 15:01
I suppose you want just this one root (no plural) – Marc van Leeuwen Mar 11 '14 at 15:23

$$x=\sqrt2+\sqrt7$$ $$x^2=9+2\sqrt{14}$$ $$(x^2-9)^2-56=0$$

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To give a non-constructive but general algebraic answer:

The numbers $\sqrt{2}$ and $\sqrt{7}$ are algebraic. As the algebraic numbers form a field, the number $\sqrt{2}+\sqrt{7}$ is algebraic, too. Thus there is a polynomial with rational coefficients which has $\sqrt{2}+\sqrt{7}$ as root. Now multiply by the common divisor to obtain a polynomial with integer coefficients.

If you require the polynomial to be monic, you are asking whether $\sqrt{2}+\sqrt{7}$ is an algebraic integer. Here, too, it is easy to see that both $\sqrt{2}$ and $\sqrt{7}$ are algebraic integers and as the algebraic integers form a ring, $\sqrt{2}+\sqrt{7}$ is an algebraic integer.

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$$x = \sqrt{2} + \sqrt{7} \\ (x-\sqrt{2})^2=7 \\ x^2-2\sqrt{2}x+2=7 \\ x^2-5=2\sqrt{2}x \\ (x^2-5)^2=8x^2 \\ x^4-18x^2+25=0$$

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