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We are given that a polynomial f(x) has integer coefficients. The coefficient of x^4 being 1. One root of it is ($\sqrt{2}+\sqrt3$). How do we find the other roots?

I tried using long division, it was so long so i just did it until $x^2$, like I divided $x^4+ax^3+bx^2+cx+d$ by $(x-(\sqrt2+\sqrt3))$. The only good result that I got was that the coefficient of $x^3$ of the function obtained after dividing would be $1$.

Arbi
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  • Try multiplying out $$(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))(x-(-\sqrt{2}+\sqrt{3}))(x-(- \sqrt {2}-\sqrt{3}))$$ – Gregory Grant Mar 03 '15 at 20:36
  • Why that exactly? – Arbi Mar 03 '15 at 20:50
  • Because if $\sqrt 2 + \sqrt 3$ is a root of a polynomial with coefficients in $\mathbb Q$ then the other three $\pm\sqrt 2\pm\sqrt 3$ must also be roots. This is perhaps not trivial to see, but there are four automorphisms of $Q[\sqrt 2, \sqrt 3]$ that must take a root of the polynomial for $\sqrt 2 + \sqrt 3$ to another root of the same polynomial. – Gregory Grant Mar 03 '15 at 20:53
  • Gregory is right. Put simpler: You will not get integer coefficients in any other way than to multiply with what is lacked to make an integer. For square roots of integers, we need squares to build integers, so each square root must occur twice in any product (or an "even" number of times). – mathreadler Mar 03 '15 at 21:30
  • Algebraically, there's no difference between $\sqrt2$ and $-\sqrt2$. They both are square roots of two. They both satisfy the equation $x^2-2=0$. (You can't say the difference is that only one of them satisfies $x-\sqrt2=0$. You can't use $\sqrt2$ to define $\sqrt2$, that's cheating. Also, you can't say that only one of them satisfies $x>0$. When I said that you can't tell the difference between them algebraically, I meant that there's no polynomial with integer coefficients that has $\sqrt2$ as a root and not $-\sqrt2$.) – Akiva Weinberger Mar 04 '15 at 02:27
  • Similarly, there's no algebraic difference between $\sqrt3$ and $-\sqrt3$. So it makes sense that there should be no difference between all the different numbers of the form $\pm\sqrt2\pm\sqrt3$. – Akiva Weinberger Mar 04 '15 at 02:29

2 Answers2

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$x^4 - 10x^2 + 1$ works. I got it by multiplying out $$(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))(x-(-\sqrt{2}+\sqrt{3}))(x-(- \sqrt {2}-\sqrt{3}))$$

Gregory Grant
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  • Thank you Gregory, using another method I also just knew that the last term must be 1, so yours must be correct. – Arbi Mar 03 '15 at 21:12
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To elaborate on the above answers, maybe it would be helpful to see how we come up with the polynomial without talking about automorphisms. Say we are trying to solve your problem for $\sqrt{a}+\sqrt{b}$ for $a,b$ integers. Set

$$x = \sqrt{a}+\sqrt{b}$$ and after squaring both sides we get: $$x^2 = a + 2\sqrt{ab} + b$$ Move the $a+b$ to the left and square the equation again:

$$x^4 + 2(a+b)x^2+ (a+b)^2 = 4ab$$

Therefore the polynomial $f(x) = x^4+2(a+b)x^2 + (a+b)^2 - 4ab $ will have the desired properties.

Setting $a=2$ and $b=3$ gives you the same result as the previous answers.

Sergio Da Silva
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  • Just a small comment on your last sentence. If you add more square roots it is not so clear that this procedure would work. Already the square of the sum of three square roots can have three terms that are square roots of numbers that need not be perfect squares. Try $(\sqrt{2}+\sqrt{3}+\sqrt{5})^2$. The problem is even worse if you consider other roots besides square roots. – Nathanson Mar 20 '15 at 19:12
  • You are absolutely right Nathanson. I will remove the last sentence from answer. Thanks for catching that. – Sergio Da Silva Mar 22 '15 at 19:42