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I dont know how to solve the problem below.

(1) $p[1]$, $p[2]$, $\ldots$, $p[n]$ are distinct primes, where $n = 1,2,\ldots$ Let $a[n]$ be the sum of square root of those primes, that is, $a[n] = \sqrt{p[1]}+\ldots+\sqrt{p[n]}$. Show that there exists a polynomial with integer coefficients that has $a[n]$ as a solution.

(2) Show that $a[n]$ is irrational.

I can solve this without using (1) for (2). But my teacher said (1) is a hint for (2). Help please?

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    $\begingroup$ Can you edit this to make the notation easier to understand? I can solve the problem if I can state it, but the notation is not very clear. $\endgroup$ – Jeffrey Young Oct 9 '14 at 4:10
  • $\begingroup$ Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. $\endgroup$ – Joao Oct 9 '14 at 4:12
  • $\begingroup$ @Winther: Good editing job! $\endgroup$ – Jeffrey Young Oct 9 '14 at 4:18
  • $\begingroup$ @user182002 See this answer for a proof. $\endgroup$ – Winther Oct 9 '14 at 11:02
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Hint:

For 2): Use the rational root theorem. If $x = \frac{p}{q}$ is a rational root of a polynomial equation $f(x) = c_mx^m + \ldots c_1 x + c_0$ with integer coefficients $c_i$ then $p$ is a factor of $c_0$ and $q$ is a factor of $a_m$.

For 1):

For $n=1$ we have:

$$(x-\sqrt{p_1})(x+\sqrt{p_1}) = x^2 - p_1$$

For $n=2$ we put $x = \sqrt{p_1} + \sqrt{p_2}$ then:

$$(x -\sqrt{p_1})^2 = p_2 = x^2 + p_1 - 2\sqrt{p_2}$$

so

$$(x^2 + p_1 - p_2)^2 - 4p_2 = 0$$

For $n=3$ we put $x = \sqrt{p_1}+\sqrt{p_2} + \sqrt{p_3}$ then

$$(x -\sqrt{p_1})^2 = (\sqrt{p_2} + \sqrt{p_3})^2 = p_2+p_3+2\sqrt{p_2p_3}$$

so

$$x^2 +p_1-p_2-p_3 = 2x\sqrt{p_1} +2\sqrt{p_2p_3}$$

squaring again gives

$$(x^2 +p_1-p_2-p_3)^2 = 4x^2p_1 + 4p_2p_3 + 4x\sqrt{p_1p_2p_3}$$

and a final rearrangement and squaring gives

$$((x^2 +p_1-p_2-p_3)^2 - 4x^2p_1 - 4p_2p_3)^2 - 16x^2p_1p_2p_3 = 0$$

which is the desired polynomial.

The general case is more tricky, see this MSE answer for a proof.

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  • $\begingroup$ Thank you, Winther. But I still need your help. I actually tried to construct the polynomial for n=3 in a way you showed me. What is the constant term of the polynomial when n=3? Is it right to start with (x-sqrt(p[1])-sqrt(p[2])-sqrt(p[3]))(x-sqrt(p[1])-sqrt(p[2])+sqrt(p[3]))? Or, (x-sqrt(p[1])-sqrt(p[2])-sqrt(p[3]))(x-sqrt(p[1])+sqrt(p[2])-sqrt(p[3]))? $\endgroup$ – user182002 Oct 9 '14 at 9:30
  • $\begingroup$ @user182002 See my update for $n=3$. $\endgroup$ – Winther Oct 9 '14 at 10:46

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