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?
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:
$$(x -\sqrt{p_1})^2 = p_2 = x^2 + p_1 - 2x\sqrt{p_1}$$
so
$$(x^2 + p_1 - p_2)^2 - 4x^2p_1 = 0$$
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
$$(x^2 +p_1-p_2-p_3)^2 = 4x^2p_1 + 4p_2p_3 + 8x\sqrt{p_1p_2p_3}$$
and a final rearrangement and squaring gives
$$((x^2 +p_1-p_2-p_3)^2 - 4x^2p_1 - 4p_2p_3)^2 - 64x^2p_1p_2p_3 = 0$$
which is the desired polynomial.
The general case is more tricky, see this MSE answer for a proof.
