# Is there a way to show the sum of any different square root of prime numbers is irrational? [duplicate]

Is there a way to show the sum of any different square root of prime numbers is irrational? For example, $$\sqrt2+\sqrt3+\sqrt5 +\sqrt7+\sqrt{11}+\sqrt{13}+\sqrt{17}+\sqrt{19}$$ should be a irrational number.

One approach I used is to let the sum be a solution of an even polynomial $f(x)$with integer coefficients and prove by induction that by adding another $\sqrt{p_{k+1}}$. The new polynomial can be written as $$f(x+\sqrt{p_{k+1}})f(x-\sqrt{p_{k+1}})$$

where $$f(x+-\sqrt{p_{k+1}})=P(x)+- Q(x)\sqrt{p_{k+1}},$$

where $P(x)$ is an even plynomial and $Q(x)$ is an odd polynomial.

The new polynomial can be written as $$P^{2}(x)- Q^{2}(x)p_{k+1}.$$

Assume it has a rational solution $a$, we must have$$P(a)=Q(a)=0.$$

My calculation stopped here since I can't find any contradiction result from this. Can anyone continue this proof, or has other better way to solve this? Thanks!

## marked as duplicate by Ross Millikan, JMoravitz, dxiv, erfink, projectilemotionMay 28 '17 at 21:28

• – Tob Ernack May 28 '17 at 17:32
• "For example... should be a prime number." You mean "should be an irrational number?" – Thomas Andrews May 28 '17 at 17:42
• – Jack D'Aurizio May 28 '17 at 17:58
• Ok, then. Thanks! – Zhenyuan Lu May 28 '17 at 21:06

For $n\ge 0$ let $E_n=\Bbb Q(\sqrt {p_1},\ldots, \sqrt{p_n})$ be the smallest field extension of $\Bbb Q$ containing $\sqrt{p_k}$ for $1\le k\le n$.
Claim. For every $(\epsilon_1,\ldots, \epsilon_n)\in\{-1,1\}^n$, there is an automorphism $\phi$ of $E_n$ with $\phi(\sqrt{p_i})=\epsilon_i\sqrt{p_i}$, $1\le i\le n$.
Proof. [By induction]. The claim is vacuously true for $n=0$.
Let $n\ge 0$ and assume $\sqrt{p_{n+1}}\in E_{n}$. Clearly, $E_{n}$ is spanned as a $\Bbb Q$-vector space by all products of some of the $\sqrt{p_i}$ (including the empty product, $1$). Thus we can write $$\tag1\sqrt{p_{n+1}}=\sum_{S\subseteq\{1,\ldots,n\}}q_S\prod_{i\in S}\sqrt{p_i}$$ with $q_S\in \Bbb Q$. Among all such representations, pick one with the minimal number of non-zero coefficients. Assume there are at least two non-zero coefficients $q_A, q_B$. Pick $k\in A\mathop{\Delta}B$. By induction hypothesis, there is an automorphism $\phi$ of $E_n$ that maps $\sqrt{p_k}\mapsto-\sqrt{p_k}$ and for $i\ne k$ maps $\sqrt{p_i}\mapsto\sqrt{p_i}$. Also, $\phi(\sqrt{p_{n+1}})=\pm\sqrt{p_{n+1}}$ so that one of $\frac{\sqrt{p_{n+1}}\pm \phi(\sqrt{p_{n+1}})}2$ equals $\sqrt{p_{n+1}}$. This way, we obtain another representation of $\sqrt{p_{n+1}}$ of the form $(1)$, but with less non-zero coefficients because at least either $q_A$ or $q_B$ (depending on the "$\pm$") is replaced with a $0$. From this contradiction, we conclude that in a minimal representation, at most one $q_S$ in $(1)$ is non-zero. Thus $\sqrt{p_{n+1}}=q_S\prod_{i\in S}\sqrt{p_i}$, contradicting the irrationality of $\prod_{i\in S}\sqrt{p_i}\sqrt{p_{n+1}}$. We conclude that $\sqrt{p_{n+1}}\notin E_n$, hence $E_{n+1}$ is a quadratic extension of $E_n$. Also, $\sqrt{p_{n+1}}\mapsto -\sqrt{p_{n+1}}$ is an automorphism of $E_{n+1}$ over $E_n$ and the claim follows for $n+1$. $\square$