Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.

Question

Note that my attempted method below is distinct from the solutions in this question.

I also know this is generally true for $\sqrt p + \sqrt q + \sqrt r$ where $p,q,r$ are prime, but I am asking if my particular method works.

Prove that $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.

Assume that $\sqrt 2 + \sqrt 3 + \sqrt 5 = \frac pq$ for some $p,q \in \mathbb Z$, in lowest terms, with $q \neq 0$. Then at most one of $p$ and $q$ can be even. Then

\begin{align*} \sqrt 2 + \sqrt 3 + \sqrt 5 &= \frac pq\\ \\ (\sqrt 2 + \sqrt 3 + \sqrt 5)^2 &= \left(\frac pq \right)^2\\ \\ 2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=\frac{p^2}{q^2}\\ \\ 2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=p^2 \end{align*}

which shows that $p^2$ is even, so $p$ is even and let $p=2k$ for some $k\in \mathbb Z$. Substituting this we get

\begin{align*} 2q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=(2k)^2\\ \\ q^2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})&=2k^2\\ \\ q^2 &= 2 \left(\frac{k^2}{5+\sqrt 6 + \sqrt{10} + \sqrt{15}}\right), \end{align*} which shows as well that $q$ is even, a contradiction. Hence $\sqrt 2 + \sqrt 3 + \sqrt 5$ is irrational.

Answer 1

$$2(5+\sqrt 6 + \sqrt{10} + \sqrt{15})=\frac{p^2}{q^2}$$

This step, you cannot conclude the LHS is an even integer.

You can try this: Assume $\sqrt{2}+\sqrt{3}+\sqrt{5}=r_1$ be rational.

$$\sqrt{2}+\sqrt{3}=r_1-\sqrt{5}\Leftrightarrow 2\sqrt{6}=r_1^2-2\sqrt{5}r_1\Leftrightarrow \sqrt{6}+\sqrt{5}r_1=\frac{r_1^2}{2}=r_2 \\ \\ \Leftrightarrow \sqrt{6}=r_2-\sqrt{5}r_1\Leftrightarrow 6=r_2^2+5r_1^2-2r_1\sqrt{5}\Leftrightarrow \sqrt{5}=\frac{r_2^2+5r_1^2-6}{2r_1}$$

$\sqrt{5}$ is irrational , but RHS is rational, so you get contradictions.

Answer 2

No. What if $5+\sqrt{6}+\sqrt{10}+\sqrt{15} = \frac{17}{2}$? (I don't think it is, but you certainly haven't shown that it isn't.)

Answer 3

There are infinite primes of the form $p=120k+61$ by Dirichlet's theorem. For any of them we have that $3$ and $5$ are quadratic residues, since $p\equiv 1\pmod{15}$, while $2$ is not a quadratic residue since $p\equiv 5\pmod{8}$. Assume that $\sqrt{2}+\sqrt{3}+\sqrt{5}=\frac{a}{b}$ and take a prime $p$ of the previous form with $p> \max(5,b)$. Since $\left(\frac{2}{p}\right)=-1$ while $\left(\frac{3}{p}\right)=\left(\frac{5}{p}\right)=1$, the minimal polynomial of $\sqrt{2}+\sqrt{3}+\sqrt{5}$ over $\mathbb{F}_p$ has degree $2$. In particular $\sqrt{2}+\sqrt{3}+\sqrt{5}$ cannot be an algebraic number with degree $1$ over $\mathbb{Q}$, i.e. a rational number.

Alternative approach. Always assuming $\sqrt{2}+\sqrt{3}+\sqrt{5}=q\in\mathbb{Q}$ we have $$ 8+2\sqrt{15}=(\sqrt{3}+\sqrt{5})^2 = (q-\sqrt{2})^2 = (q^2+2)-2q\sqrt{2},$$ $$ 2\sqrt{15}+2q\sqrt{2} = (q^2-6),$$ $$ (60+8q^2) + 8q \sqrt{30} = (q^2-6)^2, $$ $$ \sqrt{30} = \frac{(q^2-6)^2-(60+8q^2)}{8q} $$ but the last line implies $\sqrt{30}\in\mathbb{Q}$ which we know not to be the case.

Both approaches can be easily extended for proving that $\sum_{k=1}^{n}\sqrt{p_k}\not\in\mathbb{Q}$ for any collection of primes $p_k$.