<s>[INST] Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational. 
Question:
Using the fact that $\sqrt{n}$ is an irrational number whenever $n$ is not a perfect square, show $\sqrt{3} + \sqrt{7} + \sqrt{21}$ is irrational.

Following from the question, I tried:

Let $N = \sqrt{3} + \sqrt{7} + \sqrt{21}$. Then,
$$
\begin{align}
N+1 &= 1+\sqrt{3} + \sqrt{7} + \sqrt{21}\\
    &= 1+\sqrt{3} + \sqrt{7} + \sqrt{3}\sqrt{7}\\
    &= (1+\sqrt{3})(1+\sqrt{7}).
\end{align}
$$
Using the above stated fact, $\sqrt{3}$ and $\sqrt{7}$ are irrational. Also, sum of a rational and irrational number is always irrational, so $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational. Similarly, if we prove that $N+1$ is irrational, $N$ will also be proved to be irrational.

But, how do I prove that product of $1+\sqrt{3}$ and $1+\sqrt{7}$ are irrational.
 [/INST] Hint $\,\sqrt{21}+\sqrt{7}+\sqrt{3}=q\,\Rightarrow\ \sqrt7(\sqrt 3+1) = q-\sqrt3  \,\Rightarrow\, \sqrt 7\in \Bbb Q(\sqrt3)$, contradiction as below.
Lemma $\rm\ \ [K(\sqrt{a},\sqrt{b}) : K] = 4\ $ if  $\rm\ \sqrt{a},\ \sqrt{b},\ \sqrt{a\,b}\ $  all are not in $\rm\,K\,$ and $\rm\, 2 \ne 0\,$ in the field $\rm\,K.$
Proof $\  $  Let  $\rm\ L = K(\sqrt{b}).\,$ Then $\rm\,  [L:K] = 2\,$  via  $\rm\,\sqrt{b}  \not\in K,\,$  so it suffices to prove $\rm\, [L(\sqrt{a}):L] = 2.\,$ It fails only if  $\rm\,\sqrt{a} \in L = K(\sqrt{b})\, $ and then $\rm\, \sqrt{a}\ =\  r + s\, \sqrt{b}\ $  for $\rm\ r,s\in K.\,$ But that's impossible,
since squaring  $\Rightarrow \rm(1)\!:\ \ a\ =\ r^2 + b\ s^2 + 2\,r\,s\  \sqrt{b},\, $ which contradicts hypotheses as follows:
$\rm\qquad\qquad rs \ne 0\ \ \Rightarrow\ \  \sqrt{b}\ \in\  K\ \ $ by solving $(1)$ for $\rm\sqrt{b}\,,\,$ using  $\rm\,2 \ne 0$
$\rm\qquad\qquad\  s = 0\ \ \Rightarrow\ \  \ \sqrt{a}\ \in\  K\ \ $  via  $\rm\ \sqrt{a}\ =\ r+s\,\sqrt b = r \in K$
$\rm\qquad\qquad\  r = 0\ \ \Rightarrow\ \  \sqrt{a\,b}\in K\ \ $  via  $\rm\ \sqrt{a}\ =\ s\, \sqrt{b},\ \ $times $\rm\,\sqrt{b}\quad$
Remark $ $  The Lemma generalizes to any number of sqrts). See the citations there for generalizations to $n$'th roots.
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