# Show that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q \implies \sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$

Assume that $a，b，c, \sqrt{a}+ \sqrt{b}+\sqrt{c} \in\mathbb Q$ are rational，prove $\sqrt{a},\sqrt{b},\sqrt{c} \in\mathbb Q$,are rational.

I know that can be proved, would like to know that there is no easier way

$\sqrt a + \sqrt b + \sqrt c = p \in \mathbb Q$,
$\sqrt a + \sqrt b = p- \sqrt c$,
$a+b+2\sqrt a \sqrt b = p^2+c-2p\sqrt c$,
$2\sqrt a\sqrt b=p^2+c-a-b-2p\sqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)\sqrt c$,
$\sqrt c=\frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}\in\mathbb Q$.

• Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$. – André Nicolas Apr 25 '12 at 1:29
• Exactly; how do you know $p^2+c-a-b\ne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it. – Gerry Myerson Apr 25 '12 at 1:42

[See here and here for an introduction to the proof. They are explicitly worked special cases]

As you surmised, induction works, employing our prior Lemma (case $$\rm\:n = 2\:\!).\:$$ Put $$\rm\:K = \mathbb Q\:$$ in

Theorem $$\rm\ \sqrt{c_1}+\cdots+\!\sqrt{c_{n}} = k\in K\ \Rightarrow \sqrt{c_i}\in K\:$$ for all $$\rm i,\:$$ if $$\rm\: 0 < c_i\in K\:$$ an ordered field.

Proof $$\:$$ By induction on $$\rm n.$$ Clear if $$\rm\:n=1.$$ It is true for $$\rm\:n=2\:$$ by said Lemma. Suppose that $$\rm\: n>2.$$ It suffices to show one of the square-roots is in $$\rm K,\:$$ since then the sum of all of the others is in $$\rm K,\:$$ so, by induction, all of the others are in $$\rm K$$.

Note that $$\rm\:\sqrt{c_1}+\cdots+\sqrt{c_{n-1}}\: =\: k\! -\! \sqrt{c_n}\in K(\sqrt{c_n})\:$$ so all $$\,\rm\sqrt{c_i}\in K(\sqrt{c_n})\:$$ by induction.

Therefore $$\rm\ \sqrt{c_i} =\: a_i + b_i\sqrt{c_n}\:$$ for some $$\rm\:a_i,\:\!b_i\in K,\:$$ for $$\rm\:i=1,\ldots,n\!-\!1$$.

Some $$\rm\: b_i < 0\:$$ $$\Rightarrow$$ $$\rm\: a_i = \sqrt{c_i}-b_i\sqrt{c_n} = \sqrt{c_i}+\!\sqrt{b_i^2 c_n}\in K\:\Rightarrow \sqrt{c_i}\in K\:$$ by Lemma $$\rm(n=2).$$

Else all $$\rm b_i \ge 0.\:$$ Let $$\rm\: b = b_1\!+\cdots+b_{n-1} \ge 0,\:$$ and let $$\rm\: a = a_1\!+\cdots+a_{n-1}.\:$$ Then
$$\rm \sqrt{c_1}+\cdots+\!\sqrt{c_{n}}\: =\: a+(b\!+\!1)\:\sqrt{c_n} = k\in K\:\Rightarrow\:\!\sqrt{c_n}= (k\!-\!a)/(b\!+\!1)\in K$$

Note $$\rm\:b\ge0\:\Rightarrow b\!+\!1\ne 0.\:$$ Hence, in either case, one of the square-roots is in $$\rm K.\ \$$ QED

Remark  Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $$\:\sqrt{2} + (-\sqrt{2})\in \mathbb Q\:$$ but $$\rm\:\sqrt{2}\not\in\mathbb Q.\:$$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $$\,0$$.

Suppose that $a,b,c$ are all non zero. Let $K=\mathbb{Q}(\sqrt{a},\sqrt{b},\sqrt{c})$ and $n = [K: \mathbb{Q}]$. Then since $Tr_{K/\mathbb{Q}}(\sqrt{a}) = Tr_{\mathbb{Q}(\sqrt{a})/\mathbb{Q}} \circ Tr_{K/\mathbb{Q}(\sqrt{a})} (\sqrt{a})$, we have $$Tr_{K/\mathbb{Q}}(\sqrt{a}) = \begin{cases} 0,& \text{if } \sqrt{a} \notin \mathbb{Q} \\ n\sqrt{a}, &\text{if } \sqrt{a} \in \mathbb{Q}, \end{cases}$$ and same for $\sqrt{b}$ and $\sqrt{c}$.
By hypothesis $\sqrt{a} + \sqrt{b} +\sqrt{c} \in \mathbb{Q}$, so $$Tr_{K/\mathbb{Q}}(\sqrt{a}) + Tr_{K/\mathbb{Q}}(\sqrt{b}) + Tr_{K/\mathbb{Q}}(\sqrt{c}) = n\sqrt{a} + n \sqrt{b} + n \sqrt{c}.$$ It is easy to conclude that $\sqrt{a},\sqrt{b},\sqrt{c} \in \mathbb{Q}$.