How can we show that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is a rational number only if $a,b,c$ are perfect squares?

Question

Question

I saw a lot of problems that assume this: $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is a rational number only if $a,b,c$ are perfect squares. I wonder how can we demonstrate it because I saw a lot of people using it. Also, can I use it without a demonstration? Hope one of you can help me. Thank you!

My idea

$\sqrt{a}+\sqrt{b}=n$ where n is a natural number

$\sqrt{a}=n-\sqrt{b}$

$a=n^2+b-2n\sqrt{b}$

$\sqrt{b}=\frac{n^2+b-a}{2n}$

$\sqrt{b}=$ a rational number which means b is a perfect square.

I need to prove this, just so I can solve this problem Are there nonzero natural numbers such that $\sqrt{4n+5}+\sqrt{5n+1}+\sqrt{9n+4}= \frac{nx}{y}$?.

If you also find a way to solve this problem without using the fact that they must be perfect squares. I will be greatful.

Answer 1

Followed by the notation above, let $r=\sqrt{a}+\sqrt{b}+\sqrt{c},$ then $$r^2-(a+b+c)=2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\in\mathbb{Q}.$$ Let $q=\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\in\mathbb{Q}$, then $$q^2-(ab+bc+ac)=2c\sqrt{ab}+2a\sqrt{bc}+2b\sqrt{ac}\in\mathbb{Q}.$$ after deleting $\sqrt{ab}$, we have $$(a-c)\sqrt{ac}+(b-c)\sqrt{bc}:=\alpha\in\mathbb{Q}.$$ then $$\alpha^2-(a-c)^2ac-(b-c)^2bc=2(a-c)(b-c)c\sqrt{ab}\in\mathbb{Q}.$$ This implies $\sqrt{ab}\in\mathbb{Q}$, and similarly we have $\sqrt{bc},\sqrt{ac}\in\mathbb{Q}$. Suppose $a,b,c$ are square-free integers, then $a=b=c$ and the equality holds when $a=b=c=1$. On the other hand, if one of them is not square-free integer, you can modify the arguments by taking out the square factor.

Edited: There's another way to get the result. After we prove that $\sqrt{ab},\sqrt{bc},\sqrt{ac}\in\mathbb{Q}$, let $\sqrt{ab}=m,\sqrt{bc}=n$, then $\sqrt{ac}=\frac{abc}{mn}$, so we have $$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{a}\left(1+\dfrac{m}{a}+\dfrac{abc/mn}{a}\right).$$ If $\sqrt{a}\in \mathbb{Q}$, then $\sqrt{b}+\sqrt{c}\in\mathbb{Q}$, and you've proved that only if $b,c$ are perfect squares and we're done. Otherwise suppose $\sqrt{a}$ is irrational, then $\sqrt{a}+\sqrt{b}+\sqrt{c}=0$ only if $a=b=c=0$ and we're also done.

Answer 2

Let's prove a more general statement: If $a$, $b$ and $c$ are positive rational numbers and $\sqrt a + \sqrt b + \sqrt c$ is rational then $a$, $b$ and $c$ are squares (of rational numbers).

Let $r = \sqrt a + \sqrt b + \sqrt c$ rational and assume that $\sqrt a$, $\sqrt b$ and $\sqrt c$ are irrationals.

Then $r-\sqrt a = \sqrt b + \sqrt c$. Squaring we get $r^2+a-b-c=2\sqrt{b c}+2r\sqrt a$. Squaring again, we get $(r^2+a-b-c)^2 -4bc-4r^2a=8r\sqrt{abc}$.

Since $r\ne 0$ and the LHS is rational we get that $abc = s^2$ for a positive rational $s$.

Then $b c=s^2/a$, substitute in the previous equation we have $r^2+a-b-c = 2\sqrt{s^2/a}+2r\sqrt a = 2s/\sqrt{a} + 2r\sqrt a$. Multiply by $\sqrt a$ to obtain $(r^2+a-b-c)\sqrt a = 2s+2ra$. Since the RHS is rational and $\sqrt a$ is irrational we have $r^2+a-b-c=0$, so $r^2=b+c-a$.

By symmetry, an analogous argument proves that $r^2=c+a-b$.

Adding this two equations we get $2r^2 = 2c$, so $r=\sqrt c$ which is absurd since we have $\sqrt a + \sqrt b >0$.

This proves that at least one of $\sqrt a$ or $\sqrt b$ is rational, thus reducing the problem to the case of two roots which is already solved.