Square numbers related to square root

Question

Given whole numbers $a$, $b$, $c$ satisfying $\sqrt{a}+\sqrt{b}+\sqrt{c}$ are also whole numbers. Prove that $a$, $b$, $c$ are square numbers.

Answer 1

First, if $\sqrt x\pm\sqrt y=n$ is an integer, then $x, y$ are both squares (see this by squaring the equation $\sqrt x=n\mp\sqrt y$ - observe $\sqrt y$ is rational implies $\sqrt y$ is integer).

Now if $\sqrt a+\sqrt b+\sqrt c=m$, then square $\sqrt a+\sqrt b=m-\sqrt c$, we see $\sqrt{4ab}+\sqrt{4m^2 c}$ is an integer, thus $c$ is a quare. Then use the above again we see $a, b$ are squares as well.

Answer 2

https://everything2.com/title/The+square+root+of+any+positive+integer+is+either+integral+or+irrational

In the above link, it proves that the square-root of a number is either an integer or irrational.

Case 1: If at-least one of the square roots is irrational, then the sum must be irrational, and cannot be a whole number, contradicting the question.

So case 2 is true

Case 2: If the square-roots are not irrational they must be integral (see proof) Thus, $a,b,c$ are all perfect squares as their square roots are all integers

Answer 3

Because $\sqrt{a}+\sqrt{b}+\sqrt{c}\in\mathbb N$, it follows that $\sqrt a, \sqrt b, \sqrt c\in\mathbb N$. If $\sqrt{a}\in\mathbb N$, then $a$ is a square number. Because the definition is that $x$ is a square number if and only if there exists a $y\in\mathbb N$ with $y^2=x$, which is the case here for $y=\sqrt n$ and $x=n$ (similar for $b$ and $c$).