In this problem you are given that

a

b

and

c

are natural numbers. You have to show that if

\sqrt{a} + \sqrt{b}

is rational then it is a natural number.

Perhaps you can crack this Tough Nut if you use the fact that if $\sqrt{a} + \sqrt{b}$ is rational then so is its square which means that $\sqrt{a} b$ is also rational. Knowing this the next step is to use

$\sqrt{a} (\sqrt{a} + \sqrt{b}) = a + \sqrt{ab}$

to show that $\sqrt{a}$ is rational and to do likewise for $b$ .

This is all you need because it has already been proved that if $\sqrt{a}$ is rational it must be an integer.

See the problem The Root Cause in the June 2000 15+ Challenges.

Try to apply this method and then to extend it to three variables for the last part. You will need to consider $\sqrt{a} b + \sqrt{b} c + \sqrt{c} a$ and $\sqrt{a} bc .$