Find conditions on positive integers $a, b, c$ so that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is irrational.

My solution: if $ab$ is not the square of an integer, then the expression is irrational. I find it interesting that $c$ does not come into this at all.

My solution is modeled (i.e., copied with modifications) from dexter04's solution to Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational .

Suppose $\sqrt{a}+\sqrt{b}+\sqrt{c} = r$ where $r$ is rational. Then, $(\sqrt{a}+\sqrt{b})^2 = (r-\sqrt{c})^2 \implies a+b+2\sqrt{ab} = c+r^2-2r\sqrt{c}$.

So, $a+b-c-r^2+2\sqrt{ab} =-2r\sqrt{c}$. Let $a+b-c-r^2 = k$, which will be a rational number. So, $(k+2\sqrt{ab})^2 = k^2+ 4ab+4k\sqrt{ab} = 4cr^2$ or $4k\sqrt{ab} = 4cr^2-k^2- 4ab$.

If $ab$ is not a square of an integer, then the LHS is irrational while the RHS is rational. Hence, we have a contradiction.

  • 2
    $\begingroup$ Whats the question? $\endgroup$
    – Jonny
    Mar 23 '15 at 20:07
  • $\begingroup$ That's because it's not a complete characterization: $a=b=1$, $c=2$ $\endgroup$
    – WimC
    Mar 23 '15 at 20:07
  • $\begingroup$ @Jonny I think he's asking if his answer is correct. $\endgroup$
    – John
    Mar 23 '15 at 20:09
  • $\begingroup$ Sufficient conditions here : math.stackexchange.com/questions/136556 $\endgroup$
    – Watson
    Dec 25 '16 at 14:18
  • $\begingroup$ See the Lemma in the dupe. $\endgroup$ Jan 2 '21 at 9:16

The answer is that if any of $a, b, c$ is not a square of an integer, then $\sqrt{a} +\sqrt{b} +\sqrt{c}$ must be irrational.

The proof of the general case is not very easy. The paper Square roots have no unexpected linear relationships by Qiaochu Yuan at https://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/ explains this non-trivial theorem.

  • 4
    $\begingroup$ Note however that $\sqrt a+\sqrt b-\sqrt c$ could be "unexpectedly" rational, such as $\sqrt 2+\sqrt 8-\sqrt{18}$. $\endgroup$ Mar 23 '15 at 21:36
  • $\begingroup$ See also Hagen von Eitzen's answer here: math.stackexchange.com/a/437374/43288. $\endgroup$ Mar 23 '15 at 21:55

For $\sqrt{a}+\sqrt{b}+\sqrt{c}$ to be rational, $ab$ being a perfect square is only a necessary condition, but it is not sufficient (your proof is correct, only the conclusion that it is sufficient is wrong). The sufficient and necessary condition is $a,b,c$ are all perfect squares (easy to see that it is sufficient and I'm going to prove it is necessary).

Symmetry lets us conclude $ab=A^2,bc=B^2,ca=C^2$.

Let $(a,b)=d$. Then $a=da_1, b=db_1$ and since $d^2a_1b_1=A^2$, we have $a=da_2^2, b=db_2^2$.

$bc=B^2\implies db_2^2c=B^2\implies dc=b_3^2\tag{1}$

Let $(c,d)=D$. Then $c=Dc', d=Dd'$. $(1)\implies c=Dc_1^2, d=Dd_1^2$.

So $(a,b,c)=(D(d_1a_2)^2,D(d_1b_2)^2,Dc_1^2)$

$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\sqrt{D}(|d_1a_2|+|d_1b_2|+|c_1|)\in\mathbb Q\iff \sqrt{D}\in\mathbb Q$$$$\iff \sqrt{D}\in\mathbb Z\iff D=D'^2$$

I used the fact that $\sqrt{a},a\in\mathbb Z$ is either an integer or irrational. If it is rational but not an integer, then $\sqrt{a}=\frac{a'}{b'}, a'\nmid b'\implies a=\frac{a'^2}{b'^2}$.
Contradiction, since LHS is an integer and RHS is not ($a'\nmid b'\iff a'^2\nmid b'^2$).

$(a,b,c)=((D'd_1a_2)^2,(D'd_1b_2)^2,(D'c_1)^2)\ \ \ \square$


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