TITLE: Are square root binomials unique?
QUESTION [9 upvotes]: In Euclid we find the notion of a binomial, its simply a sum $s = \sqrt{a}+\sqrt{b}$ of two square roots $\sqrt{a}$ and $\sqrt{b}$. Lets say such a sum is simple iff $a$ and $b$ are positive non-zero rational numbers, and normal iff:

$a < b$  
$\not \exists r \in \mathbb{Q} \,a = r^2$  
$\not \exists r \in \mathbb{Q}\,b = r^2$    
$\not \exists r \in \mathbb{Q}\,a/b = r^2$

So for being normal we require that the summands are ordered, i.e. $a < b$. We also require that both radicands $a$ and $b$ are not squares of other rational numbers, and neither their quotient is.
I wonder whether this representation is unique. Assume we have two simple and normal square root sums with the same value:
$s = \sqrt{a_1} + \sqrt{b_1}$
$s = \sqrt{a_2} + \sqrt{b_2}$
Can we conclude without further assumptions that simultaneously $a_1=a_2$ and $b_1=b_2$?

REPLY [3 votes]: Using the same Lemma as in the answer of Fimpellizieri. I start with the following equation:
$$\sqrt{a_1}+\sqrt{b_1} = \sqrt{a_2} + \sqrt{b_2}$$
To eliminate the parameter $a_1$ I divide both sides by $\sqrt{a_1}$ and I get:
$$1+\sqrt{\frac{b_1}{a_1}} = \sqrt{\frac{a_2}{a_1}} + \sqrt{\frac{b_2}{a_1}}$$
Let us use new variables $μ$, $γ$ and $δ$, the irrationality Lemma also applies to $\sqrt{μ}$ and $\sqrt{γ δ}$:
$$1+\sqrt{μ} = \sqrt{γ} + \sqrt{δ}$$
Now square both sides to get:
$$1+μ+2\sqrt{μ} = γ+δ+2\sqrt{γ δ}\tag{1}$$
We can equate the rational and the irrational parts, so we get two equations:
$$1+μ = γ+δ\tag{2}$$
$$μ = γ δ\tag{3}$$
Using the last equation (3) in the first equation (2) we get:
$$γ δ - γ - δ + 1 = (γ - 1)(δ - 1) = 0$$
Hence we arrive at $γ=1$ or $δ=1$. Recall $γ=\frac{a_2}{a_1}$ and $δ=\frac{b_2}{a_1}$ so that we get as desired:
$$a_1 = a_2 \quad \vee \quad a_1 = b_2$$