Are square root binomials unique? 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 $b_1=b_2$?
 A: 
Lemma: $\sqrt{ab}\not\in\mathbb{Q}$

Proof: If $\sqrt{ab}\in\mathbb{Q}$, then $\frac1b\cdot\sqrt{ab}=\frac{\sqrt a}{\sqrt b}=\sqrt{\frac{a}b} \in \mathbb{Q}$, a contradiction. $\square$
Consider this answer. By the lemma, we have that $Q_i=\mathbb{Q}(a_i,b_i)$ has degree $4$ over $\mathbb{Q}$ and that $\beta_i=\{1,\sqrt{a_i},\sqrt{b_i},\sqrt{a_ib_i}\}$ is a basis for $Q_i$ over $\mathbb{Q}$.
Now, since $s=\sqrt{a_i}+\sqrt{b_i}\in Q_i$ for $i\in\{1,2\}$, we have
$${\left(1+s\right)}^2=\underbrace{(1+a_i+b_i)}_{\in\,\mathbb{Q}}+2\underbrace{(\sqrt{a_i}+\sqrt{b_i})}_s+\sqrt{a_ib_i}$$
and it follows that $\sqrt{a_1b_1}\in Q_2$ and vice-versa. With this, we can expand
\begin{align}
{(s+\sqrt{a_ib_i})}^2
&=(a_i+b_i+a_ib_i)+2\sqrt{a_ib_i}+2a_i\sqrt{b_i}+2b_i\sqrt{a_i}\\
&=\underbrace{(a_i+b_i+a_ib_i)}_{\in\,\mathbb{Q}}+2\underbrace{\sqrt{a_ib_i}}_{\in \,Q_1,Q_2}+2a_i\underbrace{(\sqrt{a_i}+\sqrt{b_i})}_{s\,\in\,Q_1,Q_2}+2(b_i-a_i)\sqrt{a_i}\tag{1}\\
\end{align}
and conclude that $\sqrt{a_1}\in Q_2$ and vice-versa. It follows that the same holds for the $b_i$ and hence $Q_1=Q_2$.

Let $v=(x,y,z,w)$ represent some element of $Q_2$ in the basis $\beta_2$. We have that, by hypothesis,
$$s=(0,1,1,0).$$
By expanding $s^2$, we find $a_1+b_1+2\sqrt{a_1b_1}=a_2+b_2+2\sqrt{a_2b_2}$, so
$$\sqrt{a_1b_1}=\left(\frac12\cdot (a_2+b_2-a_1-b_1),0,0,1\right).$$
One can check directly that
$$v^2=\Big(x^2+a_2y^2+b_2z^2+a_2b_2w^2,2(xy+zwb_2),2(xz+ywa_2),2(xw+yz)\Big).\tag{$*$}$$
Let $d=a_2+b_2-a_1-b_1$. Using $(*)$ on $(1)$, we find that
\begin{align}
2(b_1-a_1)\sqrt{a_1}
=&\left(\frac{d^2}4+a_2+b_2+a_2b_2,d+b_2,d+a_2,d+2\right)\\
-&(a_2+b_2+a_1b_1,2a_1,2a_1,2)\\
=&\left(\frac{d^2}4+a_2b_2-a_1b_1,a_2+2b_2-3a_1-b_1,2a_2+b_2-3a_1-b_1,d\right)
\end{align}
By applying $(*)$ above, we should get relations between the two representations of the number $4(b_1-a_1)^2a_1$ which may point out to the answer. I am running out of time right now but will check on this later.
A: 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:
$$a_1 = a_2 \quad \vee \quad a_1 = b_2$$
