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:

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

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 $b_1=b_2$?

  • $\begingroup$ If at least one of them is equal, then both are. Indeed, suppose $b=b_1=b_2$; the analogous case for $a$ can be dealth with similarly. Then $$\frac{s}{\sqrt{b}}=\sqrt{\frac{a_1}{b}}+1=\sqrt{\frac{a_2}{b}}+1,$$ so that $\sqrt{\frac{a_1}{b}}=\sqrt{\frac{a_2}{b}}\implies a_1=a_2$. $\endgroup$ – Fimpellizieri May 24 '17 at 16:01
  • $\begingroup$ $a_1$, $b_1$, $a_2$ and $b_2$ are arbitrary to begin with. $\endgroup$ – j4n bur53 May 24 '17 at 16:02

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


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,


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


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.


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: $$a_1 = a_2 \quad \vee \quad a_1 = b_2$$

  • $\begingroup$ I think the part where you equate rational and irrational parts needs to be better explained. Clearly it does not apply to this particular question, but we have $2+\pi=1+(1+\pi)$. $\endgroup$ – Fimpellizieri May 25 '17 at 6:25
  • $\begingroup$ I don't think that's as clear because we have two square roots here. You can have $$1+\mu-\gamma-\delta=2(\sqrt{\gamma\delta}-\sqrt{\mu})$$ then square both sides to get that a single root $\sqrt{\mu\gamma\delta}$ must be rational, that is, $\mu\gamma\delta=q^2$ for some $q\in\mathbb{Q}$. More generally, we get that $a_1,a_2,b_1,b_2$ satisfy a degree four diophantine homogenous diophantine equation, but I don't know exactly how to deal with this. $\endgroup$ – Fimpellizieri May 25 '17 at 14:53
  • $\begingroup$ Just reorder a little bit different. Take x=4μ, y=1+μ-γ-δ and z=4γδ. From sqrt(x)+y = sqrt(z) and y ≠ 0, i get by squaring both sides: x+2*sqrt(x)*y+y^2 = z, or sqrt(x) = (z-x-y^2)/(2*y), i.e. that sqrt(x) would be rational, a contradiction to the condition 4 in the question. $\endgroup$ – j4n bur53 May 25 '17 at 15:06
  • $\begingroup$ Similar reasoning as here: math.stackexchange.com/a/2262605/4414 by Yujie Zha. $\endgroup$ – j4n bur53 May 25 '17 at 15:11
  • $\begingroup$ Yeah, that's good. We get that $a_2+b_2-a_1-b_1\neq 0$ leads to a contradiction with the lemma, and that $a_2+b_2-a_1-b_1=0$ leads to both $a_2+b_2=a_1+b_1$ and $a_2b_2=a_1b_1$. At this point the condition $a_i<b_i$ takes care of ensuring that $a_1=a_2$ and $b_1=b_2$. Good job; I would just write it out in the answer to make it clearer to any who might skip the comments. $\endgroup$ – Fimpellizieri May 25 '17 at 15:23

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