# Are $a=3, b=7$ the only solutions to $\sqrt{3}+\sqrt[3]{7}=\sqrt{a}+\sqrt[3]{b}$ for $a,b \in \mathbb{Q}$?

So I had a lesson about calculating surds e.g. $$\sqrt{5+2\sqrt{6}}$$, then the teacher wrote the steps like that:

For some $$a,b\in\mathbb{Q}_{\ge 0}$$ $$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} \\ 5+2\sqrt{6}=a+b+2\sqrt{ab} \\$$ which directly implies $$\begin{cases} a+b=5 \\ ab=6\end{cases}$$. The teacher didn't know the proof and I eventually found a proof of it:

Rewrite the equation, we get: $$5-a-b+2\sqrt{6}=2\sqrt{ab}$$ Then let $$2c=5-a-b$$, where $$c$$ should be rational. Then we continue the equation: $$2c+2\sqrt{6}=2\sqrt{ab} \\ c+\sqrt{6}=\sqrt{ab} \\ (c+\sqrt{6})^2=ab \\ c^2+6-ab=-2c\sqrt{6}$$ As $$c^2+6-ab$$ is rational, $$-2c\sqrt{6}$$ is also rational, which implies $$c=0$$, i.e. $$a+b=5$$. Substituting it back to the original equation, we get $$ab=6$$. Q.E.D.

He found the proof quite interesting and asked a question as follows:

For rational numbers $$a,b$$, is $$(3,7)$$ the only solution to the equation $$\sqrt{a}+\sqrt[3]{b}=\sqrt{3}+\sqrt[3]{7}$$?

I try to prove this but it is harder than it looks. The cubic root is one of the annoying part of proving this. What I am able to prove is that $$\sqrt{3}+\sqrt[3]{7}$$ is irrational. Other than that, I have no idea. To prove this, I think it may probably consist of algebraic fields, which I totally have no idea. I would like to know if there are any ways to solve this as this problem is quite interesting. If you guys have any ideas, please feel free to send it out. Thanks :)

Rewrite the equation as $$\sqrt {a} - \sqrt{3}=\sqrt [3]{7}-\sqrt[3]{b}\tag{1}$$ and if each side is irrational then the identity cannot hold as left hand side is of degree $$2$$ or $$4$$ over $$\mathbb {Q}$$ and the degree of right hand side is a multiple of $$3$$.

Thus the only way the identity can hold is if both sides are rational. But if left hand side is rational we have that either $$a=3$$ or $$\sqrt{a} +\sqrt {3}=(a-3)/(\sqrt{a}-\sqrt{3})$$ is rational. For the second option we see that $$\sqrt{3}$$ is rational which is absurd. Thus $$a=3$$ and then $$b=7$$.

The analysis of right hand side of $$(1)$$, say $$x$$, requires a bit more analysis based on algebraic number theory. If $$\sqrt [3]{b}$$ is a rational multiple of $$\sqrt[3]{7}$$ then either $$x$$ is $$0$$ or an irrational of degree $$3$$. But otherwise it is an algebraic number of degree $$9$$. The detailed proof uses arguments similar to those used here or here.

I have finally managed to avoid algebraic number theory. Let $$x=\sqrt[3]{7}-\sqrt [3]{b}$$ be irrational. Then $$x$$ is a root of $$p(t) =t^3+3\sqrt[3]{7b}t+b-7\tag{2}$$ If $$c=\sqrt [3]{7b}\notin\mathbb {Q}$$ then $$[\mathbb {Q} (c) :\mathbb {Q}] =3$$. Further note that because of $$(2)$$ we have $$c\in\mathbb {Q} (x)$$ and hence $$\mathbb {Q} (c) \subseteq \mathbb {Q} (x)$$. Therefore by tower law degree of $$\mathbb {Q} (x)$$ over $$\mathbb {Q}$$ is a multiple of $$3$$.

The case when $$\sqrt[3]{7b}\in\mathbb {Q}$$ requires a bit more care. In this case $$p(t) \in\mathbb {Q} [t]$$ and if the polynomial $$p(t)$$ is irreducible then $$x$$ is of degree $$3$$ over $$\mathbb {Q}$$. Otherwise since $$x$$ is irrational it must be root of a quadratic factor of $$p(t)$$ and thus it lies in some quadratic extension $$\mathbb {Q} (\sqrt {d})$$ with $$d$$ being a positive rational number and not the square of another rational number.

Thus we have $$\sqrt[3]{7}-\sqrt [3]{b}=r+s\sqrt{d}$$ or $$\sqrt[3]{7}-r=s\sqrt{d}-\sqrt [3]{b}\tag{3}$$ and $$r, s\in\mathbb {Q}, s\neq 0$$. The left hand side is of degree $$3$$ over $$\mathbb {Q}$$ and we prove that the right side is of degree $$2$$ or $$6$$ over $$\mathbb {Q}$$ and get a contradiction (and conclude that $$p(t)$$ is irreducible over $$\mathbb {Q}$$).

If $$\sqrt[3]{b}$$ is rational then right side of $$(3)$$ is of degree $$2$$ otherwise $$K=\mathbb {Q} (\sqrt{d}, \sqrt[3]{b})$$ is of degree $$6$$ over $$\mathbb {Q}$$. We can prove that $$s\sqrt{d} +\sqrt[3]{b}$$ is a primitive element of $$K$$ and hence also of degree $$(6)$$ over $$\mathbb {Q}$$.

• Thanks for your solution! I get the main idea of the solution. Yet, I am not sure about the degree of $\sqrt[3]{7}-\sqrt[3]{b}$. May you explain briefly about why? Sep 10, 2021 at 12:33
• @MafPrivate: wait a few minutes for updated answer. Sep 10, 2021 at 12:34
• @MafPrivate: unfortunately I don't have a short proof independent of algebraic number theory and I refer you to typical proofs for such problems. One of the simpler proofs uses the notion of trace. Sep 10, 2021 at 13:04
• @MafPrivate: I did manage to avoid algebraic number theory and added a proof based on theory of field extensions. Sep 10, 2021 at 15:24
• Thanks! Now I get the idea. Sep 12, 2021 at 6:59

By doing some calculations you can see that $$\sqrt{3}+\sqrt[3]{7}$$ is a root of the polynomial $$p(x) = x^6-9 x^4-14 x^3+27 x^2-126 x+22.$$ Furthermore, $$\sqrt{a}+\sqrt[3]{b}$$ is a root of $$q_{a,b}(x) = x^6 -3 a x^4 -2 b x^3 +3 a^2 x^2 -6 a b x -a^3+b^2.$$ The polynomial $$p$$ is irreducible. This means that it is the only monic degree 6 polynomial with rational coefficients of which $$\sqrt{3}+\sqrt[3]{7}$$ is a root. This means that if $$a,b$$ are rational such that $$\sqrt{a}+\sqrt[3]{b}$$ is equal to $$\sqrt{3}+\sqrt[3]{7}$$ that the polynomials $$p$$ and $$q_{a,b}$$ must be the same. Now you can look for example at the coefficients in front of the $$x^3$$ and $$x^4$$ terms to find that this implies that $$a=3$$ and $$b=7$$.

• Thanks for your solution! Sep 10, 2021 at 13:02
• This is a rather smart solution using basic idea of algebraic numbers and their minimal polynomials. +1 Sep 10, 2021 at 15:40