That is, are there positive integers greater than 1 satisfying the following equation?


My inspiration for this problem was the following problem:

How many integer solutions are there to the equation $\sqrt{a}+\sqrt{b}=12\sqrt{3}$?

The key to the problem would be to show that a and b must both have and odd power of 3 in their prime factorization. The outline of my proof for the above specific case is as follows:

  1. Note that $a,b$ cannot both be perfect squares.
  2. We write $a=3^{k_1}\alpha, b=3^{k_2}\beta$, where $3∤α,β$
  3. By considering congruences, prove that $k_1, k_2$ must be of the same parity.
  4. Considering the case where $k_1, k_2$ are both even
    1. $\alpha,\beta$ cannot both be perfect squares.
    2. By squaring both sides of the original equation, we find that $\sqrt{\alpha\beta}$ must be rational, and thus it must be an integer too.
    3. Expressing ${\alpha, \beta}$ as products of perfect squares and square free integers, we find that their square free components must be equal. Denote this square free component as $f$.
    4. $12\sqrt{3}=C\sqrt{f}$, where $C$ is an integer. Then $f=3$, which is a contradiction.

An alternative, shorter proof by a friend:

  1. Note that $ab$ is perfect square.
  2. $a=0, b=0$ each give 1 solution. Otherwise, $a,b>0$.
  3. Let $d=gcd(a, b)$. Since $ab$ is a perfect square, $a={x^2}d, b={y^2}d$ for some relatively prime positive integers x, y.
  4. ${(x+y)^2}d=432$, allowing one to calculate the number of positive integer solutions.

Will it be possible to generalize either proof to surds of higher (and not necessarily equal) orders?

  • 1
    $\begingroup$ Just so you know: I asked T. Y. Lam about canonical forms in the "constructible numbers" $\mathbb E,$ meaning the smallest subfield of the reals such that $x \in \mathbb E, \; x > 0$ implies $\sqrt x \in \mathbb E.$ He said there was no canonical form. And, well, he is the one who would know. ams.org/bookstore-getitem/item=GSM-67 $\endgroup$
    – Will Jagy
    Feb 5, 2013 at 4:01
  • $\begingroup$ About the $\sqrt{a}+\sqrt{b}$ problem, there are other approaches. $\endgroup$ Feb 5, 2013 at 4:06
  • $\begingroup$ @AndréNicolas Thanks for pointing that out! I'm working on a shorter proof and will post it when it's properly written up. :) $\endgroup$ Feb 5, 2013 at 4:18
  • $\begingroup$ @WillJagy thanks for the comment, but I'm not sure how that can be applied to this question! would you care to elaborate? $\endgroup$ Feb 5, 2013 at 4:55
  • $\begingroup$ Hm $12 \sqrt{3} = 432 ... $ just wondering ... $\endgroup$
    – Calvin Lin
    Feb 5, 2013 at 6:03

1 Answer 1


We start from the equation $n_1^{\frac{1}{e_1}}+n_2^{\frac{1}{e_2}}=n_3^{\frac{1}{e_3}}$.

Write it as $((n_1)^{e_2e_3})^{\frac{1}{e_1e_2e_3}}+((n_2)^{e_1e_3})^{\frac{1}{e_1e_2e_3}}=((n_3)^{e_1e_2})^{\frac{1}{e_1e_2e_3}}$.

This now reduces to the Inverse Fermat equation $x^{\frac{1}{n}}+y^{\frac{1}{n}}=z^{\frac{1}{n}}$, for which the solution is found here: http://www.math.leidenuniv.nl/~hwl/PUBLICATIONS/1992d/art.pdf

The crux is to show that $(\frac{x}{y})^{\frac{1}{n}}$ is rational. This will then imply that $x=a^nd, y=b^nd, z=(a+b)^nd$, where $gcd(a, b)=1$.

Now for the question $\sqrt{a}+\sqrt{b}=12\sqrt{3}$, $a=0, b=0$ contribute 2 solutions, otherwise $a, b>0$.

We get $a=x^2d, b=y^2d, x, y>0, gcd(x, y)=1, 12^2(3)=(x+y)^2d$, so $x+y$ is a factor of 12 which is $>1$. For each such factor $f=x+y$, each number $\leq f$ that is relatively prime to $f$ gives a solution, so we get $\sum\limits_{i \mid 12, i>1}{\varphi (i)}=11$. Thus total number of solutions is $11+2=13$.


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