Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

${n_1}^{1/e_1}+{n_2}^{1/e_2}={n_3}^{1/e_2}$


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?

share|improve this question
    
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 –  Will Jagy Feb 5 '13 at 4:01
    
About the $\sqrt{a}+\sqrt{b}$ problem, there are other approaches. –  André Nicolas Feb 5 '13 at 4:06
    
@AndréNicolas Thanks for pointing that out! I'm working on a shorter proof and will post it when it's properly written up. :) –  Vincent Tjeng Feb 5 '13 at 4:18
    
@WillJagy thanks for the comment, but I'm not sure how that can be applied to this question! would you care to elaborate? –  Vincent Tjeng Feb 5 '13 at 4:55
    
Hm $12 \sqrt{3} = 432 ... $ just wondering ... –  Calvin Lin Feb 5 '13 at 6:03
show 3 more comments

1 Answer

up vote 5 down vote accepted

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 $11+2=13$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.