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:
- Note that $a,b$ cannot both be perfect squares.
- We write $a=3^{k_1}\alpha, b=3^{k_2}\beta$, where $3∤α,β$
- By considering congruences, prove that $k_1, k_2$ must be of the same parity.
- Considering the case where $k_1, k_2$ are both even
- $\alpha,\beta$ cannot both be perfect squares.
- 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.
- 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 $f$.
- $12\sqrt{3}=C\sqrt{f}$, where $C$ is an integer. Then $f=3$, which is a contradiction.
An alternative, shorter proof by a friend:
- Note that $ab$ is perfect square.
- $a=0, b=0$ each give 1 solution. Otherwise, $a,b>0$.
- 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.
- ${(x+y)^2}d=432$, 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?