Interesting Power Equality

Extension to my comment. I claim the following:

Suppose $a_2, a_3, a_4, \ldots, a_k$ are positive integers such that:

• $a_i \in F_i$ for all $i$
• $\frac{a_m}{a_n}$ is the $\gcd(m,n)$-th power of some rational, for all $m,n$

Then $F_2 \cap F_3 \cap \ldots \cap F_k$ is non-empty.

The idea is that we represent $a_i$ as the sum of $i$ numbers that are $i$-th powers, and then multiply them with some $i$-th power so that all of them are equal. Of course, because there are things like $a_2$ multiplied by a square and $a_4$ multiplied by a fourth power (which is also a square), we cannot pick the $a_i$ randomly, otherwise we might get stuck (say, $a_2$ has a single factor of $2$ and $a_4$ has two, if you try the obvious $2 = 1^2 + 1^2$ and $4 = 1^4 + 1^4 + 1^4 + 1^4$; you can't make them equal).

The second condition is the formalization of "we can't pick them randomly". By Bezout's identity, we can find $p,q$ so that $qn-pm = \gcd(m,n)$; we make sure that the quotient $\frac{a_m}{a_n}$ is a $\gcd(m,n)$-th power, so that we can multiply $a_m$ by a $pm$-th power and $a_n$ by a $qn$-th power, which by canceling is equivalent to multiplying $a_n$ by a $\gcd(m,n)$-th power, what we desire to make $a_m$ and $a_n$ equal.

This way, we only need to find $a_2, a_3, \ldots, a_k$; this is more freedom than finding a single member of $F_2 \cap F_3 \cap \ldots \cap F_k$ directly. I believe that it should be enough to show there always exists a solution, but I don't have a proof yet. For example, we can use this to find a member of $F_2 \cap F_3 \cap F_4$:

$\begin{aligned} a_2 &= 25 &=& 3^2 + 4^2 \\ a_3 &= 10 &=& 1^3 + 1^3 + 2^3 \\ a_4 &= 2500 &=& 5^4 + 5^4 + 5^4 + 5^4 \end{aligned}$

The only possible problem from the second condition is $m = 2, n = 4$. (For $(m,n) = (2,3), (3,4)$, both of them give $\gcd(m,n) = 1$, and all rationals are the first power of some rational, namely themselves, so this condition is trivial.) The quotient $\frac{a_2}{a_4}$ is $\frac{1}{100}$, which is a $\gcd(2,4) = 2$-nd power of a rational $\frac{1}{10}$. So the second condition is satisfied; we should be able to find the multipliers.

We can indeed multiply $a_2$ by $2500000^2$, $a_3$ by $25000^3$, $a_4$ by $500^4$ to get:

$\begin{aligned} 2500000^2 a_2 &= 156250000000000 =& 7500000^2 + 10000000^2 \\ 25000^3 a_3 &= 156250000000000 =& 25000^3 + 25000^3 + 50000^3 \\ 500^4 a_4 &= 156250000000000 =& 2500^4 + 2500^4 + 2500^4 + 2500^4 \end{aligned}$

Thus $156250000000000 \in F_2 \cap F_3 \cap F_4$.

Finding the multipliers above ($2500000^2, 25000^3, 500^4$) can be done by solving simple modular equations. Suppose the multipliers are $m_2^2, m_3^3, m_4^4$ for $a_2, a_3, a_4$ respectively. Then,

• $m_2 = 2^{m_{2,2}} 5^{m_{2,5}} k_2$
• $m_3 = 2^{m_{3,2}} 5^{m_{3,5}} k_3$
• $m_4 = 2^{m_{4,2}} 5^{m_{4,5}} k_4$
• $2 m_{2,2} + 0 = 3 m_{3,2} + 1 = 4 m_{4,2} + 2$ (the exponents of the $2$: $25, 10, 2500$ has $0, 1, 2$ as the exponent of the $2$ respectively)
• $2 m_{2,5} + 2 = 3 m_{3,5} + 1 = 4 m_{4,2} + 4$ (the exponents of the $5$)
• $k_2^2 = k_3^3 = k_4^4$ (the remaining factor)

Note that for each prime factor, we have one degree of freedom. We can use modular arithmetic to figure out what values that any one particular variable can take; we simply choose any of them, which will propagate to the rest. In this case, we happen to select $m_{2,2} = 5, m_{3,2} = 3, m_{4,2} = 2$ and $m_{2,5} = 7, m_{3,5} = 5, m_{4,5} = 3$ as the smallest solution. Also, we can simply choose $k_i = 1$ that satisfies everything; we can in fact choose $k_i = a^{k!/i}$ for any $a$, as given in my previous comment to give infinitely many solutions.

(Yes, I'm still not that good in LaTeX. Someone can fix up the aligning?)

Because the notation is so awful, let me rephrase it: $F_k$ is the set of positive integers that can be represented as $a_1^k + a_2^k + a_3^k + \ldots + a_k^k$ where $a_1, a_2, a_3, \ldots, a_k$ are positive integers. The conjecture is thus $F_2 \cap F_3 \cap F_4 \cap \ldots \cap F_k$ is infinite for any positive integer $k$.

The easier part is that if said intersection is non-empty, then it is infinite. If $n \in F_2 \cap \ldots \cap F_k$, then $a^{k!}n \in F_2 \cap \ldots F_k$ for all positive integer $a$, because if $n = a_1^i + a_2^i + \ldots + a_i^i$ then $a^{k!}n = (a^{k!/i}a_1)^i + (a^{k!/i}a_2)^i + \ldots + (a^{k!/i}a_i)^i$. Thus now the problem is reduced to proving that said intersection is non-empty.

Vikram, this sounds like the kind of problem which proof you've written down somewhere in the margin of a book, right? This is similar to, but unfortunately too different from, Waring's Problem.