Let \( f(n) = { a }_{ 1 }^{ n }+{ a }_{ 2 }^{ n }+\ldots+{ a }_{ n }^{ n }\) where all variables are positive integers. For instance

\[\begin{eqnarray} f(2) &=& {a}_{1}^{2} + {a}_{2}^{2} \\ f(3) &=& {a}_{1}^{3} + {a}_{2}^{3} + {a}_{3}^{3} \end{eqnarray} \]

and so on.

I propose the following Conjecture:

There are infinitely many valid solutions for \( f(2) = f(3) = f(4) = \ldots \).

For instance for \(f(2) = f(3)\), we have \[ 73 = {3}^{2} + {8}^{2} = {1}^{3} + {2}^{3} + {4}^{3}.\] Similarly for \(f(2) = f(3) = f(4)\) we have \[11378 = {67}^{2} + {83}^{2} = {1}^{3} + {9}^{3} + {22}^{3}\ = {1}^{4} + {3}^{4} + {6}^{4} + {10}^{4}.\]

In my opinion there are infinitely many solutions for each case. Can anyone share insights on how to prove this conjecture?

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TopNewestExtension to my comment. I claim the following:

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

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 \(F_2 \cap F_3 \cap F_4\):

\[\begin{align*} a_2 &= 25 &=& 3^2 + 4^2 \\ a_3 &= 10 &=& 1^3 + 1^3 + 2^3 \\ a_4 &= 2500 &=& 5^4 + 5^4 + 5^4 + 5^4 \end{align*}\]

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:

\[\begin{align*} 2500000^2 a_2 &= 156250000000000 =& 7500000^2 + 10000000^2 \\ 25000^3 a_3 &= 156250000000000 =& 25000^3 + 25000^3 + 50000^3 \\ 500^4 a_4 &= 156250000000000 =& 2500^4 + 2500^4 + 2500^4 + 2500^4 \end{align*}\]

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,

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?) – Ivan Koswara · 1 year, 9 months ago

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this video on Numberphile and ended up thinking about this problem. Since as mentioned in that video you can not represent numbers of form 9k+4 or 9k+5 as sum of 3 cubes, I was thinking whether there is some way we can narrow down solution set for my problem for higher values of n, by eliminating such numbers which are not possible. – Vikram Pandya · 1 year, 9 months ago

Thanks for your analysis, I recently had seenLog in to reply

– Michael Mendrin · 1 year, 9 months ago

Really nice analysis!Log in to reply

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. – Ivan Koswara · 1 year, 9 months ago

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– Vikram Pandya · 1 year, 9 months ago

Thanks, I am not good with notations. That's why I had given examples to make it more clear. Nice rephrasing but I fail to understand how proving non-empty set would help in proving infinitely many solutions. I mean even if there is a single solution, the set would be non-empty, right? I may be missing something. May be a stupid question.Log in to reply

– Ivan Koswara · 1 year, 9 months ago

I've shown how to generate infinitely many solutions from a single solution. Thus if we can find any one solution, we can use my method to generate infinitely many solutions from that. This is similar to, for example, we have the Pythagorean triple \((3,4,5)\) and we can generate infinitely many from that: \((6,8,10), (9,12,15), (12,16,20), \ldots\).Log in to reply

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. – Michael Mendrin · 1 year, 9 months ago

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this video on Numberphile and ended up thinking about this problem. I liked that subtle reference to Fermat :) – Vikram Pandya · 1 year, 9 months ago

Nice to see you here. Sorry for such a delayed response as I was bit busy. I recently had seenLog in to reply

– Michael Mendrin · 1 year, 9 months ago

I think it's safe to say that if we modified your conjecture to include the condition that all of the numbers had to be different (like your posted problem related to this), this becomes a much more difficult feat to prove.Log in to reply

EDIT: Actually, no, it's not much more difficult. Multiply the number with a big \(a^{k!}\) number. For example, for the solution \(11378\) above, multiply it by \(2^{24}\). The expressions are now \((2^8)^3 + (9 \cdot 2^8)^3 + (22 \cdot 2^8)^3\) and \((2^6)^4 + (3 \cdot 2^6)^4 + (6 \cdot 2^6)^4 + (10 \cdot 2^6)^4\); the \(1\)s are no longer equal. – Ivan Koswara · 1 year, 9 months ago

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Interesting-Equality-and-Sum-of-powers – Michael Mendrin · 1 year, 9 months ago

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– Vikram Pandya · 1 year, 9 months ago

I agree, with the condition that all the numbers must be distinct positive integers, this problem becomes insane. That's the reason I filed that other question under Computer Science but this under number theory.Log in to reply

On the other hand, finding the smallest solution is almost always a CS problem. Most constructions in math aren't specifically engineered to find the smallest solution. – Ivan Koswara · 1 year, 9 months ago

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– Vikram Pandya · 1 year, 9 months ago

Sorry, I had missed your earlier comment. So let's see how can we tackle this problem in its reduced form. Thanks again for your analysis.Log in to reply