Interesting Power Equality

Let f(n)=a1n+a2n++ann f(n) = { a }_{ 1 }^{ n }+{ a }_{ 2 }^{ n }+\ldots+{ a }_{ n }^{ n } where all variables are positive integers. For instance

f(2)=a12+a22f(3)=a13+a23+a33\begin{aligned} f(2) &=& {a}_{1}^{2} + {a}_{2}^{2} \\ f(3) &=& {a}_{1}^{3} + {a}_{2}^{3} + {a}_{3}^{3} \end{aligned}

and so on.

I propose the following Conjecture:

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

For instance for f(2)=f(3)f(2) = f(3), we have 73=32+82=13+23+43. 73 = {3}^{2} + {8}^{2} = {1}^{3} + {2}^{3} + {4}^{3}. Similarly for f(2)=f(3)=f(4)f(2) = f(3) = f(4) we have 11378=672+832=13+93+223 =14+34+64+104.11378 = {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?

Note by Vikram Pandya
4 years, 1 month ago

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Extension to my comment. I claim the following:

Suppose a2,a3,a4,,aka_2, a_3, a_4, \ldots, a_k are positive integers such that:

  • aiFia_i \in F_i for all ii
  • aman\frac{a_m}{a_n} is the gcd(m,n)\gcd(m,n)-th power of some rational, for all m,nm,n

Then F2F3FkF_2 \cap F_3 \cap \ldots \cap F_k is non-empty.

The idea is that we represent aia_i as the sum of ii numbers that are ii-th powers, and then multiply them with some ii-th power so that all of them are equal. Of course, because there are things like a2a_2 multiplied by a square and a4a_4 multiplied by a fourth power (which is also a square), we cannot pick the aia_i randomly, otherwise we might get stuck (say, a2a_2 has a single factor of 22 and a4a_4 has two, if you try the obvious 2=12+122 = 1^2 + 1^2 and 4=14+14+14+144 = 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,qp,q so that qnpm=gcd(m,n)qn-pm = \gcd(m,n); we make sure that the quotient aman\frac{a_m}{a_n} is a gcd(m,n)\gcd(m,n)-th power, so that we can multiply ama_m by a pmpm-th power and ana_n by a qnqn-th power, which by canceling is equivalent to multiplying ana_n by a gcd(m,n)\gcd(m,n)-th power, what we desire to make ama_m and ana_n equal.

This way, we only need to find a2,a3,,aka_2, a_3, \ldots, a_k; this is more freedom than finding a single member of F2F3FkF_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 F2F3F4F_2 \cap F_3 \cap F_4:

a2=25=32+42a3=10=13+13+23a4=2500=54+54+54+54\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=4m = 2, n = 4. (For (m,n)=(2,3),(3,4)(m,n) = (2,3), (3,4), both of them give gcd(m,n)=1\gcd(m,n) = 1, and all rationals are the first power of some rational, namely themselves, so this condition is trivial.) The quotient a2a4\frac{a_2}{a_4} is 1100\frac{1}{100}, which is a gcd(2,4)=2\gcd(2,4) = 2-nd power of a rational 110\frac{1}{10}. So the second condition is satisfied; we should be able to find the multipliers.

We can indeed multiply a2a_2 by 250000022500000^2, a3a_3 by 25000325000^3, a4a_4 by 5004500^4 to get:

25000002a2=156250000000000=75000002+100000002250003a3=156250000000000=250003+250003+5000035004a4=156250000000000=25004+25004+25004+25004\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 156250000000000F2F3F4156250000000000 \in F_2 \cap F_3 \cap F_4.

Finding the multipliers above (25000002,250003,50042500000^2, 25000^3, 500^4) can be done by solving simple modular equations. Suppose the multipliers are m22,m33,m44m_2^2, m_3^3, m_4^4 for a2,a3,a4a_2, a_3, a_4 respectively. Then,

  • m2=2m2,25m2,5k2m_2 = 2^{m_{2,2}} 5^{m_{2,5}} k_2
  • m3=2m3,25m3,5k3m_3 = 2^{m_{3,2}} 5^{m_{3,5}} k_3
  • m4=2m4,25m4,5k4m_4 = 2^{m_{4,2}} 5^{m_{4,5}} k_4
  • 2m2,2+0=3m3,2+1=4m4,2+22 m_{2,2} + 0 = 3 m_{3,2} + 1 = 4 m_{4,2} + 2 (the exponents of the 22: 25,10,250025, 10, 2500 has 0,1,20, 1, 2 as the exponent of the 22 respectively)
  • 2m2,5+2=3m3,5+1=4m4,2+42 m_{2,5} + 2 = 3 m_{3,5} + 1 = 4 m_{4,2} + 4 (the exponents of the 55)
  • k22=k33=k44k_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 m2,2=5,m3,2=3,m4,2=2m_{2,2} = 5, m_{3,2} = 3, m_{4,2} = 2 and m2,5=7,m3,5=5,m4,5=3m_{2,5} = 7, m_{3,5} = 5, m_{4,5} = 3 as the smallest solution. Also, we can simply choose ki=1k_i = 1 that satisfies everything; we can in fact choose ki=ak!/ik_i = a^{k!/i} for any aa, 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 - 4 years ago

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Really nice analysis!

Michael Mendrin - 4 years ago

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Thanks for your analysis, I recently had seen 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 - 4 years ago

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Because the notation is so awful, let me rephrase it: FkF_k is the set of positive integers that can be represented as a1k+a2k+a3k++akka_1^k + a_2^k + a_3^k + \ldots + a_k^k where a1,a2,a3,,aka_1, a_2, a_3, \ldots, a_k are positive integers. The conjecture is thus F2F3F4FkF_2 \cap F_3 \cap F_4 \cap \ldots \cap F_k is infinite for any positive integer kk.

The easier part is that if said intersection is non-empty, then it is infinite. If nF2Fkn \in F_2 \cap \ldots \cap F_k, then ak!nF2Fka^{k!}n \in F_2 \cap \ldots F_k for all positive integer aa, because if n=a1i+a2i++aiin = a_1^i + a_2^i + \ldots + a_i^i then ak!n=(ak!/ia1)i+(ak!/ia2)i++(ak!/iai)ia^{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 - 4 years ago

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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.

Vikram Pandya - 4 years ago

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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)(3,4,5) and we can generate infinitely many from that: (6,8,10),(9,12,15),(12,16,20),(6,8,10), (9,12,15), (12,16,20), \ldots.

Ivan Koswara - 4 years ago

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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 - 4 years ago

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Nice to see you here. Sorry for such a delayed response as I was bit busy. I recently had seen this video on Numberphile and ended up thinking about this problem. I liked that subtle reference to Fermat :)

Vikram Pandya - 4 years ago

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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.

Michael Mendrin - 4 years ago

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@Michael Mendrin Depending on which numbers that are different, my work above might still hold. If you only require numbers that make up a single expression to be different (like 1,9,221,9,22 making up 13+93+2231^3+9^3+22^3 and 1,3,6,101,3,6,10 making up 14+34+64+1041^4+3^4+6^4+10^4 to be different), but numbers between expressions to be allowed to be equal (11 is shared between them), simply modify the definition of FiF_i appropriately (so that it contains numbers that are the sum of ii distinct positive ii-th powers). If you don't even allow numbers between expressions to be equal, that becomes more difficult.

EDIT: Actually, no, it's not much more difficult. Multiply the number with a big ak!a^{k!} number. For example, for the solution 1137811378 above, multiply it by 2242^{24}. The expressions are now (28)3+(928)3+(2228)3(2^8)^3 + (9 \cdot 2^8)^3 + (22 \cdot 2^8)^3 and (26)4+(326)4+(626)4+(1026)4(2^6)^4 + (3 \cdot 2^6)^4 + (6 \cdot 2^6)^4 + (10 \cdot 2^6)^4; the 11s are no longer equal.

Ivan Koswara - 4 years ago

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@Ivan Koswara Okay, I gotta tip my hat to your response to this one. Have you tried solving Pandya's posted problem related to this one? I think I'll try using your technique to solving this one.

Interesting-Equality-and-Sum-of-powers

Michael Mendrin - 4 years ago

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@Michael Mendrin 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.

Vikram Pandya - 4 years ago

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@Vikram Pandya Nope, I don't agree; I already stated how to fix through that condition above.

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 - 4 years ago

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@Ivan Koswara 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.

Vikram Pandya - 4 years ago

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