Show that any given set of sixteen consecutive integers {$x+1,x+2,\ldots,x+16$} can be divided into two eight element subsets with the properties that they have the same sum, the sums of the squares of the elements are the same, and the sums of the cubes of the elements are the same. I know that each eight element subset has to have a sum of $$\frac{\text{total sum of sixteen integers}}{2} = 8x + 68.$$

I also assumed that $x = 0,$ and this simplified things so greatly that I actually found the answer, but by equating $x$ to $0$ you are assuming that you can do this with any sixteen consecutive integers, which is what you're trying to prove. And in a proof, you cannot use the information that you are trying to prove, so this is not a valid approach.

There must be a clever and simple solution to this problem other than brute forcing a general case, especially since computations would get exceedingly complicated with the expansion of quadratic and cubic polynomials.

I appreciate any and all help.




One has to use two theorems from equal sums of like powers. Assume the notation,

$$[a_1,a_2,\dots,a_m]^k = a_1^k + a_2^k +\dots +a_m^k$$

I. Tarry-Escott Theorem

(Escott, Quarterly Journal of Mathematics, 1910, pp. 141-167; Tarry, L’Intermediaire des Mathematiciens, vol. 19, 1912, pp. 219-221.)


$$[a_1,a_2,\dots,a_m]^k = [b_1,b_2,\dots,b_m]^k,\;\; \text{for}\, k = 1,2,3,\dots,n$$

then for any constant $c$,

$$[a_1,\dots,a_m,\,c+b_1,\dots, c+b_m]^k = [b_1,\dots,b_m,\,c+a_1,\dots, c+a_m]^k\\ \text{for}\, k=1,2,3,\dots, n+1$$

This doubles the number of terms, but you can climb the ladder of powers as high as you like. For example, starting with,

$$[1,4]^k = [2,3]^k,\;\; \text{for}\, k = 1$$

using the theorem with $c = 4$, we get,

$$[1,\,4,\,6,\,7]^k = [2,\,3,\,5,\,8]^k,\;\; \text{for}\, k = 1,2$$

Using the theorem again with $c = 8$, we have,

$$[1,\,4,\,6,\,7,\,10,\,11,\,13,\,16]^k = [2,\,3,\,5,\,8,\,9,\,12,\,14,\,15]^k,\;\; \text{for}\, k = 1,2,3\tag1$$

and so on.

II. Frolov's Theorem

(M. Frolov, Bulletin de la Societe Math de France, vol. 17, 1888, pp 69-83; vol. 20, 1892, pp.69-84.)


$$[a_1,a_2,\dots,a_m]^k = [b_1,b_2,\dots,b_m]^k,\;\; \text{for}\, k = 1,2,3,\dots,n$$

then for any constant $c$,

$$[c+a_1,\,c+a_2,\dots,c+a_m]^k = [c+b_1,\,c+b_2,\dots,c+b_m]^k \\ \text{for}\, k=1,2,3,\dots,n$$

Hence, using $(1)$,

$$[c+1,\,c+4,\,c+6,\,c+7,\,c+10,\,c+11,\,c+13,\,c+16]^k = [c+2,\,c+3,\,c+5,\,c+8,\,c+9,\,c+12,\,c+14,\,c+15]^k,\\ \text{for}\, k = 1,2,3$$

for any $c$. If we use integer $c$, then the terms will be $16$ consecutive integers.

  • $\begingroup$ Wow... this was absolutely brilliant. Thank you for making me a better mathematician. $\endgroup$ – A is for Ambition May 2 '15 at 12:08
  • $\begingroup$ @AisforAmbition: A similar problem, but not multi-grade (valid for multiple $k$), is discussed in this post. $\endgroup$ – Tito Piezas III May 2 '15 at 13:05

Hint: Show that if you have eight numbers $x+a_i$ and another eight $x+b_i$ which work for $x$, then the sets $y+a_i$ and $y+b_i$ work for any $y$.

Further suggestion: $y$ need not be an integer. So you could try $y=-7.5$ so that the sum, and the sum of cubes are both zero, and you can choose each set to be four pairs of the form $\pm a$ because $a-a=a^3-a^3=0$. Then you just need to make the squares work.


I will first start by writing my $16$ positive integers as $x+b$ with $-7\leq b\leq 8$ and from here we have: $$(x+a)^2=x^2+2ax+a^2, (x+a)^3=x^3+3ax^2+3a^2x+x^3$$

so from here one can see that if $a_1,\cdots,a_8$ are the integers we are looking for then: $$\begin{align}a_1+a_2+\cdots+a_8&=4 \\ a_1^2+a_2^2+\cdots+a_8^2&=172\\ a_1^3+a_2^3+\cdots+a_8^3 &=256 \end{align}$$

These are $3$ equations with a lot of variables $a_i\in\{-7,-6,\cdots,8\}$, in order to solve this equation we need to add an increasing order $a_1<a_2<\cdots<a_8$, because we are dividing the initials set into two subsets we can take WLOG the subset which contains $8$ so that we can assume without loss of generality that $a_8=8$ and the equations become: $$\begin{align}a_1+a_2+\cdots+a_7&=0 \\ a_1^2+a_2^2+\cdots+a_7^2&=108\\ a_1^3+a_2^3+\cdots+a_7^3 &=0 \end{align}$$ where $a_i\in\{-7,-6,\cdots,7\}$ and in order to continue in solving the equation we have to add some assumptions (or we can try all values using a computer) for example because the sum of the elements and their cubes is $0$ one can see that we can assume that $a_1=-a_7,a_2=-a_6,a_3=-a_5,a_4=-a_4=0$ (this assumption is not justified but it's going to help us eliminate two equations). now we have the remaining equation after simplification: $$a_1^2+a_2^2+a_3^2= 54$$

and this equation has the only solution (with order $a_1<a_2<a_3$), $a_1=-7,a_2=-2,a_3=-1$ and here we have found our solutions: $$x-7,x-2,x-1,x,x+1,x+2,x+7,x+8\\ y+1,y+6,y+7,y+8,y+9,y+10,y+15,y+16$$

Here we have found a solution so we answered the question but we did not find all solutions and this is a problem in this method


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