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I'm looking at examples of numbers that can be written as the sum of integers from $j$ to $k$ and from $k+1$ to $l$. For example $15$ which can be written as $4+5+6$ or $7+8$. Or $27 = 2+3+4+5+6+7 = 8+9+10$. I have been able to find a few numbers which have two ways to satisfy the above equations. For example,

$$\begin{aligned}105 &= 1+2+\dots +14 = 15+16+\dots+20\\ &= 12+13+\dots+18 = 19+20+\dots+23 \end{aligned}$$

However, I have not been able to find any numbers that can be written as the sum in three ways of consecutive sums. That is, I have not been able to find an $X$ such that,

$$\begin{aligned}X &= (a+1)+(a+2)+\dots +b = (b+1)+(b+2) +\dots +c\\ &= (d+1)+(d+2)+\dots +e = (e+1)+(e+2) +\dots +f\\ &= (g+1)+(g+2)+\dots +h = (h+1)+(h+2) +\dots +i\\ \end{aligned}$$

Does any such number $X$ exist? If so can you provide an example? If no such number exists can you provide a proof?

Thanks

Tito Piezas III
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Ryan
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  • Write the problem statement correctly. It is not clear which system of equations must be solved. There should is how much? 6 equations? – individ Aug 25 '15 at 06:45
  • @Ryan: Note that chenyuandong's result implies that for two-ways, $$(6 + i)^4 - (6 - i)^4 = (5 + 2i)^4 - (5 - 2i)^4 = 2^4\times \color{brown}{105}, i$$ while for three-ways, $$(77 + 38i)^4 - (77 - 38i)^4 = (78 + 55i)^4 - (78 - 55i)^4 = (138 + 5i)^4 - (138 - 5i)^4 \= 2^4\times \color{brown}{6561555}, i$$ using David's example. – Tito Piezas III Aug 25 '15 at 15:37

2 Answers2

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Following chenyuandong's answer, you need to find various $x,y$ with the same value of $xy(x^2-y^2)$. A search with Maple (I know, boring) gives $$x=77,\ y=38\ ;\quad x=78,\ y=55\ ;\quad x=138,\ y=5$$ which leads to $$\eqalign{ 684+\cdots+3686=3687+\cdots+5168&=6561555\cr 2761+\cdots+4554=4555+\cdots+5819&=6561555\cr 8820+\cdots+9534=9535+\cdots+10199&=6561555\ .\cr}$$ There is also a solution with sum $531485955$ and $y<x\le500$ but I don't have the details. Will post later if I have time ;-)

Tito Piezas III
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David
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  • I changed your notation to be consistent with chenyuandong's answer (since he already uses $a,b$). I hope you don't mind. – Tito Piezas III Aug 25 '15 at 13:42
  • I find the sum $531485955$ involves $x,y= 213,114;; 234,165; ; 414,15$ so using the formulas for $a,b,c$ that I added for easy reference, then, $$6152+\dots+33178=33179+\dots+46516=531485955$$ $$24845+\dots+40990=40991+\dots+52375=531485955$$ $$79376+\dots+85810=85811+\dots+91795=531485955$$ – Tito Piezas III Aug 25 '15 at 15:47
  • Thank you so much David, Tito and Chenyuandong. This is exactly what I was looking for. Very helpful. – Ryan Aug 26 '15 at 03:41
  • @TitoPiezasIII Thank you Tito for correcting my answer. – chenyuandong Aug 26 '15 at 14:50
  • @chenyuandong: I only edited it for clarity. :) It was brilliant how you transformed the problem to the Pythagorean triple $$(2a-2c)^2+(2a+2c+2)^2=(4b+2)^2$$ – Tito Piezas III Aug 26 '15 at 14:57
  • @TitoPiezasIII I think that you meant 231 not 213 in your second solution above. This makes your second solution of X,Y just 3x the first solution. In fact one could multiply the first set of solutions by any odd number to get a different solution. – Ryan Feb 22 '16 at 22:16
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Suppose,

$$X=(a+1)+(a+2)+\dots+b=(b+1)+(b+2)+\dots+c$$

which implies,

$$\Leftrightarrow \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2}$$

$$2b(b+1)=a(a+1)+c(c+1)$$

or the special Pythagorean triple,

$$(2a-2c)^2+(2a+2c+2)^2=(4b+2)^2$$

where $a,b,c\in\mathbb Z$. We need to find integer solutions of the system,

$$(2c-2a)=s^2-t^2,\quad (2a+2c+2)=2st,\quad (4b+2)=s^2+t^2$$

So,

$$a=\frac{|s^2-t^2-2st|-2}{4},\quad b=\frac{s^2+t^2-2}{4},\quad c=\frac{2st+(s^2-t^2)-2}{4}$$

and just let $s=(2m+1),t=(2n-1)$, where $m\geq n$. Since,

$$X=\bigg(c+\frac{1}{2}\bigg)^2-\bigg(b+\frac{1}{2}\bigg)^2=\bigg(\frac{2st+(s^2-t^2)}{4}\bigg)^2-\bigg(\frac{s^2+t^2}{4}\bigg)^2\\ =\frac{st(t+s)(s-t)}{4}=(2m+1)(2n-1)(m-n+1)mn$$

If the equation below has 3 or more integer solutions then the $z$ is exactly what you want:

$$z=(2m+1)(2n-1)(m-n+1)mn$$

where $m\geq n$. Alternatively,

$$\begin{aligned} a &= \tfrac{1}{2}\Big(-1+\sqrt{(x^2-2xy-y^2)^2}\Big)\\ b &= \tfrac{1}{2}\big(-1+x^2+y^2\big)\\ c &= \tfrac{1}{2}\big(-1+x^2+2xy-y^2\big) \end{aligned}$$

where $x>y$ and sign chosen so $a$ is positive. Then,

$$X = \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2} =\tfrac{1}{2}xy(x^2-y^2)$$

so it suffices to find three pairs of $x,y$ with the same $X$.

Tito Piezas III
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chenyuandong
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  • For this equation the formula can be found there. http://math.stackexchange.com/questions/794510/curves-triangular-numbers – individ Aug 25 '15 at 04:30
  • This gives solutions for $(a+1)+\cdots+b=(b+1)+\cdots+c$. However the OP wants multiple solutions for the same value of $X$, if I understood the question correctly. – David Aug 25 '15 at 05:38