<s>[INST] Is difference of two consecutive sums of consecutive integers (of the same length) always square? I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks.
If $x$ is the sum of $y$ integers, and $z$ is the sum of the next $y$ integers, then is it always true that $z$ minus $x$ equals $y$ squared? Perhaps only starting at one.
For example:
$y = 3$
$x = (1+2+3) = 6$
$z = (4+5+6) = 15$
$z - x = 9 = y^2$
Brad
 [/INST] If you start at 1, then yes.
The sum of the first $n$ integers is $\frac{n(n+1)}{2}$.
Note that the next $n$ integers are simply $n^2 + \frac{n(n+1)}{2}$. That is, you can remove $n$ from each of the subsequent n integers, and just wind up with the sum of the first $n$ integers again.
For example:
$y = 5$
$x = 1 + 2 + 3 + 4 + 5 = 15$
$z = 6 + 7 + 8 + 9 + 10 = (1+5) + (2+5) + (3+5) + (4+5) + (5+5)$
$ = (1 + 2 + 3 + 4 + 5) + (5 + 5 + 5 + 5 + 5) = 15 + 5\times 5 $
So $z-x = (15 + 5\times 5) - 15 = 5\times 5$
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