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
 A: Let the first $y$ consecutive integers be $m,m+1,\ldots,m+y-1$; then the second $y$ integers are $m+y,m+y+1,\ldots,m+2y-1$. Thus, if we subtract the first sum from the second, we have this:
$$\begin{array}{ccc}
&(m+y)&+&(m+y+1)&+&(m+y+2)&+&\ldots&+&(m+2y-1)\\
(-)&m&+&(m+1)&+&(m+2)&+&\ldots&+&(m+y-1)\\ \hline
&y&+&y&+&y&+&\ldots&+&y
\end{array}$$
There are $y$ columns, so the difference is indeed $y\cdot y=y^2$.
A: It is true in general. For each of the $y$ numbers in the first bunch  of integers, the corresponding number in the second bunch is obtained  by adding $y$ to the number in the first bunch. We do this $y$ times, so the difference is $(y)(y)$.
A: Hint $\ \ (A_{ 1}\!+\color{#0a0}K) + \cdots +(A_{\color{#c00}{\large N}}\!+\color{#0a0}K)\, =\, (A_1\!+\cdots +A_N)\!\!\!\!\!\!\!\! \underbrace{\, +\ \color{#c00}N\cdot \color{#0a0}K}_{\quad\ \ \ \large =\ \color{#c00}{N^{\Large 2}}\ {\rm if}\ K = N} $ 
Remark $\ $ The terms $\,A_i\,$ need not be consecutive integers. Rather, what makes the total increment $\,NK\,$ square is that the term increment $\,\color{#0a0}K\,$ equals the number of terms $\,\color{#c00}N$.
A: For an algebraic proof:
In general we have $1+2+\cdots+n=\frac{1}{2}n(n+1)$ so in this case
$x=\frac{1}{2}n(n+1)$ and $y=\frac{1}{2}(2n)(2n+1)-\frac{1}{2}n(n+1)$.
Thus $y-x=\frac{1}{2}(2n)(2n+1)-n(n+1)=2n^2+n-n^2-n=n^2$ as desired.
A: 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 $z-x = (15 + 5\times 5) - 15 = 5\times 5$
A: Consider the $n$th triangular number $T(n)$, which is the sum of $n$ natural numbers $1$ to $n$, and which has the closed form  $T(n)=\tfrac{1}{2}n(n+1)=\binom{n+1}{2}$ where $\binom{n+1}{2}$ is a binomial coefficient (see https://en.wikipedia.org/wiki/Triangular_number). Then we can look at the differences of triangular numbers to prove the hypotheses and from which formulate a binomial identity that always gives $n^2$. 
Starting at integer $k\ge1$ and adding $n$ consecutive integers gives:
\begin{align}
k+(k+1)+(k+2)+\dotsb+(k+n-1)&=\\
\tfrac{1}{2}(k+n-1)(k+n)-\tfrac{1}{2}(k-1)k&=\tfrac{1}{2}(2kn+n^2-n)\qquad(A)
\end{align}
Next starting at integer $k+n$ and adding $n$ consecutive integers gives:
\begin{align}
(k+n)+(k+n+1)+(k+n+2)+\dotsb+(k+2n-1)&=\\
\tfrac{1}{2}(k+2n-1)(k+2n)-\tfrac{1}{2}(k+n-1)(k+n)&=\tfrac{1}{2}(2kn+3n^2-n)\qquad (B)
\end{align}
Then equation (B) minus equation (A) gives $n^2$. 
Written out as a sum of triangular numbers gives us a sum of binomial coefficients which we can explicitly write out and simplify to give us $n^2$ showing the equivalence between the two (and maybe inspiring a binomial proof):
\begin{align}
\big(T(k+2n-1)-T(k+n-1)\big)-\big(T(k+n-1)-T(k-1)\big)&=\\
T(k+2n-1)-2T(k+n-1)+T(k-1)&=\\
\binom{k+2n}{2}-2\binom{k+n}{2}+\binom{k}{2}&=\\
\frac{(k+2n)!}{2!(k+2n-2)!}-2\frac{(k+n)!}{2!(k+n-2)!}+\frac{k!}{2!(k-2)!}&=\\
\tfrac{1}{2}\big((k+2n)(k+2n-1)-2(k+n)(k+n-1)+k(k-1)\big)&=\\
\tfrac{1}{2}\cdot 2n^2&=n^2.
\end{align}
Hence we see that whatever value of $k\ge1$ we take as our starting point is irrelevant to the final value of $n^2$.
A: We all recall the story of Gauss and his reported class room adventure of having to sum the numbers from 1 to 100.  And the approach he took. I wonder how he would have approached the teachers next question, "What do you get when you have for, example, the 100 numbers from 100 to 199 and I ask you subtract the  the sum of the first 50 from the sum of the last 50?"
Young Gauss would say 2500 and when questioned he might have responded:
I could have done it my old way twice, summing 1-50 and 51-199 and then subtract but that would be boring and I think not.  I can think of the numbers 100 thru 199 as being a set of one-hundred 99's (1 less than my starting number) with the numbers 1 through 100 tacked in sequence to again have the original set 100-199.   Then if I take the first 50 from the last 50 all the 99’s would fall away and I’d be left with 1 through 100 to deal with. I would need to subtract the first set (1-50) from the second set 51-100 like this: 51-1 = 50, 52-2=50, 53-3=50. I see the pattern! Bingo! (or the like) Fifty 50’s is 2500.
In general if one starts with an even number, k, of consecutive integers beginning with y, think of them being built built by adding 1 to k to y-1.  Then when you subtract the first half from the second all the (y-1) will fall away and you will be left with 1 to k which is easy to deal with (i.e subtracting the first half from the second. (k/2)^2
A: First, assume that your starting number is 0.  Then you have sum from i= y to 2y-1 of i, less sum from i = 0 to y-1 of i.  Comparing terms, you see that the difference in each case is y.  Since you have y terms, the result is y^2.   
If you have a different starting number, w, then it is added to each term in both summations, so it cancels.   
I wish knew how to get past plain vanilla text and post math notation in here.  I guess it is time to look into mathjax.
A: Another way to look at your problem is to consider a starting number of 0.  Then you have your first summation the difference between two triangle numbers, T(2y)-T(y).  The second summation is just the triangle number T(y).  So the difference between these two summations is T(2y)-2T(y).  Recall that the formula for a triangle number T(y) is y(y+1)/2, and do the algebra.  You will find that the difference is y^2.   
For starting number other than zero the result is the same because the starting number goes equally into the two summations and is cancelled.  
Always remember that the triangle number is the sum of consecutive integers, and is given by n(n+1)/2, as discovered by Gauss.  You can prove this to yourself by writing the upper half of the integers backward under the lower half.  You have n/2 pairs that add up to n+1.   See https://nrich.maths.org/2478.   Then go to the bowling alley to see T4 in action.
