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

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 $n^2$.
