Is difference of two consecutive sums of consecutive integers (of the same length) always square?

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 $y\cdot y=y^2$.


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 $(y)(y)$.


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 $\,\color{#c00}N$.

Tags:

Summation

Elementary Number Theory

Square Numbers

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