How many squares in the $m \times n$ grid

Question

Is there a formula to evaluate the number of all squares in the $m \times n$ grid? Well, I'm just curious, I've seen the question like this somewhere at the university, to solve this they were dividing the grid with $m - 1$ and $n - 1$ lines...I don't know what's next.

Answer 1

Suppose $n\ge m$.

• Number of squares of size 1: $m\cdot n$
• Number of squares of size 2: $(m-1)\cdot (n-1)$
• ...
• Number of squares of size m: $1\cdot (n-m+1)$

Result: $$\begin{align} \sum_{k=1}^m k \cdot (n-m+k) & =(n-m)\sum_{k=1}^m k +\sum_{k=1}^m k^2 \\ & = (n-m) m(m+1)/2 + m(m+1)(2m+1)/6 \\ & = \frac{m(m+1) (3n-m+1)}{6}\end{align}$$

Answer 2

Here's an induction approach. Let's pretend we don't know the formula as yet. Let $f(n, m)$ be the number of squares in a $n \times m$ rectangle. WLOG $n \geq m$

When $n =m$, we know that there are $ f(m, m) = \sum_{k=1}^m k^2 = \frac{ m(m+1)(2m+1)}{6} $ squares.

Suppose we have the value $f(n, m)$. Consider adding another row at the bottom. How many squares do we introduce, that involve that bottom row? If we pick any 2 vertical grid lines, they uniquely define a square that involves the bottom row, hence we're adding ${ m +1 \choose 2 } = \frac{ m(m+1) } { 2}$ squares. Thus, there are $f(n+1, m) = f(n, m) + \frac{ m(m+1) } { 2}$ squares.

This tells us that, for a fixed $m$,

• $f(n+1, m)$ is linear, with a slope of $\frac{m(m+1) } { 2}$.
• $f(m, m ) = \frac{ m(m+1)(2m+1)}{6}$.
• Using point slope form, we conclude that $f(n, m) = \frac{ m(m+1) ( 3n-m + 1 ) } { 6}$.

Answer 3

Rectangles in rectangle $$\frac{(n^2+n)(m^2+m)}{4}$$

Rectangles in square $$\frac{(n^2+n)^2}{4}$$

Squares in rectangle $$m≥ n-1,\frac{(n^2+n)}{2}m-\frac{(n^3-n)}{6}$$

Squares in square $$\frac{(n^2+n)(2n+1)}{6}$$