*Spoiler for New Scientist Enigma 1723: “Four Lots of Dots” *(Follow the link to see the puzzle.)

A square is formed by four line segments; if you go around the perimeter of the square in one direction then each segment is in a different quadrant of the plane (where the first quadrant includes the *+x* axis and goes up to but doesn’t include the *+y* axis, and so on). The segment in the first quadrant starts at (*x0*, *y0) *and ends at (*x0*+*∆1*, *y0*+*∆2*). Being in the first quadrant, and the square having sides of nonzero length, means *∆1*>0 and *∆2*≥0. Then the other two vertices are (*x0*+*∆2*, *y0–**∆1*) and (*x0*+*∆1*+*∆2*, *y0*+*∆2 –∆1*). The width (in the

*x*direction) of the square is

*∆1*+

*∆2*as is its height (in the

*y*direction). This means that if we have dots in an

*n*by

*n*array the largest allowable value for

*∆1*+

*∆2*is (

*n*–1). In that case there is only one way to make a square of that size in the array. If

*∆1*+

*∆2*is (

*n*–2) then there are two horizontal and two vertical positions for the square, or a total of four such squares; if (

*n*–3) there are nine such squares, and so forth.

So for instance, if *n*=4, we have 9 squares with *∆1*=1, *∆2*=0:

4 squares with *∆1*=2, *∆2*=0:

4 squares with *∆1*=1, *∆2*=1:

and 1 square each with *∆1*=3, *∆2*=0; *∆1*=1, *∆2*=2; and *∆1*=2, *∆2*=1:

for a total of 20.

In fact if you think about it, there will be:

- 1 square each with
*∆1*=*n*–1,*∆2*=0;*∆1*=*n*–2,*∆2*=1;*∆1*=*n*–3,*∆2*=2; …*∆1*=1,*∆2*=*n*–2 - 4 squares each with
*∆1*=*n*–2,*∆2*=0;*∆1*=*n*–3,*∆2*=1;*∆1*=*n*–4,*∆2*=2; …*∆1*=1,*∆2*=*n*–3 - 9 squares each with
*∆1*=*n*–3,*∆2*=0;*∆1*=*n*–4,*∆2*=1;*∆1*=*n*–5,*∆2*=2; …*∆1*=1,*∆2*=*n*–4

and so on. For *n*=2 this is just 1 x 1; for *n*=3, 1 x 2 + 4 x 1; for *n*=4, 1 x 3 + 4 x 2 + 9 x 1; and in general:

sum (

i= 1 ton-1) [i^{2}x (n–i)]=

nx sum (i= 1 ton-1) [i^{2}] – sum (i= 1 ton-1) [i^{3}]

As everyone knows (well, as everyone can Google), the first sum is (*n*–1)*n*(2*n*–1)/6 and the second is [(*n*-1)*n*/2]^{2}. So the number of squares for an array of *n* by *n* dots is

N(n) = (n–1)n^{2}(2n–1)/6 – [(n-1)n/2]^{2}

For *n* = 2 to 10 this is:

*N*(2) = 1

*N*(3) = 6

*N*(4) = 20

*N*(5) = 50

*N*(6) = 105

*N*(7) = 196

*N*(8) = 336

*N*(9) = 540

*N*(10) = 825

and, to answer the Enigma, *N*(*n*) = 4*n*^{2} for *n* = 7.