It may be instructive to point out that the

queen positions

in

a

regular

solution are

part

of a lattice

that

extends to

the

entire

plane.

This can be seen

algebraically by

first

observing

that the

mapping

k:

Z

x

Z

->

Z

x

Zn

defined

by ((x, y))4

=

([x], [y])

is a

group homomorphism.

Let H

be the

cyclic

subgroup

of

Z x Z

generated by

(u, v).

Then the elements of

(HO)4f`

may

be

interpreted

geometrically

as the

positions

of the

queens

obtained

by tiling

the entire

plane

with

copies

of

the

chessboard having queens

located

on

the

squares

Ho.

However,

(HO)O`

is a

subgroup of Z

x

Z, and

therefore it

is a lattice of dimension

two, having

a fundamental

region

of area

equal

to the

absolute

value of the determinant of the matrix

gotten by expressing the lattice basis in terms of

the canonical

basis

of Z

x

Z. Now

an

arbitrary regular

solution

to the

n-queens problem

is

a

translation

of

Ho

for

some cyclic group

H

of

Z

x

Z,

and

this

corresponds

to

the

same

translation

in Z

x

Z

of the

subgroup

(H4)f

'.

Fermat's Two Square

Theorem

In

order to prove Fermat's

result,

we need to show that there

is

a

regular,

doubly symmetric

solution

to

the p-queens problem

whenever

p

is

a

prime

of the form

4k

+

1. To do

this,

we

will

count

the

total

number

of

regular

solutions

for the

p

x

p

board

in

two different

ways.

LEMMA. The number of regular solutions

to

the

p-queens problem,

where

p

is a

prime,

is

p(p

-

3).

Proof. We know that regular solutions have the form

(p, c)

+

((1, d)). Clearly

we do not

get

a

solution

when d

is

p, p

-

1,

or 1. But

any

of

the

other

p

-

3

possibilities

for

d,

in

Zp,

will

produce

regular solutions,

since in these cases

d

-

1, d,

and

d

+

1 will

each be

relatively prime

to

p.

Since

c can

take

on

any

of

p val ues,

the total number of

regular

solutions

is

p(p

-

3).

A

second

way

of

counting

the

regular

solutions

is

to

partition

them into three

classes, depending

upon

their

symmetry

-

doubly

symmetric, symmetric (invariant

under

a

1800 rotation but not a 900

rotation),

or

nonsymmetric (no

symmetry).

The

symmetries

of the

square

consist of

four reflections

and four

rotations,

and these form

a

group G,

under

composition.

If x

denotes a

regular

solution and

U

E G,

the

regular

solution which results from

x

by applying

the

transformation

U will be denoted

by

xU.

Two

regular

solutions

x

and

y

are

essentially

the

same

if and

only

if

there exists

a

U

E

G such that

xU

=

y.

This is an

equivalence

relation

on the set of all

regular

solutions. The

equivalence

class

of a

solution x consists of all those

solutions

that can be

obtained

from

x

by

rotation

and

reflection. For

each

regular

solution

x,

let

HI-

denote the set of

all

symmetries

of

x;

that

is,

HI

=

{U

E

G

I

xU

=

x}.

HI-

is a

subgroup

of

G.

Furthermore,

if

U,

V

E

G,

xU

=

xV

if

and

only

if

xUV-

=

x,

and

this

happens

if and

only

if

UV-

E

HI-.

It

follows that the number of elements in the

equivalence

class

of

x is

equal

to the index of

HI-

in

G,

and

this

is

2, 4,

or 8

depending upon

whether

x

is

doubly symmetric,

symmetric,

or

nonsymmetric

respectively.

Thus

we

have

the

following

lemma:

LEMMA. The number of regular

solutions to the n-queens problem is 2x

+

4y

+

8z,

where x, y, z are,

respectively,

the number

of essentially

different doubly symmetric, symmetric, and nonsymmetric

regular

solutions to the

n-queens problem.

Now

suppose

that

p

is a

prime

of the

form

4k

+

1.

Combining

the results

of the

preceding

two

lemmas,

we know

that

p(p -3)

=

2x

+

4y

+

8z.

Since,

in this

equation, p

1

(mod

4),

it

becomes

2

2x

(mod 4).

But

this

completes the

proof,

since

this

last

equation implies that x, the number of

essentially

different

regular, doubly

symmetric solutions,

is

not zero.

Finally, we can show that the

positive integers u and v in Fermat's result are unique.

To do this, it

is

sufficient to

prove that

there

are

only two regular, doubly symmetric solutions to the

p-queens

problem

-

a

single solution, and its

horizontal reflection, both of which induce the

same positive

integers

u

and v

for

which

u2

+

v2

=

p.

So, suppose

that

(p, c)

+

((1, d))

is

a

regular,

doubly symmetric

solution

to the

p-queens problem. Since

a

queen

is

located

on

the square (1, [c

+

d]), there must also

be

(by

rotational

symmetry)

a

queen

on the square ([c

+

d], p). This means that

VOL. 50, NO. 2, MARCH 1977

73

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