Download Tiling Square Rooms with Equal Stacks of Tiles or Solution Patterns

Tiling Square Rooms
with Equal Stacks
of Tiles
or
Solution Patterns of
x2 – ny = 1
by
Donald E. Hooley
Some Problems
• A box contains beetles and
spiders. There are 46 legs in the
box. How many belong to
beetles?
• 14 October 1066. The men of
Harold stood well together and
formed sixty and one squares.
When Harold joined the Saxons
were one mighty square of men.
How many in Harold’s army?
Diophantine Equations
6x + 8y = 1
x2 = 61y2 + 1
or
x2 - 61y2 = 1
Another Problem
A mildly eccentric floor tile layer
tiles only square rooms with
square tiles which come in
stacks of equal numbers of tiles.
Unfortunately the number of
tiles in each stack is unknown
and the tile layer always leaves
himself standing on one untiled
square near the center of the
room.
Simpler Quadratic
Diophantine Equation
x2 = ny + 1
or
x2 - ny = 1
Use Excel to explore positive
integer solutions.
2
x
– 6y = 1
Solutions
Diff
2
4
2
4
2
4
x
5
7
11
13
17
19
23
y
4
8
20
28
48
60
88
Diff
4
12
8
20
12
28
2
x
– 6y = 1
Taken modulo 6 we have
x2 = 1 modulo 6
But the quadratic residues mod 6
are 1, 4, 3, 4, 1 so if
x = 1 mod 6
or
x = 5 mod 6
then
x2 = 1 mod 6
2
x
– 6y = 1
Thus
x = 6t + 1
or
x = 6t - 1
Since x2 – 6y = 1 gives
y = (x2 - 1)/6
2
x
– 6y = 1
If
x = 6t + 1
then
y = (x2 - 1)/6
= ((6t+1)2 - 1)/6
= (36t2+12t+1-1)/6
= 6t2 + 2t
x2 – 6y = 1
And if
x = 6t - 1
then
y = (x2 - 1)/6
= ((6t-1)2 - 1)/6
= (36t2-12t+1-1)/6
= 6t2 - 2t
2
x
– 6y = 1
So we have solutions
x = 6t –1, y = 6t2 – 2t
and
x = 6t +1, y = 6t2 + 2t
for any positive integer t.
2
x
– 8y = 1
Solutions
Diff
2
2
2
2
2
2
x
3
5
7
9
11
13
15
y
1
3
6
10
15
21
28
Diff
2
3
4
5
6
7
2
x
– 8y = 1
Taken modulo 8 this gives
x2 = 1 mod 8
But the quadratic residues mod 8
are 1, 4, 1, 0, 1, 4, 1 so if
x = 1, 3, 5 or 7 mod 8
then
x2 = 1 mod 8
2
x
– 8y = 1
But
x = 1, 3, 5 or 7 mod 8
is equivalent to
x = 1 mod 2
or
x = 2t + 1
t = 1, 2, 3, …
2
x
– 8y = 1
Now solving for y
if
x = 2t + 1
then
y = (x2 – 1)/8
= ((2t+1)2-1)/8
= (4t2+4t+1-1)/8
= (t2+t)/2
= T(t), t = 1, 2, 3, …
the triangular numbers
2
x
– ny = 1
In general
(nt+1)2 = n2t2 + 2nt + 1
= 1 mod n
and
(nt-1)2 = n2t2 - 2nt + 1
= 1 mod n
2
x
– ny = 1
So if
x = nt + 1
solving for y gives
y = (x2 – 1)/n
= ((nt+1)2-1)/n
= (n2t2+2nt+1-1)/n
= nt2 + 2t
2
x
– ny = 1
And if
x = nt - 1
solving for y gives
y = (x2 – 1)/n
= ((nt-1)2-1)/n
= (n2t2-2nt+1-1)/n
= nt2 - 2t
2
x
– ny = 1
So there exist an infinite number
of solutions of the form
x = nt –1, y = nt2 – 2t
and
x = nt +1, y = nt2 + 2t
for t = 1, 2, 3, ...
and there may also exist other
solutions depending on
quadratic residues mod n.
References
Alpern, Dario. (2001) Quadratic
Diophantine Equation Solver at
www.alpertron.com.ar/quad.htm
Beiler, Albert. (1964) Recreations in
the Theory of Numbers – The Queen
of Mathematics Entertains. Dover
Publications, Inc., New York.
Dudley, Underwood. (1969)
Elementary Number Theory. W. H.
Freeman Co., San Francisco.