Download mn is a lattice point if ,mn∈ (2,3)

Brahmagupta 3
All letters will denote integers
A point in the plane (m, n) is a lattice point if m, n ∈ 
Let a ∧ b =
0 are
1 . Then the lattice solutions to ax + by =
all of the form (kb, −ka ) where k ∈ 
If ax + by =
c has a solution, then a ∧ b is a divisor of c
If ax + by =
c has a solution (m, n) , then infinitely many
as given by (m, n) + k ( a−∧bb , a∧a b )
(2,3),(3,0),(−1,1) are, ( 12 ,0) is not
4x + 7 y =
0 has solutions
(4, −7),(8, −14),(−4,7),(−8,14),...
Thus 4 x + 6 y =
7 could not have a solution
4x + 6 y =
8 has (2,0) + k (−3, 2) for solutions
Actually, it is algorithmic:
1 has a solution
a∧b =
1 if and only if ax + by =
Kuttaka
Stage
Rs
Qs
0
2155
1
138
2
m
am + bn
n
1
0
2155
15
0
1
138
85
1
1
-15
85
3
53
1
-1
16
53
4
32
1
2
-31
32
5
21
1
-3
47
21
6
11
1
5
-78
11
7
10
1
-8
125
10
8
1
10
13
-203
1
ax + by =
c has a solution when and only when a ∧ b is a
a
b
c
x+
y=
divisor of c . Equivalent to
.
a ∧b
a ∧b
a ∧b
12508 x − 58823 y = 177 is solvable since
12508 ∧ 58823 =
59 and 177= 3 × 59
There were 63 equal piles of plantain fruit put
together and 7 single fruits. They were divided
evenly among 23 travelers. How many plantains
did each traveler receive?
Start with equation 23
=
x 63 y + 1 . By Kuttaka
this leads to y = 4 x = 11 . Thus solution to
23x = 63 y + 7 is given x = 77 and y = 28 .
Another solution is x = 14 y = 5
Let d be a common divisor of a and b . Then d= a ∧ b if
and only if d can be written as a linear combination of
them, namely, if and only if ax + by = d has a solution.
Sylvester’s Theorem. Let a and b be positive integers
with a ∧ b =
1 . Then c = ab − a − b is the largest value of
c for which the equation ax + by = c does not have a
nonnegative integer solution. Moreover, half of values
below c have unique solutions, and the sum of all the
c ’s for which there is no solution is:
Consider a = 30 and b = 24 . Of their common
divisors, 1, 2, 3 and 6, 30 x + 24 y =
6 is the
only one to have a solution.
( a − 1)(b − 1)(2ab − a − b − 1)
12
Consider a = 5 and b = 11 . So the last c for
which there is no solution to 5 x + 11b =
c is
c = 39 . Indeed we have no solutions for 1, 2, 3,
4, 6, 7, 8, 9, 12, 13, 14, 17, 18, 19, 23, 24, 28,
29, 34,and 39. These 20 numbers add up to 310.