HP 40gs User Manual page 287

Exercise 7
Step-by-Step Examples
×
b = 999
so,
3
×
b 1000 + c
3 3
The calculator is not needed for finding the general
solution to equation [1].
b
We started with
and have established that
So, by subtraction we have:
⋅ ( )
b x 1000 – + c
3 3
⋅ ( )
b x 1000 –
or
3
According to Gauss's Theorem,
c
is a divisor of
3
Hence there exists
( )
x 1000 – = k c
and
( )
– y + 999 = k b
Solving for x and y, we get:
×
x = 1000 + k c
and
×
y = – 999 – k b
∈
k Z .
for
This gives us:
⋅ ⋅
b x + c y =
3 3
The general solution for all
×
x = 1000 + k c
×
y = – 999 – k b
Let m be a point on the circle C of center O and radius 1.
Consider the image M of m defined on their affixes by the
F : z >
transformation
( )
c – b + 1
, or
3 3
× – ( )
999 = 1
x ⋅ y ⋅
+ c = 1
3 3
×
b 1000
3
⋅ ( )
y + 999 = 0
⋅ ( )
= – c y + 999
3
c
3
( )
x 1000 –
.
∈
k Z
such that:
×
3
×
3
3
3
× × – (
b 1000 + c
3 3
∈
k Z
is therefore:
3
3
1
2
⋅
-- - z
– – Z
. When m moves on
2
× – ( )
+ c 999 = 1
3
b
is prime with , so
3
)
999 = 1
16-13
.