Exercise 7 – HP 40gs User Manual
Page 287
Step-by-Step Examples
16-13
so, ,
or
The calculator is not needed for finding the general
solution to equation [1].
We started with
and have established that
.
So, by subtraction we have:
or
According to Gauss’s Theorem,
is prime with
, so
is a divisor of
.
Hence there exists
such that:
and
Solving for x and y, we get:
and
for
.
This gives us:
The general solution for all
is therefore:
Exercise 7
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
transformation .
When
m moves on
b
3
999
c
3
b
3
–
(
) 1
+
×
=
b
3
1000 c
3
999
–
(
)
Ч
+
Ч
1
=
b
3
x
⋅
c
3
y
⋅
+
1
=
b
3
1000
×
c
3
999
–
(
)
×
+
1
=
b
3
x 1000
–
(
) c
3
y 999
+
(
)
⋅
+
⋅
0
=
b
3
x 1000
–
(
)
⋅
c
3
–
y 999
+
(
)
⋅
=
c
3
b
3
c
3
x 1000
–
(
)
k
Z
∈
x 1000
–
(
)
k c
3
×
=
y 999
+
(
)
k b
3
×
=
–
x
1000 k c
3
×
+
=
y
999
–
k b
3
×
–
=
k
Z
∈
b
3
x c
3
y
b
3
1000 c
3
999
–
(
)
Ч
+
Ч
1
=
=
⋅
+
⋅
k
Z
∈
x
1000 k c
3
×
+
=
y
999
–
k b
3
×
–
=
F : z >
1
2
--- z
2
⋅
Z
–
–
hp40g+.book Page 13 Friday, December 9, 2005 1:03 AM