HP 40gs, Руководство пользователя

Страница: 287 / 444

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 –
( ) ×
+
×
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 –
( ) ×
+
×
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