HP 40gs - 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–()×+× 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–()
kZ∈
x 1000–()kc
3
×=
y 999+()kb
3
×=–
x 1000 kc
3
×+=
y 999– kb
3
×–=
kZ∈
b
3
xc
3
yb
3
1000 c
3
999–()×+× 1==⋅+⋅
kZ∈
x 1000 kc
3
×+=
y 999– kb
3
×–=
F : z >
1
2
---
z
2
⋅ Z––
hp40g+.book Page 13 Friday, December 9, 2005 12:27 AM