HP 40gs - 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 – –
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