hp40g+.book Page 13 Friday, December 9, 2005 1:03 AM
Exercise 7
so, b3 = 999 ⋅ (c3 – b3) + 1 , or b3 ⋅ 1000 + c3 ⋅
The calculator is not needed for finding the general solution to equation [1].
We started with b3 ⋅ x + c3 ⋅ y = 1
and have established that b3 ⋅ 1000 + c3 ⋅
b3 ⋅ (x – 1000) + c3 ⋅ (y + 999) = 0 or b3 ⋅ (x – 1000) =
According to Gauss’s Theorem, c3 is prime with b3 , so c3 is a divisor of (x – 1000) .
Hence there exists k ∈ Z such that: (x – 1000) = k ⋅ c3
and
Solving for x and y, we get:
x = 1000 + k ⋅ c3
and
y= – 999 – k ⋅ b3
for k ∈ Z .
This gives us:
b3 ⋅ x + c3 ⋅ y = b3 ⋅ 1000 + c3 ⋅
The general solution for all k ∈ Z is therefore:
x= 1000 + k ⋅ c3
y= – 999 – k ⋅ b3
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
1 | ⋅ z | 2 | – Z . When m moves on |
transformation F : z – |
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