HP 40gs hp 40gs_user's guide_English_E_HDPMSG40E07A.pdf - Page 287
Exercise 7
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hp40g+.book Page 13 Friday, December 9, 2005 1:03 AM Exercise 7 Step-by-Step Examples so, b3 = 999 × (c3 - b3) + 1 , or b3 × 1000 + c3 × (-999) = 1 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 × (-999) = 1 . So, by subtraction we have: b3 ⋅ (x - 1000) + c3 ⋅ (y + 999) = 0 or b3 ⋅ (x - 1000) = -c3 ⋅ (y + 999) 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 -(y + 999) = k × b3 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 × (-999) = 1 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 transformation F : z - >1-- ⋅ z2 - Z . When m moves on 2 16-13