HP 40gs Manual de usuario descargar pdf (Pagina 287)

100

101

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:

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

999 c

–()1+×=

1000 c

999–()×+× 1=

x⋅ c

y⋅+1=

1000× c

999–()×+1=

x 1000–()c

y 999+()⋅+⋅ 0=

x 1000–()⋅ c

– y 999+()⋅=

x 1000–()

kZ∈

x 1000–()kc

×=

y 999+()kb

×=–

x 1000 kc

×+=

y 999– kb

×–=

kZ∈

1000 c

999–()×+× 1==⋅+⋅

kZ∈

x 1000 kc

×+=

y 999– kb

×–=

F : z >

---

⋅ Z––

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