MATHmaniac aka CrUNchy

1)Find the continued fraction of

1st step:


2nd step:
Find the gcd(13,7)
13 = 1 x 7 +6
7 = 1 x 6 + 1
6 = 1 x 6 + 0
gcd (13,7)=1

3rd step:
Multiply
and
to obtain,



4th step:
Substitute it




Thus,


2)Find the continued fraction of


1st Step:


2nd step: find the gcd (23,21)
23=1 x 21 +2
21=10 x 2 +1
2=2x1+0

3rd step:
Multiply and to obtain



4th step:Substitute




Thus,

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prove that for any integer a, 3|a^3-a

let a be an integer.when we divide by 3,
we have 3 possiblities:


a=3k
a=3k+1
a=3k+2

we want to show that 3 divides a^3-a
we know that a^3=a(a^2-1)=a(a-1)(a+1)

1st case,
if a=3k,then a^3=3.k(3k-1)(3k+1)
Thus,3|a^3-a for a=3k

2nd case,
if a=3k+1,Then a^3-a=3((3k+2)(3k+1)(k))
Thus 3|a^3-a for a=3k+1

3rd case,
if a=3k+2,then a^3-a=3((3k+2)(3k+1)(k+1))
Thus, 3|a^3-a for a=3k+2


since 3|a^3-a for all cases,then we can conclude that 3|a^3-a for any integer a

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Solve .Give 3 positive solutions.

First step:
Find the continued expansion of




thus,

Second step:
Since n=2 and it is even, the solutions are
where k=1,2,3...
Thus,



.
.
.
are the positive solutions.

Third step:
Calculate the value of the numerator and denominator,







So,the answers for 3 positive solutions of are

, ,

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Definition:
An arithmetic function is a function that defined for all positive integers

Definition:
An arithmetic function f is called multiplicative if f(mn) = f(m) f(n) whenever m and n are relatively prime positive integers.It is called multilplicative if
f(mn) = f(m) f(n) for all positive integers m and n.

Let's see...
and are multiplicative function.
Let say . Find and .


Positive divisor of n=

by multiplying P,



thus,


Get it??
Let's use number,..






thus,
Let say k=5,so

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