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# number theory

Mathematics

Homework

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(1)
Consider
200
We have
12
1 1 1
1 1 ... 1
t
nn
p p p



where
12
, ,...
t
p p p
are positive integers.
The prime factors of
32
200 2 5
So
32
200 2 5n
32
200 2 5
11
200 1 1
25
80


Consider
210
The prime factors of
210 2 3 5 7  
So
210 2 3 5 7
1 1 1 1
210 1 1 1 1
2 3 5 7
48

(2)
Consider
12 28 mod20x
The congruence is
12 28 mod20x
Comparing with
ax b Mod m
, we have
12, 28, 20a b m
, 12,20 4d a m
which is not divides
28b
Therefore, the congruence
12 28 mod20x
has no solution.
(3)
Consider
3 mod9x
The congruence is
3 mod9x
Comparing with
ax b Mod m
, we have
1, 3, 9a b m
, 1,9 1d a m
which is divides
3b
And hence the solution of the given congruence exists and is unique.
Now
3 mod9x
…… (1)
Also
0 27 Mod 9
…… (2)
30 Mod 9x
Therefore,
30 Mod 9x
is a solution of the congruence
3 mod9x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(4)
Consider
5 mod7x
The congruence is
5 mod7x
Comparing with
ax b Mod m
, we have
1, 5, 7a b m
, 1,7 1d a m
which is divides
5b
And hence the solution of the given congruence exists and is unique.
Now
5 mod7x
…… (1)
Also
0 35 Mod 7
…… (2)
40 Mod 7x
Therefore,
40 Mod 7x
is a solution of the congruence
5 mod7x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(5)
Consider
2 mod5x
The congruence is
2 mod5x
Comparing with
ax b Mod m
, we have
1, 2, 5a b m
, 1,5 1d a m
which is divides
2b
And hence the solution of the given congruence exists and is unique.
Now
2 mod5x
…… (1)
Also
0 10 mod5
…… (2)
12 mod5x
Therefore,
12 mod5x
is a solution of the congruence
2 mod5x
Since
,1am
, the congruence
Mod max b
has a unique solution.
(6)
Compute the order of 5 modulo 36
That is
5 0 mod 36x
That implies
0 mod 36x
That implies
36x

(7)
Show that if
|ax
and
|by
then
|ab xy
Since
|ax
, therefore there exists an integer
d
such that
…… (1)
Since
|by
, therefore there exists an integer
c
such that
y bc
…… (2)
Multiplying (1) and (2) we get
xy abcd
xy a bc d
ab cd
ab e
Therefore,
|ab xy
( By Definition of divisibility )
(8)
Show that if
, , , 1a b b c a c
and
| , | , |a m b m c m
then
|abc m
Since
|am
Therefore, there exists an integer
d
such that
Since
|bm
Therefore, there exists an integer
e
such that
m be
Since
|cm
Therefore, there exists an integer
f
such that
m cf
Since
,1ab
, therefore there exists integers
,mn
such that
1am bn
Since
,1bc
, therefore there exists integers
,op
such that
1bo cp
Since
,1ac
, therefore there exists integers
,qr
such that
1aq cr
1am bn
on multiplying both sides by
m
amm bnm m
a be cf bn ad m
abc ef ab nd m



1
abc ef ab c m
abc ef m


Therefore,
|abc m
1bo cp
on multiplying both sides by
m
bom cpm m
ab od cb pe m



11
ab c cb a m
abc abc m
abc m



Therefore,
|abc m
1aq cr
on multiplying both sides by
m
aqm crm m
aq be cr be m
ab qe cb re m



11
ab c cb a m
abc abc m
abc m



Therefore,
|abc m
(9)
Show that
1n n n

for all positive integers
n
We have
n
is the number of divisors of
n
and
n
is the number in
1,2,3,..., 1n
relatively prime to
n
|
1
dn
n
and
|
prime
1
1
pn
p
nn
p



1n n n

Try to prove by induction.
So
1n
is true.
Fix an n. Assume that the inequality
Where the sum are taken over appropriate of
d
in the range
1 dn
Now
1|n
and
|nn
so the LHS is
11n n n n
Also
1, 1n
,so the RHS is
1n
If
n
composite there is some
d
in the range
1 dn
with
|dn
So the LHS
1n
,and equality cannot hold.
If
n
prime then the only divisors of
n
are
1
and
n
, so the LHS
1n
and every
1 dn
is co prime to
n
so the RHS
1n
(10)
Assume
p
is a prime and
a
has order
modulo kp
. Show that if
,1dk
then
d
a
also
has order
modulo kp
We have if
,1am
and
h
is the smallest positive integer such that
1 mod
h
am
then
say
h
is the order of
mod am
written as
m
h ord a

So that
a
has order
modulo kp
that implies
p
a ord k
1 mod
r
ra h
k k p
Suppose we have
d
a
also has order
modulo kp
To show
|hk
. Write
k hq r
where
k hq r
1
k
hq k
hq r
a
a
aa
1
mod p
r
r
a
a
So
d
p
a ord k

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