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NANYANG TECHNOLOGICAL UNIVERSITY
MAS 111 FOUNDATION OF MATHEMATICS
ASSIGNMENT 1
Hints
1. Prove that the square of any odd integer leaves remainder 1 upon di
vision by 8.
Hint
: Let
m
be an odd integer, and we can let
m
= 2
k
+ 1, where
k
is an integer. Then
m
2
= (2
k
+ 1)
2
= 4
k
(
k
+ 1) + 1. As
k
and
k
+ 1
are two consecutive integers, one of them is even, and hence
k
(
k
+1) is
also even. Thus 4
k
(
k
+1) is a multiple of 8. Therefore, when we divide
m
2
by 8, the remainder is 1.
2. Prove that the product of two numbers of the form 4
k
+3 is of the form
4
k
+ 1.
Hint
: As the given two numbers are both of the form 4
k
+ 3, we can
let these two numbers be
m
= 4
k
1
+ 3 and
n
= 4
k
2
+ 3, where
k
1
and
k
2
are both integers.
We need to be aware that we cannot let these two
numbers be just
4
k
+ 3
, because otherwise, these two numbers would be
the same.
3. Prove that for natural numbers
m,n
,
m
2
+
n
2
is a multiple of 3 if and
only if both
m
and
n
are multiples of 3.
Hint
: Consider the remainders when
m
or
n
is divided by 3.
4. Prove that there are inﬁnitely many
n
such that
n
2
+ 23 is divisible by
24.
Hint
: Note that
n
2
+23 is equal to (
n
2

1)+24, so to prove that there
are inﬁnitely many
n
such that
n
2
+ 23 divisible by 24, we only need
to show that there are inﬁnitely many
n
such that
n
2

1 divisible by
24. The latter is obviously true, as
n
2

1 = (
n

1)(
n
+ 1), and if we
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View Full Document let
n
= 24
k
+ 1, then we will have that
n

1 is always a multiple of 24
(hence
n
2

1 is a multiple of 24). Note that diﬀerent choices of
k
will
give diﬀerent values of
n
2

1.
5. (a) Prove that the square of any integer is of the form 4
k
or 4
k
+ 1.
(b) Prove that no integer in the sequence
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This note was uploaded on 01/23/2011 for the course MAS 111 taught by Professor Drchansongheng during the Spring '10 term at Nanyang Technological University.
 Spring '10
 DrChanSongHeng

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