Question

Use Euclid's division algorithm to show that the cube of any po
integer is of the form 4m or 4m + 1 or 4m +3.
​

Answer 1

Let a be an arbitrary positive integer. Then, by Eculid's division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that
a=4q+r,where0≤r<4
a
=
4
q
+
r
,
where
0
≤
r
<
4

⇒a3=(4q+r)3=64q3+r3+12qr2+48q2r
⇒
a
3
=
(
4
q
+
r
)
3
=
64
q
3
+
r
3
+
12
q
r
2
+
48
q
2
r
[∴(a+b)3=a3+b3+3ab2+3a2b]
[
∴
(
a
+
b
)
3
=
a
3
+
b
3
+
3
a
b
2
+
3
a
2
b
]
...(i)
⇒a3=(64q2+482r+12qr2)+r3
⇒
a
3
=
(
64
q
2
+
48
2
r
+
12
q
r
2
)
+
r
3

where, 0≤r<4
0
≤
r
<
4

CASE I When r=0, Putting r=0 in Eq. (i), we get
a3=64q3=4(16q3)
a
3
=
64
q
3
=
4
(
16
q
3
)

⇒a3+4mwherem=16q3
⇒
a
3
+
4
m
where
m
=
16
q
3
is an integer.
CASE II When r=1, then putting r=1 in Eq. (i), we get
a3=64q3+48q2+12q+1
a
3
=
64
q
3
+
48
q
2
+
12
q
+
1

=4(16q3+12q2+3q)+1
=
4
(
16
q
3
+
12
q
2
+
3
q
)
+
1

4=(16q3+12q2+3q)+1
4
=
(
16
q
3
+
12
q
2
+
3
q
)
+
1

4m+1
4
m
+
1

Where, m=(16q2+12q2+3q)
m
=
(
16
q
2
+
12
q
2
+
3
q
)
is an integer.
CASE III When r=2, then putting r=2 in Eq, (i), we get
a3=64q3+144q2+108q+27
a
3
=
64
q
3
+
144
q
2
+
108
q
+
27

=64q3+144q2+108q+24+3
=
64
q
3
+
144
q
2
+
108
q
+
24
+
3

4(16q3+36q2+27q+)+3=4m+3
4
(
16
q
3
+
36
q
2
+
27
q
+
)
+
3
=
4
m
+
3

where, m=(16q2+27q+6)
m
=
(
16
q
2
+
27
q
+
6
)
is an integer.
Hence the cube of any positive integer is of the form 4m, 4m+1, or 4m+3 for some integer m.