Question

1) prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

2) prove that the square of any positive integer is of the form 5q, 5q+4 for some integer q.

3) show that the square of an odd positive integer is of the form 8q+1,for some integer q.

4) show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer.


please solve this problem for 20 points

Answer 1

Hi friend,

1.Let positive integer a = 4m + r , By division algorithm we know here 0 ≤ r < 4 , So
When r = 0
a = 4m
Squaring both side , we get
a2 = ( 4m )2
a2 = 4 ( 4m​2 )
a2 = 4 q , where q = 4m2

When r = 1
a = 4m + 1
squaring both side , we get
a2 = ( 4m + 1 )2
a2 = 16m2 + 1 + 8m
a2 = 4 ( 4m2 + 2m ) + 1
a2 = 4q + 1 , where q = 4m2 + 2m

When r = 2
a = 4m + 2
Squaring both hand side , we get
a2 = ​( 4m + 2 )2
a2 = 16m2 + 4 + 16m
a2 = 4 ( 4m2 + 4m + 1 )
a2 = 4q , Where q = ​ 4m2 + 4m + 1

When r = 3
a = 4m + 3
Squaring both hand side , we get
a2 = ​( 4m + 3 )2
a2 = 16m2 + 9 + 24m
a2 = 16m2 + 24m ​ + 8 + 1
a2 = 4 ( 4m2 + 6m + 2 ) + 1
a2 = 4q + 1 , where q = 4m2 + 6m + 2
Hence
Square of any positive integer is in form of 4q or 4q + 1 , where q is any integer . ( Hence proved )


2.Let positive integer a = 5m+ r , By division algorithm we know here 0 ≤ r < 5 , So
When r = 0
a = 5m
Squaring both side , we get
a2 = ( 5m )2
a2 = 5 ( 5m​2 )
a2 = 5 q , where q = 5m2

When r = 1
a = 5m + 1
squaring both side , we get
a2 = ( 5m + 1 )2
a2 = 25m2 + 1 + 10m
a2 = 5 ( 5m2 + 2m ) + 1
a2 = 5q + 1 , where q = 5m2 + 2m

When r = 2
a = 5m + 2
Squaring both hand side , we get
a2 = ​( 5m + 2 )2
a2 = 25m2 + 5 + 20m
a2 = 5 ( 5m2 + 4m + 5)
a2 = 5q , Where q = ​ 5m2 + 5m + 1

When r = 3
a = 5m + 3
Squaring both hand side , we get
a2 = ​( 5m + 3 )2
a2 = 25m2 + 9 + 30m
a2 = 25m2 + 30m ​ + 10- 1
a2 = 5 ( 5m2 + 6m + 2 ) - 1
a2 = 5q -1 , where q = 5m2 + 6m + 2
Hence
Square of any positive integer is in form of 5q or 5q + 4. , where q is any integer . ( Hence proved )


3.Let a be any odd positive integer with b=4.
Therefore,
a= 4q+1 ( By Euclid's Division Lemma)
When a=4q+1
a2=(4q+1)2
=16q2 +1 + 8q
=8(2q2 +q) +1
=8q+1 where (q=2q2+q)
Hence proved.


4.Let a be a given integer.

On dividing a by 6 , we get q as the quotient and r as the remainder such that

a = 6q + r, r = 0,1,2,3,4,5

when r=0

a = 6q,even no

when r=1

a = 6q + 1, odd no

when r=2

a = 6q + 2, even no

when r = 3

a=6q + 3,odd no

when r=4

a=6q + 4,even no

when r=5,

a= 6q + 5 , odd no

Any positive odd integer is of the form 6q+1,6q+3 or 6q+5...

Hope this helps you...
Please mark it as brainliest answer...☺☺☺

Answer 2

1)question
Let 'a' be any positive integer and b=2
Then by Euclid's division lemma we get,
a=2m+r, such that 0≤r<2
Therefore,
a=2m, a=2m+1
now,
a=2m
a^2=(2m)^2
a^2=4m^2
a^2=4q ,where q=m^2...................1
now,
a=2m+1
a^2=(2m+1)^2
a^2=4m^2+4m+1
a^2=4(m^2+m)+1
a^2=4q+1. ,where q=m^2+m.................2
By eq1 and eq2 we can. conclude that
square of any positive integer is of the form 4q or 4q+1


2)question
let a be any positive integer and b=5
Then by applying Euclid's division lemma we get,
a=5m+r. such that 0≤r<5
Therefore ,a=5m. a=5m+1,a=5m+2,a=5m+3,a=5m+4
now,
a=5m
a^2=25m^2
a^2=5m(5m)
a^2=5q, where q=m(5m)................1

now,
a=5m+1
a^2=25m^2+10m+1
a^2=5m(5m+2)+1
a^2=5q+1, where q=m(5m+1)..............2

now,
a=5m+2
a^2=25m^2+20m+4
a^2=5m(5m+4)+4
a^2=5q+4 ,where q=m(5m+4).............3

now,
a=5m+3
a^2=25m^2+9+30m
a^2=25m^2+5+4+30m
a^2=5(5m^2+6m+1)+4
a^2=5q+4 ,where q=(5m^2+6m+1)..........4

now,
a=5m+4
a^2=25m^2+16+40m
a^2=5(5m^2+8m+3)+1. (as 16=15+1)
a^2=5q+1 ,where q=(5m^2+8m+3)..........4
By eq 1,2,3,4,5 we can conclude
square of any positive integer is not only in the form.of 5q,5q+4 but also 5q+1.

3)question
Let a be any positive integer and b=8
Then by Euclid's division lemma we get
a=8m+r such that 0≤r<8
Therefore ,
a=8m,a=8m+1,a=8m+2.......................a=8m+7
Odd integers :
a=8m+1 ,a=8m+3 ,a=8m+5 ,a=8m+7
Similar method as above we can follow and can show what is given.

4)question
Let a be any positive integer and. b=6
Then by Euclid's division lemma we get
a=6m+r such that 0≤r<6
Therefore,
a=6m ,a=6m+1, a=6m+2, a=6m+3 ,a=6m+4, a=6m+5
Odd integers:
a=6m+1, a=6m+3, a=6m+5
we can follow similar method as followed above to show what is given.
Here square of odd positive integer is in the form of 6q+1,6q+3 but not 6q+5.