The square of a positive integer cannot be a form/(, K is an integer) :
(a) 3 k + 1
(b) 4 k + 2
(c) 5 k + 1
(d) 5 k + 4
The square of a positive integer cannot be a form/(, K is an integer) :
(a) 3 k + 1
(b) 4 k + 2
(c) 5 k + 1
(d) 5 k + 4
Answer : Option b
Square of any number can not be of the form 4k+2
The square of any positive integer is either of the form 3k or 3k+1 for some integer m.
Hence, the square of any positive integer is either of the form 4k or 4k +1 from some integer q.
Hence, the square of any positive integer can be of the form 5k + 1 or 5k + 4 for any integer k.
Detail Proof
Proof 1: Let a and b are two positive integer such that a is greater than b, then: a=bq+r, where q and r are also positive integer and 0≤r<b
Taking b=3, we get: a=3q, 7+r, where 0≤r<3
The value of positive integer a will be 3q+0, 3q+1 or 3q+2 i.e., 3q, 3q+1 or 3q+2
Now we have to show that the square of positive integers 3q, 3q+1 and 3q+2 can be expressed as 3m or 3m+1 for some integer m
∴ Square of 3q=(3q)² = 9q²=3(3q²)=3m; where m is some integer and m=3q²
Square of 3q+1=(3q+1)²=9q²+6q+1 =3(3q²+2q)+1=3m+1 for some integer and m=3q²+2q
Square of 3q+2=(3q+2)²=9q²+12q+4 =9q²+12q+3+1 =3(3q²+4q+1)+1=3m+1 for some integer and m=3q²+4q+1
∴ The square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Proof 2
Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers a and 4, there exist non - negative integers m and r, such that,
a=4m+r, where 0≤r<4
⇒a2=(4m+r)2=16m2+r2+8mr…(i)
Where, 0≤r<4
Case I
When r = 0, then putting r= 0 in EQuation (i), we get
a2=16m2=4(4m)2=4q
Where, q=4m2 is an integer.
Case II
When r =0, then putting r = 0 in Equation (i), we get
a2=16m2=4(4m2)=4q=4(4m2+2m)+1=4q+1
Where, q=4(4m2+2m) is an integer.
Case III
When r = 2, then putting r = 2 in Equation (i) we get
a2=16m2+4+16m=4(4m2+4m+1)=4q
Where, q=(4m2+4m+1) is an integer.
Case IV
When r =3 , then putting r = 3 in Equation (i), we get
a2=16m2+924m=16m2+24m+8+1=4(4m2+6m+2)+1=4q+1
Where, q=(4m2+6m+2) is an integer.
Hence, the square of any positive integer is either of the form 4q or 4q +1 from some integer q.
Proof 3:
Let a be an arbitrary positive integer.
Then, by Euclid's division algorithm, corresponding to the positive integers a and 5, there exist non - negative integers m and r such that:
⇒a=5m+r,where0≤r<5
⇒a2=(5m+r)2=25m2+r2+10mr[∵(a+b)2=a2+2ab+b2]
⇒a2=5(5m2+2mr)+r2 ...(i)
Where, 0≤r<5
Case 1
When r = 0, putting r = 0 in eq. (i) we get,
a2=5(5m2)=5q
where, q=5m2 is an integer.
case II
When r = 1, putting r = 1 is eq. (i) we get,
a2=5(5m2+2m)+1
⇒a2=5q+1
Where, q=(5m2+2m) is an integer.
Case III
When r = 2, putting r = in eq. (i) we get,
a2=5(5m2+4m)+4=5q+4
Where, q=(5m2+4m)+4=5q+4
Case IV
When r = 3, putting r = 3 in eq (i) we get,
a2=5(5m2+6m)+9=5(5m2+6m)+5+4
=5(5m2+6m+1)+4=5q+4
Where, q=(5m2+6m+1) is an integer.
Case V
When r = 4, putting r = 4 in eq (i), we get,
a2=5(5m2+8m)+16=5(5m2+8m)+15+1
⇒a2=5(5m2+8m+3)+1=5q+1
Where, q=(5m2+8m+3) is an integer.
Hence, the square of any positive integer can be of the form 5q + 1 or 5q + 4 for any integer q.
Square of any number can not be of the form 4k+2
The square of any positive integer is either of the form 3k or 3k+1 for some integer m.
Hence, the square of any positive integer is either of the form 4k or 4k +1 from some integer q.
Hence, the square of any positive integer can be of the form 5k + 1 or 5k + 4 for any integer k.
Detail Proof
Proof 1:
Let a and b are two positive integer such that a is greater than b, then: a=bq+r, where q and r are also positive integer and 0≤r<b
Taking b=3, we get: a=3q, 7+r, where 0≤r<3
The value of positive integer a will be 3q+0, 3q+1 or 3q+2 i.e., 3q, 3q+1 or 3q+2
Now we have to show that the square of positive integers 3q, 3q+1 and 3q+2 can be expressed as 3m or 3m+1 for some integer m
∴ Square of 3q=(3q)² = 9q²=3(3q²)=3m; where m is some integer and m=3q²
Square of 3q+1=(3q+1)²=9q²+6q+1 =3(3q²+2q)+1=3m+1 for some integer and m=3q²+2q
Square of 3q+2=(3q+2)²=9q²+12q+4 =9q²+12q+3+1 =3(3q²+4q+1)+1=3m+1 for some integer and m=3q²+4q+1
∴ The square of any positive integer is either of the form 3m or 3m+1 for some integer m.
Proof 2
Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers a and 4, there exist non - negative integers m and r, such that,
a=4m+r, where 0≤r<4
⇒a2=(4m+r)2=16m2+r2+8mr…(i)
Where, 0≤r<4
Case I
When r = 0, then putting r= 0 in EQuation (i), we get
a2=16m2=4(4m)2=4q
Where, q=4m2 is an integer.
Case II
When r =0, then putting r = 0 in Equation (i), we get
a2=16m2=4(4m2)=4q=4(4m2+2m)+1=4q+1
Where, q=4(4m2+2m) is an integer.
Case III
When r = 2, then putting r = 2 in Equation (i) we get
a2=16m2+4+16m=4(4m2+4m+1)=4q
Where, q=(4m2+4m+1) is an integer.
Case IV
When r =3 , then putting r = 3 in Equation (i), we get
a2=16m2+924m=16m2+24m+8+1=4(4m2+6m+2)+1=4q+1
Where, q=(4m2+6m+2) is an integer.
Hence, the square of any positive integer is either of the form 4q or 4q +1 from some integer q.
Proof 3:
Let a be an arbitrary positive integer.
Then, by Euclid's division algorithm, corresponding to the positive integers a and 5, there exist non - negative integers m and r such that:
⇒a=5m+r,where0≤r<5
⇒a2=(5m+r)2=25m2+r2+10mr[∵(a+b)2=a2+2ab+b2]
⇒a2=5(5m2+2mr)+r2 ...(i)
Where, 0≤r<5
Case 1
When r = 0, putting r = 0 in eq. (i) we get,
a2=5(5m2)=5q
where, q=5m2 is an integer.
case II
When r = 1, putting r = 1 is eq. (i) we get,
a2=5(5m2+2m)+1
⇒a2=5q+1
Where, q=(5m2+2m) is an integer.
Case III
When r = 2, putting r = in eq. (i) we get,
a2=5(5m2+4m)+4=5q+4
Where, q=(5m2+4m)+4=5q+4
Case IV
When r = 3, putting r = 3 in eq (i) we get,
a2=5(5m2+6m)+9=5(5m2+6m)+5+4
=5(5m2+6m+1)+4=5q+4
Where, q=(5m2+6m+1) is an integer.
Case V
When r = 4, putting r = 4 in eq (i), we get,
a2=5(5m2+8m)+16=5(5m2+8m)+15+1
⇒a2=5(5m2+8m+3)+1=5q+1
Where, q=(5m2+8m+3) is an integer.
Hence, the square of any positive integer can be of the form 5q + 1 or 5q + 4 for any integer q.
