Question

show that square of any positive integer cannot be of the form 5q+2 or 5q +3 for some integer m
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Answer 1

Step-by-step explanation:

Let 'a' be a positive integer and b=5.

Using Euclid's Division Lemma,

a=bq+r , 0<=r<b,

• where a is any integer,
• b is divisor,
• q is quotient and
• r is remainder

Here, a=5q+r, 0<=r<5

So, r can be 0, 1, 2, 3, or 4.

0, 1, 2, 3, and 4 are the possible values of remainder when a is a positive integer.

There can be no other value.

For r = 0,

a = 5q+0 = 5q

(a) ^2 = (5q)^2

= 25q^2

= 5(5q^2)

= 5m ( where m=5q^2)

For r = 1 ,

a = 5q+1

(a) ^2= (5q+1)^2

=25q^2 + 10q + 1

=5(5q^2+2q)+1

=5m+1 ( where m = 5q^2+2q)

For r = 2 ,

a = 5q+2

(a) ^2= (5q+2)^2

=25q^2 + 20q + 4

=5(5q^2+4q)+4

=5m+4 ( where m = 5q^2+4q)

For r = 3 ,

a = 5q+3

(a) ^2= (5q+3)^2

=25q^2 + 30q + 9

=25q^2 +30q + 5 + 4

=5(5q^2 +6q + 1) + 4

=5m+4 ( where m = 5q^2 + 6q+1)

For r = 4 ,

a = 5q+4

(a) ^2= (5q+4)^2

=25q^2 + 40q + 16

=25q^2 +40q + 15 + 1

=5(5q^2 +8q + 3) + 1

=5m+1 ( where m = 5q^2 +8q + 3)

Thus, the square of any positive integer is of the form 5m, or 5m+1, or 5m+4 for some integer m.

Hence, the square of any positive integer cannot be of the form 5m+2 or 5m+3 for some integer m.

Hence proved