1.Let the square of any posititve integer be in the form 6q + 2, 6q +5

x = 6q +2

Squaring both sides

Xsquare = 36(q)square + 4 + 24q

x square = 36(q) square - 6+ 2 + 24q

xsquare = 6 [ 6(q) square - 1 + 4q] + 2

put [ 6(q) square - 1 + 4q] = m

Therefore x = 6m + 2

2 .let a be any +ve integerb=6

by euclid 's division lemma,

a=bq+r, 0=< r < ba= 6q+r,0=< r

when

r=0,a = 6q= (6q)3 ---> a3 = 216q3 =6(36q3 )=6q(where m is = 6q3 )

by similar manner u can prove for

r=1,2,3,4,5

and u will get the proof..(if you need it as 1,2,3,4,5ask me)...

3.If n = 5q

n is divisible by 5

Now, n = 5q

⇒ n + 4 = 5q + 4

The number (n + 4) will leave remainder 4 when divided by 5.

Again, n = 5q

⇒ n + 8 = 5q + 8 = 5(q + 1) + 3

The number (n + 8) will leave remainder 3 when divided by 5.

Again, n = 5q

⇒ n + 12 = 5q + 12 = 5(q + 2) + 2

The number (n + 12) will leave remainder 2 when divided by 5.

Again, n = 5q

⇒ n + 16 = 5q + 16 = 5(q + 3) + 1

The number (n + 16) will leave remainder 1 when divided by 5.

Case II:

When n = 5q + 1

The number n will leave remainder 1 when divided by 5.

Now, n = 5q + 1

⇒ n + 2 = 5q + 3

The number (n + 2) will leave remainder 3 when divided by 5.

Again, n = 5q + 1

⇒ n + 4 = 5q + 5 = 5(q + 1)

The number (n + 4) will be divisible by 5.

Again, n = 5q + 1

⇒ n + 8 = 5q + 9 = 5(q + 1) + 4

The number (n + 8) will leave remainder 4 when divided by 5

Again, n = 5q + 1

⇒ n + 12 = 5q + 13 = 5(q + 2) + 3

The number (n + 12) will leave remainder 3 when divided by 5.

Again, n = 5q + 1

⇒ n + 16 = 5q + 17 = 5(q + 3) + 2

The number (n + 16) will leave remainder 2 when divided by 5.

Similarly, we can check the result for 5q + 2, 5q + 3 and 5q + 4.

In each case only one out of n, n + 2, n + 4, n + 8, n + 16 will be divisible by 5.

4.Let a be any positive integer and b = 6.

Then, by Euclids algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 1, where k 1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k 2 + 1, where k 2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k 3 + 1, where k 3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5....

thankyou but it seems to hard plez make it simple form and easy to understand