Solution:

Let, a = 2q + 1 and b = 2m + 1, where, q and in are some whole numbers

2q + 1 > 2m + 1

2q > 2m

q > m

Therefore ,

Thus, a–b2 is a positive integer.

Now we need to prove that one of the two numbers

Consider ,

Also, we know that from the proof that

We know that the difference of two integers is an odd number if one of them is odd and another is even. (Also, difference between two odd and two even integers is even)

Hence it is proved that if a and b are two odd positive integers is even.

Hence, it is proved that if a and b are two odd positive integers such that a > b then one of the two number a+b2 and a–b2 is odd and the other is even.

Solution:

To Prove: that the product of two consecutive integers is divisible by 2.

Proof: Let n – 1 and n be two consecutive positive integers.

Then their product is n (n – 1) = n

We know that every positive integer is of the form 2q or 2q + 1 for some integer q.

So let n = 2q

So, n2 – n = (2q)

n2 – n = (2q)

n2 – n = 4q

n2 – n = 2q (2q – 1)

n2 – n = 2r [where r = q (2q – 1)]

n2 – n is even and divisible by 2

Let n = 2q + 1

So, n

n

n

n

n

Hence it is proved that that the product of two consecutive integers is divisible by 2.

Solution:

To Prove: the product of three consecutive positive integers is divisible by 6.

Proof: Let n be any positive integer.

Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5.

If n = 6q,

n (n + 1) (n + 2) = 6q (6q + 1) (6q + 2), which is divisible by 6

If n = 6q + 1

n (n + 1) (n + 2) = (6q + 1) (6q + 2) (6q + 3)

n (n + 1) (n + 2) = 6 (6q + 1) (3q + 1) (2q + 1) Which is divisible by 6

If n = 6q + 2

n (n + 1) (n + 2) = (6q + 2) (6q + 3) (6q + 4)

n (n + 1) (n + 2) = 12 (3q + 1) (2q + 1) (2q + 3),

Which is divisible by 6.

Similarly we can prove others.

Hence it is proved that the product of three consecutive positive integers is divisible by 6.

Solution:

To Prove: For any positive integer n, n

Proof: Let n be any positive integer. n

Since any positive integer is of the form 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4, 6q + 5

If n = 6q,

Then, (n —1) n (n + 1) = (6q —1) 6q (6q + 1)

Which is divisible by 6

If n = 6q + 1,

Then, (n —1) n (n + 1) = (6q) (6q + 1) (6q + 2)

Which is divisible by 6.

If n = 6q + 2,

Then, (n – 1) n (n + 1) = (6q + 1) (6q + 2) (6q + 3)

(n – 1) n (n + 1) = 6 (6q + 1) (3q + 1) (2q + 1)

Which is divisible by 6.

Similarly we can prove others.

Hence it is proved that for any positive integer n, n

Solution:

To Prove: That if a positive integer is of the form 6q + 5 then it is of the form 3q + 2 for some integer q, but not conversely.

Proof: Let n = 6q + 5

Since any positive integer n is of the form of 3k or 3k + 1, 3k + 2

If q = 3k,

Then, n = 6q + 5

n = 18k + 5 (q = 3k)

n = 3 (6k + 1) + 2

n = 3m + 2 (where m = (6k + 1))

If q = 3k+ 1,

Then, n = (6q + 5)

n = (6 (3k + 1) + 5) (q = 3k + 1)

n = 18k + 6 + 5

n = 18k + 11

n = 3 (6k + 3) + 2

n = 3m + 2 (where m = (6k + 3))

If q = 3k + 2,

Then, n = (6q + 5)

n = (6 (3k + 2) + 5) (q = 3k + 2)

n = 18k + 12 + 5

n = 18k + 17

n = 3 (6k + 5) + 2

n = 3m + 2 (where m = (6k + 5))

Consider here 8 which is the form 3q + 2 i.e. 3 x 2 + 2 but it can’t be written in the form 6q + 5. Hence the converse is not true.

Solution:

If n = 5q + 1

Then n

n2 = (5q)

n

Hence n

Solution:

To Prove: that the square of an positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Proof: Since positive integer n is of the form of 3q , 3q + 1 and 3q + 2

If n = 3q

n

n

n

n

If n = 3q + 1

Then, n

n

n

n

n

If n = 3q + 2

Then, n

n

n

n

Hence, n