Number Theory

In number theory, we deal only with integers and their properties. Many of you may have heard about this branch of mathematics but may not know much about it. Here I am going to introduce you to some of the basic concepts of this branch of mathematics. So, let's begin :)

One of the primary concepts essential for learning number theory is the divisibility of integers. The first of this list is Euclid's division algorithm which states that:

Given integers a and b with b>0, there exist unique integers q and r such that a=qb+r.

Well, this is what we have learnt while learning division—considering an integer a and an integer b if we divide a by b. we get a quotient and a remainder. In the above theorem, q is the quotient and r is the remainder. But the exciting thing is that such a representation of an integer is always unique, which is quite apparent. 

Let's see how can we use this property to solve some interesting problems. But before that, you all must be knowing that an even integer N can be written as N=2k where k is also an integer. If N is odd, it can be represented as N=2k+1.

Problem 1: For n≥1 establish that the integer M=n(7n² +5) is divisible by6, where n is an integer.

Solution: What do you think? Isn't it interesting that a number of this form will always be divisible by 6 irrespective of what value we put in place of n, if we place 1, it's true, same for 2, same for any n? But seeing the form, you cannot tell that the number will be divisible by 6.

So, let's see how we can approach the problem.

M=n(7n² +5) =n(6n² + n² +6-1)=6n³ + n³ +6n -n = (6n³⁺ 6n) + (n³ -n). Let me tell you how this helps. We needed to prove that that M is divisible by 6. So, I decided to take out those terms which have 6 as a factor. Quite clearly  (6n³⁺ 6n)  will be divisible by 6. We just need to prove that (n³ -n)  will be divisible by 6. 

Let's factorise (n³ -n). (n³ -n)=n(n+1)(n-1) . Now suppose we have a number and we divide it by 3, the possible remainders are 0,1 and 2. So we can write any integer in the following forms: 3k,3k+1,3k+2  where k is an integer. Let's see what happens if we replace n with these forms.

If n=3k. So, (n³ -n)=3k(3k+1)(3k-1). So, (n³ -n) is divisible by 3.

If n=3k+1, (n³ -n)=(3k+1)(3k+2)3k. So, (n³ -n) is divisible by 3.

If n=3k+2, (n³ -n)=(3k+2)(3k+3)(3k+1)=3(3k+2)(k+1)(3k+3) NOTE: HERE WE TOOK 3 COOMON FROM 3K+3. WE WROTE 3K+3=3*(k+1).  So, (n³ -n) is divisible by 3.

So, for all integers,  (n³ -n) is divisible by 3.

We also know that an integer is either odd or even. So, n-2k if n is even or n=2k+1 if n is odd.

If n is even, (n³ -n)=2k(2k+1)(2k-1) .So, (n³ -n) is divisible by 2.

If n is odd, (n³ -n)=2k+1(2k+2)2k. So, (n³ -n) is divisible by 2.

So, (n³ -n)  is divisible by both 2 and 3. So, (n³ -n)  is divisible by 2*3=6.

Hence, we have proved that M is divisible by 6.