 Okay friends, so this is the problem-solving video on nucleates division lemma, so let us see the problem here. It says show that n square minus 1 is divisible by 8 if n is an odd positive integer. So in such cases, first of all let us understand what should be the approach. Approach should be first, first step is, step is, read the question very carefully. This is one step where students tend to ignore and in the hurry of solving the problems, they actually make a lot of silly mistakes. So what is that? Read the question, question thoroughly, question thoroughly. How do we do this? And this is, you know, this particular approach should be adopted in every problem-solving thing you do, right? So how do I do? I usually find out the elements given in the question. Show that what is the first element you encounter? So n square minus 1 is then the next keyword is divisible by what? 8, 8. And if n is an odd, another important thing, positive integer. So many facts, right? So n is an odd positive integer. The moment I hear the word odd, first thing which strikes my mind is, if a is odd, if a is odd, then if a is odd, then it can be expressed as either or only 2k plus 1, right? Why? Because 2k will be an even number. So a is of the form 2k plus 1. But whether this particular concept will help me or not, that depends on once we go to the next step. Next step is, now try and understand what all underlying concepts will be used. So you have to prove that something is divisible by 8. When is something divisible by, something, some number is divisible by 8. Some number is divisible by 8. When, if we divide it, your remainder is 0. Your remainder is 0. So you can, you can see that 8, 8 or you have to prove that 8 is a factor of, a factor of, a factor of n square minus 1, n square minus 1. This is what you have to establish, where n is of the form, where n is 2k plus 1. So there would be two approaches to solve again. So one is you take n as 2k plus 1 form or if you, if you, if you see, like the same thing can be expressed as the same, same, any integer can also, any odd integer for that matter can be expressed as 4k plus 1 or 4k plus 3 as well. Now you will ask, why am I taking 4 here? Because to get a factor of 8, you can start with this also. Let's adopt both the, both the techniques and see which one is better. So as you are exposed to multiple techniques, you will know which one to adopt. So let's say we will adopt these two and how do I know that any odd integer is of the form of 4k plus 1 on 4k plus 3 because any integer we just saw in the last lecture, you can visit in the last video and you understand, any integer can be expressed as 4k, 4k plus 1, 4k plus 2 and 4k plus 3. I cannot express it as 4k plus 4 because 4k plus 4 is as good as 4a, the remainder is 0, right? So hence, now clearly this is not an, not an even, not an odd number. Why? Because there is a, there is a, there is a factor of 2 here. So it can be expressed as 2 into 2k. So hence it is not an even number. Why? Because you know, any integer multiplied by 2 is always an even number. This similarly has 2 as a factor. So 2k plus 1. So again, 2 times any integer, whatever it is, whether odd or even, will give you an even number only. So hence, only two possibilities left, this one and this one. So let us adopt both the method, method number 1 where we'll take up this and method number 2 where we'll take up this, okay? And you can compare both of them. So let us first assume n is 2k plus 1. So what can I do now? So n is, let's say 2k plus 1, I am adopting method number 1. n is 2k plus 1. So n square minus 1 will be nothing but 2k plus 1 whole square, whole square minus 1, which is nothing but now you adopt this identity which one. So I'm writing this identity a plus b whole square. You have learned this before is equal to a square plus twice of a b plus b square. So using this, let us expand this. So hence what will you get? You'll get 2k square plus 2 times 2k times 1 plus 1 square. This dot represents a multiplication sign. Don't get confused as a decimal sign here. And then minus 1, okay? This minus 1 comes here. Now what is this? Expand it. It is 4k square plus 4k, 4k and 4k and then this one and this one gets cancelled. So now you can take 4 as common. In fact 4k as common and this is k plus 1. This is what it is. So hence clearly n, so if you can write what? 4 divides this n square minus 1. How do I know that? Because you can see clearly there is a factor of 4. Why it is a factor? Because k and k plus 1. k was an integer. So k and k plus 1 will be an integer itself. k times k plus 1 will be an integer. So if 3k was 3, so 3 into 4 will give you 12. k was 7, so 7 into 8 will give you 56. So all are integers. Isn't it? Now, so hence clearly 4 divides n square minus 1. But this was not the objective. We have to prove that 8 divides n square minus 1. So how do I do it? We again apply logic. If you see, these two numbers are consecutive numbers. Yes or no? So if k is 1, k plus 1 will be 2. If k is 2, k plus 1 will be 3. If k is 3, k plus 1 will be 4. So there are always two consecutive numbers. If you see and this is the logic you'll apply, what will you write? You'll write k and k plus 1 are consecutive, consecutive integers. And in this case we are talking about positive integers because the question asks so. So k and k plus 1 are consecutive integers. And we know that one of any two consecutive, consecutive integers, one of any two consecutive integers is always even. How do I know that? You can check. How? So let's say if k was odd, then the next k plus 1 will be even. So hence k plus 1 will be even. And if k is even, so k is even, then you know k is even anyways. So if k is even, then no problem. If k is odd, then k plus 1 will be even. So in this case, so k times k plus 1 will be and this is equal to an even number. Why? Because one of k and k plus 1 is even. So hence if you multiply any even number, you will get an even number. Even number with any number, you'll get an even number. So this is an even number. And hence this can be expressed as 2 times, let's say, q, where q is another integer. q is another integer. What is its value? We don't require it. We don't require it. So another positive integer. Another positive integer. Because we are dealing with positive integers, isn't it? So hence, what can I write? n square minus 1 becomes now what? 4 into k into k plus 1. So 4 into k into k plus 1. Now k plus 1, k into k, I have found out 2q. So it can be written as 4 into 2q. So hence it is 8q. So 8 times an integer leads to n square minus 1. So hence 8 divides n square minus 1. Hence proved. Okay. Let us see the other method where we consider, we consider this one. Which one? This one. Case number 2, this one. So let us try and understand through this. So what would I do? We would consider now here. So let us say, let us say n is of the form of 4k plus 1. In this case, we will be having 2 cases. So this is case a, let's say. And parallely we'll do case b as well. Where n is of the form of 4k plus 3. Now what is it? So n square minus 1 will be again 4k plus 1 whole square minus 1. And here n square minus 1 will be 4k plus 3 whole square minus 1. Okay. Again use the same identity a plus b whole square is a square plus twice a b plus b square. Expand it. It will give you 4k square plus 2 times 4k times 1 plus 1 square minus 1. And it will give you 4k square plus 2 times 4k times 3 plus 3 square minus 1. If you simplify this one, you will get what? 16k square plus 8k plus 0 plus 0. 1 square minus 1 is 0. Here it we will get 16k square plus 24k plus 3 square is 9 minus 1 is 8. Isn't it? So if you see here directly, you can write 8 is common between the two terms. So it is 2k square n plus k. I am not interested in pulling out k because I am interested in showing that 8 is a factor. So you can see this item 2k square plus 1k is an integer. Why? Because you have k as an integer. So k square will be an integer and twice of k square will be an integer. So 2k square plus k is an integer. So hence we can clearly say 8 divides n square minus 1. Here again take 8 common. So 8 is common again. So you can write it as 2k square plus 3k plus 1. Again this item, this is an integer is an integer. Why? Because k is an integer, k is an integer and hence 2k square plus 3k plus 1 has to be an integer. So in both the cases I see that 8 divides what n square minus 1. Hence proved. So what is the learning? In such number theory questions or what you need to do is you have to express the divisor in the form of or any number which is in question in the form of 2k plus 1 and 4k plus 3 and then build your case and by mathematical manipulation, mathematical simplification we got to know that. We got to see that 8 was indeed a factor and that was the objective. If you see we said this objective. So we somehow will prove that 8 is a factor here. So here 8 is a factor of n square minus 1. So we saw that in all cases 8 is a factor of n square minus 1. So you have to take care of what? Number of possibilities and each of the possibilities you will have to show that yes, whatever is being asked for is valid. So I hope you understood this question. So let us take another problem to solve or to understand Euclid's division lemma application further more. Thanks a lot.