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Exercise 3.6 Let a € Z and n € N. Prove that ged(a, a +n) divides n. Deduce that and a + 1 are always relatively primne. 3 Marks)...

Question

Exercise 3.6 Let a € Z and n € N. Prove that ged(a, a +n) divides n. Deduce that and a + 1 are always relatively primne. 3 Marks)

Exercise 3.6 Let a € Z and n € N. Prove that ged(a, a +n) divides n. Deduce that and a + 1 are always relatively primne. 3 Marks)



Answers

Use mathematical induction in Exercises $31-37$ to prove divisibility facts.
Prove that 6 divides $n^{3}-n$ whenever $n$ is a nonnegative integer.

High in this video, we are going to prove the Steadman that tree divide in Q plus two in whenever in these positive integers. So first list Look at this term, it can be a factor into in times in square plus two. Why we do this? Well, if you are familiar with this kind off subject about Modelo you you know that any positive integer in can be categorized into one of the tree type so it can Koreans to one two are zero What tree, right, Because they're all the tree marked in model a tree, is it? It will be in one off this cases and this form gonna help us showed at tree divi the term as everyone depending on what time in falls into so less do this first case, it is congruent to zero Montree So three divide in already. It would mean that tree divide the whole The whole thing's right, because days in there. So we're done with that case For the second case, it is congruent to one or two or three. Uh, list. Look at in square. It is one square. It do stupid one, right? It is too square is will become four wishes. Also, one marked tree for is one more three. So no matter what it start off as one No two enough in squired being one more tree. Why we do this? Well, it made in squire plus two asked tree So it zero more tree right? So tree divide the second term now not the first to tree Divi it gives the same result So it divide the whole thing So we had done Yes. So by dividing it into two cases like this we prove that no matter what the case, it is tree always d y the whole too full So that is actually we can end. That is what we want to prove And here we have it. Thank you.

This problem. We want to show that n cubed minus and is divisible by three for all and in the positive integers. So we check our base case space cases and equals one. It's positive Integers Ah well, that would mean one cubed minus one in the expression that are equal to zero is divisible by three zero is divisible by anything. So this is true. It means we can conclude and equals one is true. Next assume true up to ah and equals K. Then we can say that kay cubed minus k is divisible by three. Then when n is equal to K plus one, we would have k plus one cubed minus K plus one and we would try to change this so that we can clearly tell that it is divisible by three So So Ah, this can be written as ah common factor of K plus one. We could take it out So it will be, uh, que plus one times K plus one squared minus one. I love K plus one and this is K squared plus two K plus one. So we just have k squared plus two k left because of the minus one at the end and we can expand this pretty easily. Is K cubed plus to K squared plus que scores of plus three k squared plus two K and we want to find a way to use our inductive step So ah, que cubed minus K is what we want k cubed minus K plus three k squared And this last term would have to turn into three K for us to ah, right that negative k earlier. And ah, this here Recognize that if que cubed minus k is divisible by three were really saying k cubed minus K is equal to three times some integer Where, uh, call it a and hes an integer is it's perfectly divisible. So in our current step, we can write that as 38 plus three k squared plus three k where a is an integer and ah, since all the terms of a factor of three, we can pull up the three and we've just shown that, uh, K plus one cube might escape us. One is divisible by three. We have pulled out a factor of three. Um so therefore, we can conclude that ah, and cubed minus end is divisible by three. Uh, for all n in the positive integers

In this question. We want to prove that if three end plus by two is even and this implies that N is being so, we want to prove this in two different weights. First, by culture position. The Contra position means that we want to fit this in absolute implication. Do you want to prove it? If not and either and even then supplies that not three and plus by two. Even so, this is equivalent off saying, If N is odd, this implies that three and plus by to default, we're just important enough to prove. So I would just prove that if any spot and we can write us as to a plus one, where is the interview Dan Substituting that into this equation we have three n Plus y two is equal to £3 to a plus 30.1 plus. Why shoot? So then this is equal to six. A 322 56 plus by three monkey the three plus +52 six a plus. By five, we can write five. That's four plus one. Now we factor out to former 1st 2 terms, so this gives you three a plus buy to just buy one. So we know that agent into job not what an inter job multiplier I interject uses an integer and inter job. Plus, a ninja gives you the picture as well. So we can write this as two times would be suspect one where B is equal to three a plus bite you and is some integer So hence if any spot three plus two is orders. Well, so we have proven this application There were contra position We're just not three end plus five to on implies that not And oh so if three plus two is no, don't imply that three m plus five two is evil is implying that and back to fit the Irish Yeah, no visitor. So three and plus by two is even implies that end is that's approved by a contradiction. We're three and plus fight too even implies and even well, first he is. The soon three n plus 30.2 is even, but we also assume the officer that end is not even so If Mom and even then this implies that and his father. So what we do is that if anything and can be written as she, that's a. That's for one again. Where a is some interject then substituting this into this equation. We have three n plus five. Two is even too fruit husband to a plus one which dig to six a pass by, well, three plus two book +12 this plus three plus two This is six a +15 Now what we've shown in the previous slide is that 68 plus far can be written as two b plus one. So this can be written as two B plus by one where B is equal to three A plus two, three a plus two. So hence this who, sir? Hence, if any, is an odd number three interest you is all. But this contradicts the assumption that three employees to even so that this turns out. So what this does is that being cruelly implies that if n is on, this implies that three and plus while she is. But this contradicts our assumption that three plus two is even so. Therefore this assumption must be false. Hence, we must go not on. So this implies that n is either is required


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