 Trajectory for the rest of the course here. I guess what is this April 30th now? Okay So We've got two weeks and then finals that next week and the way that you see siesta's finals I'm so pretty new here I think that finals week there aren't any classes that that week at all So really we have these two weeks to finish everything up So all I'm trying to do is do chapter five the two parts of the chapter that I think are relevant And useful and that just means two sections and one of them I'm going to get done today for sure So all we have to do basically is finish up one more section in review and that that's it. Okay So Having said that I'll start giving you some idea of what to study for the final You know what the final is going to be comprised of in terms of old exams and definitions and all that kind of stuff So you have some idea where to begin I'm probably not going to do that today, but try to remind me if I forget We'll start talking about that on Thursday. Okay, I think you have plenty of time to get an idea of what what to look back over Okay, sounds like Okay, all right So Yeah, so basically here's what we're gonna do it's nice and compact this time Let's see Okay, sorry Joe. I didn't mean to close the door on you there This is Oops See that's what I need. I'm screwing it all up. Okay There we go Section 5.2 5.1 is just a history lesson about there's a geeky guy And the ma who got all excited about all this stuff and just like all these guys back then, you know They died of tuberculosis when the time of their 26 They don't tell you that though So there's my augmentation to 5.1. And so we're gonna move on just to 5.2 for ma's Little theorem is what this is called. He didn't have any pets. Nobody loved him. So he just Throwing it calling it little made him feel better Okay so Actually, yes, I mean there's he's he's actually in the number theory community sort of famous about there's another Theorem called for ma's last theorem which maybe some of you have heard of before that was a much much It's it's much bigger than this theorem. So hence. This is the little one And it basically just says this just says let P be a prime a and integer and suppose that P does not divide a Okay, that's that's the assumption. So this is very important that you don't forget this. He does not divide a Then I'm gonna okay. This P's gonna cut this off a to the power p minus one is congruent to one mod p The statement's very simple. In fact the proof is actually pretty simple And so what this says is of course you can already translate this yourselves at this point and just says that if you take an integer a And piece this is important that p does not actually divide a p's not a factor of a Then if you raise that number to the p minus one power and subtract one p divides that number Okay So this is the first kind of I don't know I think sort of surprising Results in the book, which is why I really wanted to get to this and after we do this Just do a couple of examples for you and then I'm gonna do you know like I've been trying to do some some stuff related to your homework So what we're gonna do is just suppose that again, I'm not saying that What where these things live, but we're just gonna suppose that p does not divide a of course a is an integer He does not divide a and Let's consider the integers a 2a 3a and just we're just gonna keep on going down until we get to P minus one times a okay, so here's here's the idea This is actually not very hard the proof of this theorem is not is not hard And this is the main object that you're gonna be looking at in here Okay, so I'm gonna make a couple claims so claim one Yes, yes Well What okay, so if I say claim what what I really mean is I'm sort of I'm proving something for you a Supposition depending on the context could mean this is one of your hypotheses in the statement of your theorem All I'm saying is when I say claim. I'm just going to show you something that asked it That's true, and then we're gonna use it later on Okay, so the first claim is that none of The above Integers Okay, just to be clear. I'm just gonna make a little box here That's what I mean by the above integers. I mean all of these guys right here none of the above integers are Congruent to zero mod p. Yeah, sorry Joe a 2a 3a all 4a all the way down to p minus one times a Yes, yes, none of these are congruent to zero mod P Yeah, it is it's saying the same thing I'm just gonna use this congruence language because it's gonna come up here in a second. But yes, it's exactly the same thing so Why is that true? Well Suppose one of them was congruent to zero mod p. Let's say suppose You know I have to do this in general now, okay, but let's just suppose that I a is congruent to zero mod p where high as an integer right and Well, what are the possible coefficients in front of the a's right? There's the first one is a so I could be one and the biggest it could be would be P minus one right? You see where I'm getting that Just comes from the coefficients in front they range from one to P minus one well then as Nick pointed out I a congruent to zero my P just means that P divides I times a right. That's what it means. Okay? P divides I a All right, so we're trying to get a contradiction here and so Yes I Say that again You basically yeah So the way to say it is our supposition is that P doesn't divide a and if remember if a prime divides a product of two integers It has to divide one of them, right? So since it doesn't divide this one it has to divide this one But there's no way it can do that because it's too small in some sense, right? It's between one and P minus one so it couldn't be divisible by P. That's the idea Okay, since P does not divide a I'll write this out We are forced to conclude that P divides I because of the property of primes that we've known for a while But this is impossible since Like I said, I is sort of too small one is less than or equal to I Which is less than or equal to P minus one so that can happen, right make sense Okay, so we got a contradiction that means what we assumed What's true is actually not true. So this proves the claim Oh, okay So I is less than or equal to P minus one and oh you're saying is the other part we're done And that one is less than or equal to I you mean Right, right. Yeah, I mean that yeah, you're right, but it just sort of Yeah, I mean and that's that's a good question is what would do you leave out I mean how obvious it does have to be before you actually don't state it, you know Yeah, I mean in this case It's it's just again because it's positive and it's smaller than P So P can't that goes back to a theorem way back in chapter one I think but but yeah I mean really it's the it's the fact that that I is positive and strictly less than P So P cannot divide it because they're they're both positive okay, so claim two is that none of the integers in Box hopefully this makes sense. So I'm writing on a slant here Are try to correct that here congruent mod P. Okay So let's again, and I'm running out of rooms. I'm gonna try to get this in here. Let's suppose not let's suppose that say, you know I times a is congruent to J times a mod P Okay, well, okay Then and now we're gonna actually use something that we've already known To shorten the proof a little bit the book I don't think went into this but I'm gonna I'm gonna quote this because this is something that you should have in the back Your mind to work on problems. This can certainly help you out What can we do here? Well, there is something we can do to this congruence to simplify a little bit What's that? The a's yes, and the reason for that is because A and P are relatively prime right if P does not divide a they have to be relatively prime They have to because otherwise think you think about it Okay, if I'm missing anyone if P doesn't divide a is an integer P and a are have to be relatively prime So what's what's what are the only possibilities for the GCD of P and a one and P because it has to divide P? But it can't be P because then P would divide a and we're assuming P doesn't divide a so it has to be one right So that's why they're relatively prime and so by and I'm not gonna keep writing this down But hopefully you're all with me now Then we can cancel as Nick said we can cancel the a out Okay, and reduce this congruence. I'm not gonna quote the theorem now, but Okay, so then I is congruent to J Mod P Okay, and this I'm gonna gloss over just just a bit because I'm not that interested in it So what does that mean? That means that P divides I minus J Okay. Well, what do we know about I and J? So I And J are between one and P minus one and just just try to think about this I'm hopefully if you just maybe think about this geometrically this will make sense I'm not gonna write out too much here because I feel like I'm starting to split hairs now because I want I want to really get to The more complicated stuff here. You've got two different numbers I and J that are both between one and P minus one What happens if you subtract them? Well? Okay, let's just suppose that I's bigger. Let's just suppose it doesn't really matter But I is bigger than J and they're both between one and P minus one when you subtract them you're gonna get something that's between Zero strictly between zero and P minus one when you subtract those two numbers Okay, if you think you think about it. I mean just just think you know two and seven well What happens when you subtract you're gonna get five? okay So you're gonna end up with something that now regardless one, but maybe the the smaller one comes first You know, maybe you're taking two minus nine or something like that But the point is hopefully you'll buy this an absolute value because I and J are squeezed between one and P strictly Well between zero and P and strictly between zero and P When you subtract You're an absolute value. You're still gonna be smaller. You're still gonna be small You're gonna be strictly between zero and P an absolute value because of your I and J are switched to two small together So when you subtract it can't be divisible by P because that difference would have to be too big That's that's the idea roughly Okay, okay, so I mean here's the idea roughly Okay, just just in case you want to see something like this. Let's just suppose one is less than or equal to I Which is less than J? Which is less than P right? Well, what happens if I take I minus J? You know We're not gonna worry so much about the The one minus J here, but well, okay Let's just subtract and we can certainly subtract J from all these right so One minus J is less than or equal to I minus J, which is what? less than zero Okay, right, which is less than P minus J and actually You can show that I minus J then has to be between zero and minus P So again, it's just it's too small to be divisible by P. So that's all I'm gonna say about that Well, okay, if you go back over here So J Okay, I'm just answering this question J is certainly less than P plus one right for sure That's for sure, right? Because J is some number between one and P minus one so Let's see That's not what I wanted to say. Sorry so So what do we know about J? So Yeah, yeah, okay. So no this this is all this is all amounts to so J is less than P plus one. Okay, so From here we get Yeah, so Minus if you move everything over here you get minus P then is less than One minus J. Okay, I know this is getting squished This is definitely true We're mad J is definitely less than P plus one because J is one of these numbers here Okay, it's actually less than P even but Okay, but if J is less than P plus one then just move the minus P over here and the minus J over here And you get minus P is less than one minus J Then once you've got that then you've got minus P So so then you've got I minus J squeeze between minus P and P Okay, so the only way to be divisible by P is if it's zero that's that's it. Yeah, sir It would be sufficient to say if I and J are Either I equals J in which case no problem or they differ by a factor of P and we don't have that many But that's exactly the thing. I'm trying to prove rigorously here. I'm trying to answer his question Yeah, I think you'd still need to do to demonstrate that that too which is what I'm trying to say here And this the minus P less than then one minus J is Where I throw this in here so Minus P is less than one minus J and then you've got one or sorry You've got I minus J strictly squeezed between minus P and P. So it's got to be zero. Anyways, all right No, I don't want to focus on this anymore, but that's the idea Okay, I go on to a new page now Okay, so So now the thing to note is that This is kind of the cool part of the proof the integers a to a three a on Down to P minus one times a non-zero, right? I showed you that none of them are zero mod P and Mutually incongruent So what does this say? First of all if you take all the integers mod P how many distinct Integers are there reduced modulo P. They're exactly P of them right zero one two three all the way down to P minus one P's congruent to zero right P plus one is congruent to one. They're exactly P of them So I just showed you that none of these are zero and they're all mutually incongruent mod B That and in other words neither of them is and I showed you this neither of them is congruent mod P and so A has to be so listen what I'm saying. I'm just going to say this and I'm not going to write it down Because it's just going to get too long a has to be congruent. It's certainly congruent to something mod P Right, so every integer is congruent to something mod P. It's the remainder when you divide it by P, right? A is congruent to something mod P That's not zero right Two ways also congruent to something mod P, but it has to be congruent to something different mod P Otherwise those two guys would be congruent and I just prove that they're not so the point is all of these guys all of these P minus one Scalar multiples of a are congruent to different things Mod P and none of them are zero Because I showed you that none of them is congruent to the none of them is congruent to each other mod P So I'm just okay. I'm just going to say this too. There's a theorem. This goes back into chapter 4 So let's say that a and 3a were both congruent to 5 say mod P then there's a theorem that says that okay, well a is congruent to 5 and 5 is I can't remember a and 3a okay, so a is congruent to 5 and 3a is congruent to 5 So we've got a is congruent to 5 and you can flip the congruence around so 5 is congruent to 3a and by transitivity a is congruent to 3a, which is what we just proved isn't possible That's why they all have to be congruent to different things Okay, so what do we know and I want you to think about this for a second All right, this this might require just a little bit of thought, but I again I don't want to be too wordy here a times 2a times 3a on down to P minus 1 times a is congruent to 1 times 2 times 3 times 4 all the way down to P minus 1 Hang on a second Mod oh, that's that's bad. Okay I'm gonna try to erase this, but I'm not going to Mod P there we go. All right Sorry about that. Well. Yeah, I mean So the point is I think is what you're making here It's not necessarily the case that a is congruent to 1 and 2 a is congruent to 2 and 3 is congruent to 3 That's not necessarily the case, but here's here's what you you should keep in mind and this also comes from a Induction in an application. I think I think it was there in 4.2. At least it was in section 4.2 So, okay, let me let me just say this Remember what I said before a is congruent to one of those numbers for sure It's not zero remember none of them are congruent to zero mod p so a has to be congruent to one of those guys Maybe it's not one. Maybe it's Seven who knows right? Two ways congruent to a different one of those guys as we were just talking about right three a's congruent to a different one of Those guys how many how many elements are there here? They're p-1 here They're p-1 elements here, so they all match up in some way Okay, they all match up and so because they all match up Remember this is also a theorem if a is congruent to b and c is congruent to d then ac is congruent to bd You can multiply them together. That was also in theorem 4.2 I believe so once you have these these individual congruences you can literally just multiply them down the line And that's essentially what we're doing and we're just using the fact that Multiplications commutative just to write it in the natural order on the right-hand side Thus, so let's let's rewrite the left side. Okay. There's a natural way that we can rewrite the left side All of these terms have an a in them, right? There's p-1 of them, so we get a To the p-1. What do we have with the numbers? Well, okay, you see what I did I just take I just took all the care of all the a's but now we saw two times three times four times five Down to p-1, which which is p-1 factorial, right? You might say well, but there's no one But it doesn't matter if there's a one right if you leave the one out and where you put the one on you get The same thing because you're multiplying by one. Okay, and then of course on the other side we get P-1 factorial Mod p. Okay. We're almost done and What can we say here? Yes We can do that. Yes. Why can we do that? It is relatively prime to p, right? Okay, and it's it's not because It's not as easy as saying that that p-1 factorial is smaller than p that may not be the case I mean if you take five factorial right it it's or five minus one factorial. Okay, four times three times two times one That's not smaller than five. It's still bigger, but you know that Okay, so since somebody I'm just right quizzing you today here, but somebody tell me why it is That and remember in order for this to be relatively prime to p All we need to know is that p doesn't divide it then it's automatically relatively prime from what I was saying just a little Bit ago. Why can't p divide p-1 factorial? That's basically the idea Right, right, right. Yeah The point is if p divided p-1 factorial remember it's a product right p-1 times p-2 all the way down to one So p is being prime when has to divide one of them Okay, but if p divided any one of them that's not possible because that factor is smaller than p So p camp so sorry So p can't possibly divide it because p's too big right whatever that number is p's bigger than it so p can't divide it Okay So we can now cancel the p-1 factorial to get a to the p-1 is congruent to one Mod p okay, and I don't want you to forget the assumption in the beginning of the problem was that p does not divide a Remember, that's very important. Oops. That wasn't supposed to be a D. Yeah, sorry How about that? Mod p. Yes Sure, let me just let me kind of do this as an aside over here Okay, this I think we'll make it make it simpler. Hopefully I gave myself enough space so For example, why can't you know five divide a five minus one which will be four right? Five can't divide four factorial right? Well, this is the same thing as saying five cannot divide one times two times three times four right if so Because of this product remember five will have to divide one of the factors That's one of the properties of primes is if a prime divides a product that has to divide one of the factors So five would have to divide one two three or four which it can't because it's just too big right So then five would divide I for some I between one and four which is not possible Okay All right Hopefully this is legible here. This is really horrible P. But yeah So that's what the theorem says and this is a kind of an interesting trick here If Yeah, if we if we had any actual real college bars here You could you could impress your friends by whipping out your calculator and saying hey, guess what I can I can I can show you that you know 13 To the 102nd is Divisible by minus one is divisible by 103. I'm like, how would you know that and they can ask you these questions and seem like a genius It's not gonna really no one's Now if it was that easy believe me I would I would do it but doesn't it doesn't work that easy So that now but in actual I mean in all honesty it is kind of cool that this is true You take a number that's not divisible by P and for some reason if you raise it to the P minus one power and subtract one It is divisible by P. That's a sort of a non-obvious fact Okay, I see well it sort of Sort of falls apart right away Because If you took that away then for example just plug in P for a here then certainly P It's not the case that P doesn't divide a P does divide it But then P will divide a to the P. I mean just just imagine replacing it with P Okay, then P does divide a in that case Then P is definitely not going to divide this because this will be already be a multiple of P and then we need to subtract one There's no way you'll divide it. Yeah Okay, and that same argument actually can be used for any any value of a that's divisible by P You just replace it with Pk or something and you can see that it's not going to divide that to the P minus one minus one Okay, so the corollary to this and see I may not get as far today as I wanted to but that's that's okay I mean I'll basically get at least get through this section So then we're going to have plenty of time for them for the next section and review So corollary is actually very nice very succinct very easy to state and that is that for any integer a Let's see. What is the corollary? a to the P is Congruent to a Mod P yes For any integer a and you can try some test cases here and just plug in the simple ones like zero and one And you can see that they both work, right? In fact, it works for everything Peace prime still. Yes. He's still prime and so the proof here is really pretty simple. So the proof goes like this case one P does not divide a then By I'm going to abbreviate this. Hopefully you don't mind Flt for ma's little theorem, right? a to the P minus one Is congruent to one mod P, right? Okay, that's what we just proved So and this is also something in this you should have known going to the next exam You can always multiply any congruence by any number you want and preserve it. Okay, so hence Multiplying through by a that's you want to write that down. I get a to the P is congruent to a Mod P Okay Case two P does divide a Okay well Now we're just gonna What is it we need to show I just want to remind you what you need to prove we need a To the P is congruent to a mod P Okay All right now you might think I'm being lazy, but I'm just gonna write obvious here, and there's a reason why do that Here you can write it down you can write it down. I'm gonna tell you okay. I'm just running out of room And I want you to think about Think about this. Okay case two is that P divides a he's a factor of a If P divides a Do you think that P is going to divide a to the P? Okay, just just think about what this is just think about it written out a times a times a times a times a p times Well to replace every a with pk right pk times pk times pk times p certainly going to divide it for sure And if P divides a then P divides a and Therefore P will divide their difference if it divides both things it divides their difference done That's it. That's all there is to it. Does that make sense? Okay, I didn't want to write all that out, but hopefully that's not too hard Okay, so now I'm gonna do an example and You know probably what I'll get to Please tell me if you need a little more more time on this before I go to the next page Okay, so I'm gonna get give you an example that has a flavor of of your homework, and then I'm gonna do one more Lama which might be used in the homework. I'm not sure but just to be on the states out I'm gonna go ahead and go through it or at least I'll tell you what it is I mean I prove it, but it's the proofs in the book okay, so Here's an example to verify to the 38th power is congruent to for mod 11 Okay, now the direction is actually going to say more I'm like used for my last theorem to do this Yeah, sorry not for my last theorem. Definitely not for my little theorem. Yes Okay, so Here's the thing to do and again that the principle is very similar in a sense to what you were doing before if we want to Okay, we want to use for my little theorem to do this. Okay, so what's the modulus? It's 11. So remember what for my little theorem says it says that p doesn't divide a then a to the p minus 1 is congruent to 1 mod p So that knowing what the p is and what the power is should be immediate. It's Whatever the modulus is that's what you should be starting with. I mean at least that that should be your first strategy Okay, so and I want you to think about this. Okay, five to the 11 Minus 1 is congruent to 1 mod 11 okay, so the a is five right the p is 11 and P does not divide a in other words 11 does not divide five Right, I'll write that down if it helps The a is five in this case right the p is 11 and just know it. I'll just put this in brackets P does not divide a this case right, okay So What does that tell us? Well, let's we can write this a little bit nicer, right? So this is 5 to the 10th It's congruent to 1 mod 11 38 okay. Well. Yeah, we're gonna try to get there. That's our whole goal is to get there Okay, and you'll see you'll see how this works here in a second This is this is just one way of going about doing it. So we've got 5 to the 10th We want to get up to 5 to the 38th right so Well, what can we do the nice thing is that and of course as you remember what our goal was is to get ones popping out We've got one on one side so whatever we if we raise first of all we can raise both sides Whatever power we want and the right side is always going to be one for sure and we want to get five to the 38 so It stands to reason maybe that The okay, and this is the way you should be thinking you might think I'm gonna just raise both sides to the third power That'll give me up to the 30th But you still have a lot more powers left you're better off raising both sides to the fourth power Because five to the 40th is a lot closer to five to the 38 and then work from there. Okay, so And I'll write this out so five to the 10th to the fourth is Congruent to one to the fourth mod 11 right, okay All right, so now we've got five to the 40th Congruent to of course, I don't need the fourth power here one mod 11 Okay. Well, what can I do from here? I'm kind of bigger than I want to be Okay, I want I want to know what five to the 38th is so what I can do though Since I'm not that far away what I can do is just take five to the 38 out and then I've got a five squared left, right? Make sense Okay, so we're almost done Okay, so I'm just gonna do this in one step here. Let's look at the five squared is 25, right? What's 25 mod 11? Three right right sure. There are other ways you could do it too So now let's look at where we're at Where do we want to get we want to get five to the 38 congruent to four mod 11 so Here's what I'd like what I'd like to have happen is if I just multiply this congruence through by four I like this three times four to go to one then I get exactly what I want But it does go to one because it's exactly one bigger than 11 Now we're done Okay, so this I'm gonna circle this up here and just keep in mind where everything's modulo 11, right? This is equal to one mod 11 Okay So then five to the 38 now is congruent to four mod 11. Okay, so here's what I'm gonna do Here's the last thing I'm gonna do Okay, and again, this is just for the sake of time. You're certainly welcome to use this fact I think I did I guess I forgot my book Everybody have this down The book kind of lists this as a lemma, so I'm just gonna call this a lemma Suppose Let's see. Actually, I don't want to I don't want to get this Wrong here. Actually, do I have it written? I'm gonna say yes. Okay Yeah, yeah, I've got it I still got it here. I'm just not gonna write the proof out suppose that P and Q are distinct primes a to the P Congruent to a mod Q and A to the Q Congruent to a mod P then a to the P Q is congruent to a Mod PQ and the proof is actually really easy really, but I'm not gonna do it. So What I'm gonna do instead to finish the day You may actually finish a little early today Is to do an example first before I do that? Let me before I forget. Let me give you homework from this section Okay, so this is 5-2. We'll have to talk about of course when this is due So Let's see So we have two two more sections probably you wouldn't like to have them both do at the final because then you're not gonna Get any feedback really on that We've got okay, here's what we're gonna do so we still have this week and next week All right for the final and Because I'll be talking about a bunch more problems on Thursday this assignment will be do a week from today one week. Okay, okay, so one two four six seven ten Okay, so now what I'm gonna do Start off with This is the last part of the lecture for today. We'll do number 2b Okay, now some of them do have you know quite a few parts, but once you get the hang of this you're not doing a lot of proofs You're just kind of You know modifying congruences until things come out. It's it's not really that bad. Okay, I'll show you what I mean with this example so to be And it says And most of these just boiled down to from Oz last year and they really do so if the GCD of A and 35 is equal to one show that I think this is 2b. Oh Is that what to be? Oh, it's the two two a is the GCD of A and 35 being one show that a to the 12th can grow into one month 35 Okay, sorry, it's two a then yeah to a okay, so Here's what's gonna do Keep them in mind some of your old tricks, okay, and then these will help you Going first of all I want to point this out, right? You can't directly go to from Oz last theorem you want to prove this, right? But from Oz last theorem has to do with congruences mod p, right? This is not a prime. Okay, you definitely shouldn't be thinking I apply from Oz last theorem to this You can't do it But what you can do is break 35 up into the product of two of two primes, right? Seven times five and then you can start thinking about from Oz last and you put everything together at the end So I will write that first just to make sure that you're aware of this note that 35 is definitely not So what we're gonna do instead is Well, we're gonna show two things Okay, so the first Is that eight of it and I'll if you're wondering where I'm getting this I'll explain it to you all in the end eight of the 12th is congruent to one mod Seven okay, that's the first thing we're gonna show You'll see in a second and then okay, so what I'm gonna do is I'm just kind of go through this And I'm gonna try everything in the at the very end and then if you still have questions Then I'll try to address those I think once you see how it all ties together. I think it'll make sense at the very end Okay, well, okay now think about this If we were to apply for miles last theorem here, there's this okay, so this is prime Right, so if we're gonna apply for miles last theorem It's gonna say if the other condition is met all we can say is then that a to the seven minus one in other words A to the sixth this congruent to one mod seven doesn't say anything about 12, right? So but there's still something we need to know this is important We in order to even apply the theorem we need we need to know that seven does not divide a that's one of the conditions on the theorem remember Right a to the p minus one is congruent one mod p when p does not divide a so we need for seven to not divide a We have to have that right? Yes Seven does not divide a okay, so otherwise the GCD of A and 35 I'm trying to get all this in here would would certainly be at least seven right you see that Seven divided a do also divides 35 so the GCD has got to be at least that that big But it's not it's one it's one so that can't happen Seven no So yeah, so if seven so if the GCD were not one so seven divides We don't know anything about a really so the only thing we can say is if seven actually did divide a Then seven would divide divide a and 35 and so then yes seven would have to divide the GCD It's the GCD would have to be a multiple of seven. Yes That's right. So By For ma's little theorem, right a to the seven minus one is congruent to one mod Seven right that's exactly what For modest little theorem says okay, we can we check the condition seven's prime seven does not divide a so we get this for sure and So that says that a to the sixth Is congruent to one mod seven right and now this is also a theorem from 4.2 now We can square both sides okay to get a to the 12 It's congruent to one mod seven okay, so that took care of the first one right any questions Okay, so what do we want to do for the second one? Well, it's similar right now We want to show that a to the 12th is congruent to one mod five again note that um Five does not divide a right otherwise the GCD of And 35 would have to be bigger than or equal to five right five divided a Then five divides both a and 35 It's common divisor, so the GCD has to be at least that the greatest common divisor of course has to be at least that Okay So then it's the same thing for ma's little theorem then implies that What does it say it says that a to the? Five minus one is congruent to one mod five right that's what it says right? In other words a to the fourth is congruent to one mod five and Now you can probably see where we need to go right? Just cube both sides right cube both sides I'm running out of room here, but we want to write that your notes. I'm cubing both sides Just like I squared both sides before okay, so this gives me a to the 12th is congruent to one cube which is one mod Five here's the punch line then now we're now we're done okay thus a to the 12th is congruent to one mod 35 this is something that We talked about actually before This is actually one of your homework problems from the last section remember there's a problem That's that something like you know solve 217s congruent to something mod 300 and whatever it was and the problem said break it down into these congruent You know into these separate congruences The whole point is remember that if your modular relatively prime if you have the same congruence But the modular are relatively prime that those congruences are the same then you can just multiply all the moduli together Into one congruence. I'm just going to say one thing out last thing about that before we finish Think about I just because I want you to I want to remind you of this right? What does this mean? I want you to think about this this will help you if you understand it It'll it'll help you make sense out of it later. What does this mean? This means that seven divides a squared. Sorry. This means that seven divides a to the 12th minus one What is this mean? This means that five divides a to the 12th minus one But since seven and five are relatively prime their product divides it that goes way back to a divisibility theorem for a long time ago So these are the kinds of things that you're doing you're just playing numbers games again really Okay, so I've done a couple of examples for you. I've basically done the whole section. We only have one more to do So we should have plenty of time to do that and review and do more homework problems So you really should be in decent shape for the final Thursday. I'll start trying to tell you more about what to expect on the final exam Okay, and just FYI the final exam is is May 16th So it's Thursday of finals week. So you still have a plane of time to study for the exam I believe please don't quote me until I verify this I think the the finals from 140 to 410 May 16th in this room. I'm almost positive. That's right You know, it'll basically be cumulative, but I'm gonna there's some things I'm gonna say about that But I'll start talking about that on Thursday So and if I had to say is it cumulative or is it absolutely not cumulative the answer is it is cumulative Okay, but you will be given a break in a sense, but we'll we'll talk about that on Thursday. Yes Mm-hmm. Okay this one