IMO 2026 Q5 Live Solve (plus congrats to the SG IMO team!)

[00:00:03]Hello, I am back with uh slightly less of a flu and uh by the time I'm back to look at Q5, it has transpired that the IMO results are already out. Um so of course uh congratulations to all students uh especially to our Singapore team uh who apparently did pretty well on question five among other things which uh played a big role in a set of pretty good results. Um I also did not want to um compete with the World Cup finals for programming. I don't think that uh IMO questions done four days late are as exciting as the World Cup finals. But uh well, I hope whatever transpires in this video uh will help you to pick up some interesting ideas.

[00:00:55]Uh I have tried not to have any spoilers. Uh so I still do not know the solution to this question. In fact, uh since I typed it out a week ago, I haven't really looked at this question.

[00:01:06]Uh but of course since the IMO results are out I know the cutoffs and um at least I am aware that this should be easier than the Q2.

[00:01:16]Uh, nonetheless, uh, the same thing applies, which is that, uh, I do not want this video to go on for absolutely long. And, uh, even if, let's say, that most students at the silver to gold level could solve this in 2 hours or less, that would be considered as a relatively achievable Q25. Uh, but that would make the video very long if I spend two hours. So, if I need to go off uh, camera for a while, let me pause a bit. And if I um just want to uh rest my throat a little bit, I would also just pause the video and update you what has happened. Okay, so we're going to try it together. Uh this is the problem statement which I have. Uh I probably won't use R greater than zero. This was the official one from what I saw, but I'll just use R+. I'm used to positive real numbers being R+.

[00:02:09]So you want uh functions from r plus to r plus such that this inequality is satisfied.

[00:02:22]Now uh straight up uh it obviously looks like uh f(x)= to x is going to be well it doesn't look like I do know that it is a solution because if fx is equal to x uh this would just become qmg.

[00:02:43]Um, in fact, I could say that uh for the first and the third one, it is already the QM and the GM.

[00:02:55]But in between those two, I have got a swap of FX and Y. [snorts] So, uh, or rather I've swapped the F from Y onto X. So it it's not just QMGM otherwise literally uh there would be nothing I can deduce from here.

[00:03:21]So my first thought is that uh there's always an equality case. Uh I just taught some of my uh students the squeeze theorem in doing limits. And so if I can make these two the same by squeezing uh x equals to f_sub_y uh I would end up with equality must hold throughout.

[00:03:45]Uh which is that this is f_sub_y this is f_sub_y and the thing in between has become fy + y / 2. So uh this is probably useful uh I mean that the statement is greater than or equal to but I can squeeze it in between which would tell me that uh I can rearrange this into f_sub_y - y equals to ffy - f_sub_y or uh equivalent ly uh if I change it back I can change it back to f_sub_x - x.

[00:04:31]So uh if x equals to f_sub_y then f_sub_y - y = to f_sub_x - x uh I don't know if this is useful uh or if let's say this statement here is useful but uh it's the first equality statement that I can get. So I am happy with that.

[00:04:59]Uh I am curious therefore whether uh letting y equals to fx will then be a good idea.

[00:05:15]Uh I should just try that because I'm not sure yet.

[00:05:33]Okay. So this is the statement uh the statement which I have from here uh is that well if I just rewrote that in terms of x fx = to 2 fx - x I just can put it into here and I well I mean I if x + f ofx equals to 2x that means that this would be the qm of x and f ofx. This would be the am of x and f ofx and this would be a gm. So all right uh unfortunately that means that this does not tell me anything.

[00:06:35]Okay. So interesting.

[00:06:44]Likewise, of course, if I just let y equals to x, then the same thing would happen. So, I'm not going to do that.

[00:06:49]Right? If I let y equals to x, uh I will just end up with qmg again.

[00:06:55]Okay? So um the question is kind of telling me that uh f x is somehow quite restricted by this.

[00:07:17]So I know that if x equals to f_sub_y would be the equality case that gives me uh this thing here.

[00:07:28]Uh that's sort of the best quality information I can get. So I'm actually I'm curious whether is there anything else I can say about this statement.

[00:07:43]uh ffx = to 2 f_sub_x - x um it's actually a recurrence in terms of u applying your fs or equivalently it's telling you that this sequence of applying fs gives you an arithmetic progression right so it means that uh x fx ffx is an AP.

[00:08:19]I know that uh for positive wheels uh one of the things about positive is that there cannot be a infinite descending arithmetic progression.

[00:08:36]So if this thing is an arithmetic progression uh and I know that uh it cannot go to negative.

[00:08:44]So it's a AP of uh positive reals the common difference has to be greater than or equals to zero.

[00:08:59]uh now I'm saying that uh to be clear here each x will have their own common difference potentially right we're not saying that fx minus x is a constant uh for all x this is not what we're saying we're saying that for each x uh this would be an ap of reals and therefore for each x at least there is a common difference which is greater than or equals to zero which just means that fx x is greater than or equals to x is a bound that I have. Uh that feels useful because uh I mean I suspect strongly that is fx equals to x uh if I suspect strongly that if it's fx equals to x then uh I am happy to have one direction of this.

[00:09:52]uh one thing that I also know is that there's no constraint on f being uh polomial.

[00:10:00]Uh but I am pretty certain that uh if let's say f were on the order of some x to the power of k.

[00:10:16]It's not even polomial but let's say that it were on the order of uh power of x uh this would give some issue because uh if I were to yeah maybe I can just say that if I just fix x to be a constant or fix y to be a constant and then vary the other one right so I just say that if we just let y equals to one for example Oh uh then this thing here is on the order of magnitude of x. Uh this thing is on the magnitude of square<unk> x.

[00:11:08]Uh so your f(x) is going to be quite squeezed uh between these two. Uh since I have f(x) is greater than or equals to x already.

[00:11:22]H let me just check that for a second. Uh if f_sub_x is greater than x, f_sub_y is greater than y, under what circumstances will the fx being bigger than x have more impact uh such that that gets messed up and same for the right side. Under what circumstances will that get more messed up? Uh I'm not sure yet, but uh I'm thinking that maybe just the order of magnitudes I can have some kind of reasoning.

[00:12:21]Yeah. The the condition here uh is logically equivalent to saying that these are all just arithmetic progressions.

[00:12:55]I do know that f is injective right because if because of this thing as well uh it does mean that if let's say uh f_sub_x= fx prime uh then my x would also be fixed.

[00:13:17]Now why I am thinking about this is just that uh I don't know that the function is continuous but if I know it's injective then uh increasing f is going to be kind of like increasing feels about right even though that's not yet uh something I've shown.

[00:13:44]Okay. So let me think a bit more now about what I can do. Uh it's all currently very disjointed stuff right like uh all of the things that have here I'm not sure what is useful uh but I feel I'm understanding the condition better and I feel that if let's say I needed to properly write out the justification of any of these I could but uh I'm not going to start to write that out because it may turn out that one third of these are unnecessary at the end or maybe even twothirds of these are unnecessary. So uh we just want to get the feel for what's happening. So so far I have f(x) is greater than or equ= to x I have got the fact that these are an ap for each x actually if it's uh a for each x and f is injective that means that my function is just going to be made up of like it's partitioned into separate chains All right. Uh for each value, let's say that if I just consider two of these chains, if my y is not in this chain then uh by the injectivity uh these two chains would be not will be disjointed and then the relationship that we are talking about is like this kind of a comparison between uh the QM of this pair and the am of the other pair.

[00:15:49]Of course, if f(x) equ= to x, then this chain is just of size one, right? But uh if fx is not x, then you will get an infinite chain of stuff.

[00:16:09]Um what if I use stuff in the same arithmetic progression?

[00:16:13]Like if I use stuff in the same arithmetic progression, do I get a contradiction?

[00:16:22]So let's try that. Okay, assuming that is not uh if f(x) is not x, right? Difference is greater than zero.

[00:16:39]Uh then I know that each f that I do is going to be just n copies of that difference.

[00:16:49]Now if I then just let x and y be some number of fs that are part of the same chain. uh here I want to talk about a specific value.

[00:17:19]So let's just for sum a.

[00:17:30]So for sum a if the function does not uh land on itself. So you really have this chain meaning that uh this is a + n d and this is a + md.

[00:17:52]Uh then of course one extra f will just push it forward one step uh on both of them.

[00:18:02]Let's see uh how that looks like. So root of no actually uh without log um if f of a is not a then I don't need to start from f to the n of a right I can just start from a itself.

[00:18:27]Yeah that would be less messy. I'm going to start with a.

[00:18:32]There's no need to have two of those.

[00:18:36]So again, what do I have? I have x + f_sub_y. So x² + f_sub_y 2. So, then the one in the uh is using fx and y and it is just their average plus d.

[00:19:30]Is there any information in here?

[00:19:43]Hey, actually, no, that that doesn't really help, right? Because if I have this, I can always just rewrite it into Yeah, because if they're part of the same chain, uh, I can just move the thing to the other side.

[00:19:58]Uh yeah. So if I just move it then it becomes still just qmg all the same. So I guess I don't want x and y to be part of the same chain.

[00:20:12]In order for this to be meaningful I would like my x and y to be part of different chains.

[00:20:27]Now um in some of my classes uh this will be the point where somebody will ask me that uh teacher can you confirm like whether this um like general approach of considering these chains is useful or not now depending on what is the purpose of your practice right uh if let's say you're trying to do as a mock paper for the IMO or for the SMO round You probably do not want to have that as a hint because um you want to have some practice with like knowing how to manage your time and then to be able to switch your approach. If let's say that after a while this doesn't work, maybe I try something else. Uh if let's say that you're doing it just for your own practice or I just want to practice some questions, uh it's okay to have someone tell you that things are working or not working, right? Just like I did for the job. I just took a peek to see whether like the general approach I was taking is correct. don't need to read the whole solution, but just like okay, that it does look like what is there is related to what I'm doing. Just to, you know, make your practice a bit more efficient.

[00:21:29]We all only have finite amount of time to practice. Uh for here, I'm not going to peek yet. It's only 20 minutes in.

[00:21:34]So, I'm going to continue to try uh to see if I can make something happen. Uh I'm not looking at the Q6 anytime soon.

[00:21:42]Uh so, what is the information that I have?

[00:21:51]Okay, first I'll just copy the question statement.

[00:21:58]Uh I have discovered so far that uh f(x) is greater than or equals to x. I have discovered that uh this is an AP for X.

[00:22:19]I have also discovered that uh I'm not going to get anything meaningful from here if X and Y are part of the same chain. So uh I guess I can try X and Y being part of different chains.

[00:22:42]So the thing is that um [snorts] philosophically right QM AMGM uh QM AM and AMGM uh on their own are still relatively loose.

[00:23:01]Uh but the definition of loose is also quite hard to quantify right which means that like for example sometimes we are talking about the equality case and then when we say that the thing is loose it is that there is no equality case but it could also be that when you move away from the equality case then the gap is very big i.e the difference between the left hand side and the right hand side uh is going to have sort of like a very large derivative uh when you're near the equality case. So you move a little bit, it's like an unstable equilibrium. It just jumps very far away.

[00:23:40]uh so I am trying to see if I can find the location of that unstable equilibrium meaning that uh such that uh a small shift will make them very far apart uh and then uh because of that by swapping x fy with fxy it's enough to mess it up that it changes the order of am and qm or am and gm right that is what I would hope to accomplish I am Not sure when that will happen, but I believe that there are so-called some places where it is more likely to crisscross faster.

[00:24:19]Uh, so I'm going to just let uh x and y be two chains.

[00:24:33]Uh, maybe I should just use a and b so it don't confuse with the x and y is there. So let B be not in this so-called orbit of A under F.

[00:24:47]So I'm looking at a couple of these uh starting from A and starting from B.

[00:25:02]Uh I will let the common differences be uh D1 and D2 for these two and then see what uh I can get.

[00:25:26]So in other words what I am thinking is that uh where would be the place that is going to break this the most. But in either way I am going to uh say that without log uh I can choose my one of these to be at the start of a chain and the other to be not as part of a chain.

[00:25:49]So, uh let's just uh use A and F and B as my x and y uh and we will get so x is a f_sub_x is now a + d1 y is b + n d2 to actually no point right.

[00:26:30]Yeah, actually there's no point in like writing it as this form because I'm not going to be maybe I'll just write it in that for just think about whether I want to move n along further or not. So I'll just leave it that way.

[00:26:50]But then again, if I let Y be this, it's just like as if my B is just sitting in the middle, right? It's like I could technically just start from like calling this thing B. I'm not insisting that the start of my chain has to be not in the domain of F because I'm not using the functional graph approach. I'm just writing this for effect to see what is going on. So, uh yeah, I guess I will just Use B first.

[00:27:22]And just note that Y could be replaced by B + ND2 uh and X by A + like MD1 if those are useful. Okay. Then then after that we can check and see what it does.

[00:27:43]Okay. So that it's uh a bit nicer.

[00:28:02]So you see that this is the D1 and D2 are currently not the same.

[00:28:20]Oh, all right.

[00:28:26]Yeah, that's silly of me. I am just realizing now that that means that the guess of f(x) equ= x is just wrong, right? uh because if D1 equals to D2 then this will again end up being just uh QM AMGM and D1 will equal to D2 if all of the D1's and D2s are the same throughout that just means that my yeah it's actually f(x) equals to x + d where d is greater than or equals to zero would work. So it's in fact a bit looser than fx equals to x. Uh so that's good right? uh the realization that if d1 equal to d2 there is no issue at all with this tells me that uh I'm looking for a contradiction if d1 and d2 are not the same uh somewhere uh I'm not trying to show that these are zero so that's a useful development I'm realizing that uh I just want uh these to be incompatible if my D1 and D2 are not the same.

[00:29:43]So in fact now I think it would be interesting if I just uh split into cases because this one has like sort of a D2 thing bigger than a D1 thing. This one has like a D1 thing bigger than a D2 thing.

[00:30:08]Uh, that's interesting. Um, let me think.

[00:30:33]Yeah. So what I was thinking is this right that if my D1 and D2 are both not zero uh you can think of it as like therefore like I have got two straight lines right they are increasing uh but they're not really the lines you're just taking fixed distance points uh I should be able to if let's say that they are both increasing. I should be able to pick things that are very close together.

[00:31:11]And if they are very close together, that means the let's say the A and B plus D2 or the A and uh B plus D1 or A plus D1 and B depending on which one we want to do. Uh then that would be near the equality case. um such that uh having that shift in the D is enough to uh mess things up.

[00:31:39]It's enough to make it such that it's near the equality case but I have now adjusted it away.

[00:31:47]Um there's also the case where uh right so basically what I'm saying is that uh I believe the rest of my proof if this strategy works which feels viable it's going to look something like this right it's going to look like uh I will consider the range of gx= to fx minus x right which I do know is non- negative.

[00:32:22]So it's like if the range uh I'll just call it r.

[00:32:30]If there is just one value in the range that is what I am hoping for.

[00:32:37]If there are two values in the range or more the main thing is that is one of them zero and one of them non zero. uh may give a different logic right because if one of them is zero, one of them is non zero uh I may not be able to go backwards from something and make them meet to be arbitrarily close.

[00:33:09]So it's like sort of that if one of them is constant and the other one is just going up, these two are never going to be that close to each other. So I might need to change my strategy.

[00:33:18]So I know that if my range is at least three or if it's two does not contain zero then I can uh further split it into the cases where D1 or D2 is bigger when handling this Whereas if let's say zero is inside then I should end up with a different kind of contradiction.

[00:34:03]So there are effectively three different cases which I envision myself having to handle. uh and I think that uh all of them are going to be sort of like uh not that hard by IMO standards as inequality task but therefore it may take a little while to handle. Okay. So uh let's now uh just try to do the zero case first uh and see what happens. Okay. So uh what happens if f(x) equals to x for some x uh and I'll just let f right so so one of let's say r = to 1 is okay uh case two is that zero is inside uh the range of uh this I mean when I do all these cases there's always a chance that I accidentally find another second solution that I didn't expect to find. If that happens so be it. Okay. Uh so uh zero d um let f a equals to a uh and f of b equals to b plus d meaning that I have got now this chain that starts from b.

[00:35:47][snorts] Uh I'm not sure whether I prefer to put x is a or y is a or it doesn't matter.

[00:35:54]Uh let's just maybe you get different contradictions, right, for the two types. I'm not sure.

[00:36:05]Uh I I'll try.

[00:36:23]So now all f_sub_x is also a.

[00:36:29]So what I'm thinking is that the problem might be here right because f_sub_y is more than y. So which means this part is definitely okay. Uh the problem might be here that uh uh well um a is a constant though.

[00:36:54]Um so if a is a constant if I let y be from this chain and I change this to y + d.

[00:37:18]if my y + d. Okay, I think I need to write both cases down just to see if they mean they are meaningfully different.

[00:37:44]Okay, I feel like this one may be more of an issue, right? because uh which one's more of an issue? Um it's like intuitively, right, is that if my A is somewhere here, I can basically pick two of them that are very close to each other. uh and then if they're very close to each other then the f of y being bigger uh than y that difference is enough to make all the um things break down. Uh on the other hand, if my A is very small, if A is very small and like almost negligible then actually am I allowed to go backwards? Let me just think. Am I allowed to go backwards? I know that uh f(x) - f_sub_x = to f(x) - x.

[00:39:11]Is there some sort of like f inverse?

[00:39:13]Not necessarily. I know. Uh but am I allowed to assume that b must be less than d? Uh because that may help.

[00:39:25]If I can assume B is less than D, that will help. Am I allowed to go backwards?

[00:39:40]until that happens.

[00:40:04]Actually, maybe I just forget about this first, right? I want to make sure that my main argument works. So this is sort of actually a side case. I thought the side case would be straightforward but uh maybe the side case is not instantly straightforward. So I'll deal with this later.

[00:40:19]Uh there exists sort of like uh D1 and D2 in the range which can be because the range has at least three different numbers then I don't have to care about whether there's zero or if it has two different numbers. Right? So uh we'll just handle that case first.

[00:40:38]So now I know that uh clearly A and B are not the same.

[00:40:56]Uh so which means that I should be able to uh pick these to be arbitrarily close to each other.

[00:41:32]to be part of those chains where this thing is still true.

[00:41:45]Okay. So now I'll just put it in uh D1 is more than D2.

[00:42:23]This was my statement. If I let uh x be a prime and y be b prime.

[00:42:36]uh remember that we are having D1 to be bigger than D2 which means that I think that this is the problem.

[00:42:45]Yeah, I can make them arbitrarily close and also arbitrarily large because the them being close to each other um will repeat eventually uh in the sense that uh what I want from here is that uh if a prime minus b prime is zero means that I want this difference to approximately make up the a minus b uh and I know that uh if my d1 d2 are positive uh I will always be able to get arbitrarily close. This is actually a pigeon hole argument. A classic pigeon hole argument. Uh that is more or less that make your m over k a rational approximation of d2 over d1 or is it the other way? D1 over d2. Uh then I will be able to keep on getting arbitrarily close. So, so this is almost its own side argument which uh for the purpose of a live soft I'm not going to uh write out the details of that. Uh in the IMO submission I certainly would.

[00:44:04]So if it's arbitrarily close and arbitrarily large, let's just open it up and see what the difference would be.

[00:44:39]Okay, move the two there.

[00:45:00]in fact a D2 is the smaller one, right? So, uh I think I can subtract off the I'll subtract off the this from both sides cuz I know that the comparison on the left is the QM comparison. So I know that is non- negative still. Uh so it but it will simplify into a square. So that should be nice.

[00:46:01]So they're different and there's some Yeah. So this will give me a contradiction, right? Because uh if I make this thing arbitrarily small. So in fact I can make uh not that a prime and b prime to be arbitrarily close. I actually want a prime and b prime plus d2.

[00:46:48]So I want these two to be arbitrarily closed. That's also fine, right? So I just make this thing uh go to zero uh for large enough m and k.

[00:47:08]Now uh by doing that uh this is constant and for my large m and k uh this will now go to infinity.

[00:47:19]This are still going to be constant but this one is as small as I want.

[00:47:35]Okay. So uh that does give me the contradiction that I was looking for.

[00:47:41]Uh in this general case of uh there is something that is greater than zero.

[00:47:53]Uh so there are two different common differences uh in those APS uh would be enough to break it.

[00:48:03]Uh so let's come back and look at the um the case that is left over.

[00:48:10]Um for the case that is left over uh I have got also if my A can be near to this I get the same logic right so if my actually it's not the same logic uh I do need to check this carefully uh because uh I cannot use both of them to get arbitrarily close if I've got two things I can get arbitrarily close to each other.

[00:48:43]Uh that's basically the logic behind rational numbers being dense, right? Uh but I only have one thing. I cannot say integers are dense. So so that does make things a bit more tricky than uh the previous case. But maybe that's just a simple explanation.

[00:49:06]So let me take a moment to think about how this could work.

[00:49:13]There's always the concern that this is actually the harder case. Uh but I mean if I have a function where it's like f(x)= to x somewhere and f(x) = to x plus d somewhere.

[00:49:40]Okay. There's a there is the potential issue that maybe just maybe uh it is possible that I could mix the two, right? Maybe the maybe I should try to mix the two because uh if it turns out that uh it is possible and I spend half an hour trying to prove it's impossible, I feel very silly. Uh but if it does turn out that is impossible by trying to make it happen, I might understand why.

[00:50:09]>> [snorts] >> So it's like uh mixture of these two types in like a pointwise trap kind of way. Maybe the logic I should employ is just not going to be the same as the general case. Maybe I'll have some other kind of logic. Uh so I will clean that out. I don't think that was very effective yet. And just just we'll go back to the original statement and try actually wait uh no I mean that uh there was nothing stopping me from just putting it in to here. Right? So I don't need to erase all of that. I'll just put in uh x to be a and y to be b and uh vice versa and just see how it looks like. So nothing is stopping me. I'll just because for the same a and b both of these must be true, right? So maybe that's the issue individually there's no contradiction. Uh but collectively something interesting happens. Maybe.

[00:51:55]>> [snorts] >> So these are now the pair of statements that I have. If there is this mixture of uh f a = to a and fb= to b + d uh as long as I can find let's say what pairs can coexist right because the the these ones amongst themselves are okay these ones amongst themselves are okay if I can find out what is the condition for these pairs to exist then I will know what is the potential uh reason why it breaks or what is the potential construction.

[00:52:35]So looking at this uh thing which for a pair AB to coexist this is the entirety of what it needs to do.

[00:52:46]Uh clearly the two statements here are the ones that are the tighter ones, right? The uh this is clearly tighter than that because uh the D has moved position and uh same for the other side. So I can just try to see what is the implication of these two. [snorts] ID is going to be fixed throughout.

[00:53:30]I suspect you can just uh homogenize it though like uh I'll just let Yeah.

[00:53:38]uh if I just let a= to d a prime and b = to db prime then the d will get thrown out it will just be a common factor throughout you can absorb the d uh normalize it uh so without log I can let d equals to one uh is okay and that will make things a lot better as long as I don't forget later to change it back if there are solutions. If there's no solution, then I don't need to change it back. I've just got a little bit easier algebra to do. Okay, so now let me just try both uh a + b / 2 greater than or equals to this.

[00:54:30]Okay, so this is uh the first statement.

[00:54:33][snorts] Okay, this one I'm not so sure. I do need to expand to see how it looks like.

[00:54:58]Uh, there will be a b square, there will be a squ, there's a minus1 - 2 b - 2 a - 2 a b.

[00:55:09]Uh, oh wait, that's just okay. That's not exactly a difference of that's like b + like b minus a².

[00:55:37]um is greater than equals to 2 b + 2 a + 1.

[00:55:43]I'll just put it that way.

[00:56:11]is these are all if and only if right.

[00:56:13]So, uh, if it's all if and only if wait a second.

[00:56:30]Oh, yeah.

[00:56:34]I almost want to talk about this as like a cautioning it out, but that's a very undergrad level description. So uh I'm realizing that actually uh I can just focus on the interval from let's say 0 to D first.

[00:56:53]Now for all the stuff from zero to D I they need to fall into the category of uh the plus nothing or plus D.

[00:57:13]The moment that something does plus zero, uh, it's fine. But the moment something does plus D, it means that henceforth all the equal stuff mod D, I know I'm talking real numbers, but mod D, you know what I mean? Uh, all the stuff will continue to be that way.

[00:57:34]So it's sort of like uh potentially it's like for this can I start the plus D later and begin with just zeros uh is the question right so alluding to what I have written just now it basically means that if B uh can always be Uh, a it depends and B being big.

[00:58:22]Yeah, because I use D equals to one, right? You can almost think of it as just I am working on the different decimal parts.

[00:58:31]So looking at this here, it's like will it be true that my B can be big?

[00:58:45]Uh is enough to break this?

[00:58:48]Not necessarily.

[00:58:52]Uh I should actually write [clears throat] it from the perspective of quadratic in B right because B can move.

[00:59:01]I can increment B by one any time in this.

[00:59:15]So I should actually write it as B minus A here. And this b minus a minus one squ if I completed the square.

[00:59:28]Okay, that's not very different for the one on top actually.

[00:59:52]But wait a second. If I'm thinking of the number line starting from let's say zero, which is not included.

[01:00:06]If I have a two, there will just be a and b right next to each other, right?

[01:00:10]um that's just uh by yeah I mean that if I have got both types of A and B uh it's then there is a transition point right there cannot be some big gap in between yeah so that is like if I were to shade I bound this thing in two colors colors. There will clearly exist two arbitrarily close points of the two different colors.

[01:00:58]Yeah, I can just hit this arbitrarily close A and B.

[01:01:05]uh for convenience of course is like when I want to say arbitrarily close A and B.

[01:01:24]I can make it as small as I want.

[01:01:31]So why would that be a contradiction?

[01:01:34]uh I need to say this very carefully.

[01:01:42]I can find a to be less than epsilon.

[01:01:48]It depends on uh whether Okay. So I can sort of um say that if the set of places that fa equals to a I think I want to say that. Yeah. So if this set is actually if this is upper bounded now I I have to be careful in uh in mathematical terms. I cannot say that consider it's maximum right because uh an upper bounded set on real numbers may not have a maximum uh but there will be a supreum like let's say if the set is of uh n / n +1 where n is a positive integer right that's a set that's upper bounded uh that isn't going to be going to have upper a maximum but I can consider the supreum a supreum is the least upper bound and so if let's say that my S is here there will be an arbitrarily closed thing of the type B down Here there will be a a that is arbitrarily close to this uh and then that would force that for my small enough epsilon this would get broken.

[01:04:10]Whereas uh if let's say that this does not have an upper bound.

[01:04:41]Then I can always just adjust my B to make this thing here uh less than one and that will die.

[01:04:57]Right? Because if A is not upper bounded in your choice, then I just pick an arbitrarily big A, I just take the B and just jump jump all the way there because uh if my FB is equal to B + one, that means that F of B + N is also okay.

[01:05:23]So I can just increase my B by a suitable integer value to make B close enough to A + one.

[01:05:32]My issue earlier was that if my B already started bigger uh but if my B already started bigger then I realized that the reason why I was forgetting is that everything must fall into one of these two categories.

[01:05:46]So uh you can use this um logic here uh to say that they are close. So yeah now that if a is not upper bounded if there is a big a I can make my b get close to it and by getting close enough to it will break this one because there's a two down here. So uh you also break the one above but uh yeah it just uh is enough to confirm that uh certainly uh this case is also resolved in a slightly different way and thus I am pretty sure that uh you cannot have these to coexist. It will just be f(x) = x + d where d is greater than or equals to zero being your final answer.

[01:06:39]Right? Uh that took slightly over an hour. But this time I didn't have any pauses. So uh it is a legitimate 1 hour solve which will probably require a good 20 minutes at least to write out properly because this is a bit of a mess. Uh but overall nice question. Uh I'm not sure whether this is the best solution but uh hey a lot of solutions now exist in the while because of the fact that the IMO is essentially over.

[01:07:06]So uh if you want to see how it's written out you can look at AOPS or you can look at other people's source. This is just my attempt and uh after this video I will see whether my attempt is actually similar or very different from what everyone else has done.

[01:07:21]Okay. So uh I don't expect to live life solve three and six. Uh apart from the fact that uh I will be busy pretty soon.

[01:07:29]Uh also I would prefer to make three and six a bit more organized. So either I won't do it so soon or if I do it uh I will do it as like I solve it on my own first and then I'll explain my thought process back to you. Okay. But that is going to be it for question five for now. So thank you for watching again and see you soon.