Note: my equation labelling does not seem to work. Someone please help me fix it. Thanks in advance.

**The problem: IMO 2017 Q2**

Find all functions such that for all we have the following:

**Solution and Discussion**

Denote the given equation by . Cmon. It isn’t easy to see what solution would work save for . An observation is that if we simply substitute into the equation and we get for all So now we can assume . Where to go?

It’s a medium so preliminary investigations can guide us quite a long way most likely. (Even for the monstrous 2015 Q5 one should at least get to showing or before getting helplessly stuck.) So suppose , which is to say that , we shall have

for all . Now suppose that for a certain we have (you see such just has to exist, say ) we will then have and thus which is bad…. Unless .

That is to say,

As a side, for all . Ergo, . Also, gives us . Combining these two equations we have either

or vice versa.

Do they really correspond to solutions at all? At present stage it seems that the equation (*) is suggesting that , and the ‘vice versa’ demands . Substitute them in … they work! We want to only pursue one line of reasoning though.

Observation: if is a solution, is also a solution. This is because . So we can really WLOG assume (*) is true. (why not the vice versa? Because is an involution, that is, which could simplify our work a lot potentially.)

First of all,

And here you go,

which is to say that So it remains to show is surjective!

Sounds nice, doesn’t work. Sad!

If surjection doesn’t work, one might still try injection. So let’s try to show injection. Doesn’t seem easy… Suppose , and gives us that, provided that , . So let’s pause for a moment, and see if can be shown…

As a matter of fact, your OP got exceedingly lucky when trying this problem. He somehow made a bad mistake that gave him true conclusions. Don’t ask him how. The OP just somehow figured out that

To show (4), we look back into the original equation .

Combining the above two pieces of computation we get

by referencing (2) and (3).

Whooosh. That was precarious. But now we have (4), which gives us not only that , but also that, if , , and if we both sides, . Suppose we could show that , we would be able to ‘undress’ the RHS and get and thus getting . Sounds like a plan. Well, (3) says if , and (2) says if , . Now so it works.

Hooray! We have just shown that if with , we must have and thus . Reference (2), we have that . Recall that is the *only* root to , we have . Thus . YAY!!! is indeed injective!!!

The rest is easy. (3) shows that and by injectivity and by (3) we have for all .

The solutions are and

**Comment**

What a tedious problem…. Took me two hours + to solve. It’s easy to leave gaps in the solution here and there, for instance the OP believed he got the identity that but in fact it demands to hold. It helped him discovering (4) though. (4) is really important (and somewhat natural to think of after we have discovered (3)).

The entire solution shows the power of setting instead of because (3) will take some much more deplorable forms had we chosen to work with, so will be (4) — in my humble opinion. I am happy to be shown wrong and be presented a proof using as starting point — I have not yet tried, but it looks disastrous to me.

OP was surprised to see that the key to the solution is injectivity really. The equation looks nothing like something for which injectivity would be a step towards solution — no naked variable anywhere. The point that is the only root to must be made for it is the essential cornerstone of any effort towards injectivity.

One might be tempted to use analysis, but I personally do not believe that analysis will be helpful. Analysis required inequality whereas this problem only gives equality that is difficult to be transformed into inequalities.

]]>**.**

Answer: .

Let the equation be .

*All right… Surjective… That means we will be able to let be whatever we want…*

* obviously works, and don’t work. don’t work either. Since the right hand side is symmetric in and , we’d expect the LHS to be basically symmetric as well (hmm we might be able to twiddle take this, at some point), thus we expect to be the only solution. (Constant solutions work but they are not surjective.*

*Turns out that term is hard to control, and we don’t find that easily.*

*We want to show it is injective. Because if it was injective we can twiddle take the equation and show by switching and , from which the solution follows straightaway.*

So LHS, as function of only depends on , and the RHS has only . Thus we see that , i.e. only depends on .

Now, let , then as . Let . We then have . *(This uses surjectivity!!!! It is only because the function is surjective that we can make range over the entire real number.)*

Seems like our injectivity scheme is not going well… Well we will twiddle take the equation anyway.

Define as . That is, is a period of . It is easy to check that this is an equivalence relation that is preserved by addition of equal term.

, or,

.

*This is like a pseudo …. Let’s sub in values and see… *

Let . (by surjectivity it exists.)

In (*) let , we see that .

. LHS, RHS. Thus .

. . Since , . Thus .

.

But since , thus . But .

*So, we tried to show that is injective, but ended up showing . That is still great though, because now we have . All these substitutions are found by trial and error, which makes the problem quite challenging indeed. The idea behind, though, is that if the “” sign is replaced by equal, we will be able to determine by equating coefficients of LHS and RHS of . We are just doing in modulo the set of periods.*

Since , all values in the image of are fixed points of .

But is surjective. Thus , thus every real number is a fixed point. Or, .

Comment: the middle part of finding the is tricky, and it is useful to notice the structure that all difference between pair of number giving off the same value are periods of the function. We used the surjectivity twice, once when showing that they are periods, the other when showing that the image of is the entire real number. The second time, despite that the equation could be reduced to additive Cauchy function, the zero function can still work, so we need to use the surjectivity to do this.

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**form the three sides of a non-degenerate triangle. (IMO 2009 Q5)**

Answer: .

Let the condition be .

*It’s obvious that works, and that or don’t work. With the term which is linear in the functional inequality, we expect only linear solutions to exist.*

*Because the triangle condition implies strict inequalities, we wish to use the discreteness of integers to reduce it. *

. The two inequalities reduce to

If , we have and is a period of the function . Since is defined on , that implies that is bounded. Suppose that .

. This is absurd. Thus .

*We did this because there is one unbounded term!*

Using the same trick as above,

.

Or, is an involution. This automatically shows that it’s bijective.

We already know a lot about the function now… If we could show any extra condition the conclusion should be within reach. Monotonicity, or all work. Let’s see what happens if . Since . WLOG as otherwise one can replace with . We have . Now, . Reiterate this inequality, we have

.

*What does this imply? , and we thus have that grows ‘more slowly than . We wish to make to create a contradiction. But how? We want to move forward with as the length of our ‘pace’$. Thus if we exhaust a complete system of remainder modulo , we will be able to show that for large enough. That *should *be able to finish the problem off: as will be bad, so we only have to make sure is ‘large enough’ to ensure a contradiction is reached.*

Let . Letting , we have (upon reducing into ,

Let exhaust the set , we see that for large enough we indeed have .

We now claim that as . This is because is injective, and thus there are only finitely many such that for any fixed . Thus, if we let be such a number that , there exists such that . We let . Contradiction. Thus . We conclude that .

Comment: Easy problem. We used proof by contradiction because the problem’s condition is ultimately an *inequality *for which it is easier to assume certain inequalities to compare the values. One should be able to solve it rather quickly: Andrew Elvey Price only took half an hour to finish it up.

The following problem is basically a spin-off the previous one, and is left to the reader as an exercise.

**Find all function such that **

**1) **

**2) **

** **

**form the three sides of a non-degenerate triangle. (AUS&UNK 2016 IMO Final Training)**

**Determine whether there exists such that for all pairwise different ,**

** **

**are the three sides of an acute-angled triangle.(Renhui Jiuzhang 2015 Prac Exam 6 Q2)**

**Find all function such that , pairwise different, **** form the three sides of a non-degenerate triangle if and only if form the three sides of a non-degenerate triangle. (China TST 2016 Q18)**

Solution to be published gradually….

]]>*All right, looks old-fashioned. For old-fashioned functional equation whose solutions are probably only the obvious ones, we aim to characterise its properties instead of its specific values. Sleeves up, let’s get to work. *

Answer: and .

(Checking is omitted for it is trivial.)

Let , and .

That is, are sets of second-order fixed points, zeroes, and periods of the function, respectively. We will see why we mentioned these sets very soon.

Let the statement of the question be . The condition looks rather symmetric, but not quite. This gives us space for manipulation.

or .

Having conjectured that only the two aforementioned solutions to the functional equation exist, we aim to rule out the second case.

. Substitution shows that this is NOT a solution. Thus .

Keep going, we are on track.

.

Look, look, these two switch between each other. Better take this information into account. .

But how are we going to exploit the second order fixed points? We have to substitute it into somewhere and compare it with the original value, so we’d better do it with something where and are both present, as in such occasion substituting will make this expression comparable with the original one. Such expression exists in the LHS.

,

as and . If , as and . Thus . If , . Thus in whatever case, .

Look at that. That looks like quite some information isn’t it, especially after we substitute in…

.

That wasn’t a big surprise, since the and are supposed to either cancel or both give vanishing function value anyway. But this expression has an advantage….

, and the LHS of this expression is equal to . Thus,

or .

Once we are here normally there are many ways to finish the problem off.

Claim: .

If , .

If . . If , we can divide both sides by and conclude . If , . Thus .

If only vanishes on , we recall the statement (*), in which , thus . But anyway, thus .

If , then is a period of the function. .

But now, . Thus . Which, since , forces us to have .

Comment: Boring problem, just a few substitutions to crack it open. But the substitutions are not that easy to find, especially the middle part where one needs to play around with the second order fixed point set. The part proving that all zeroes are periods makes use of the ubiquity of on both sides (since it is ubiquitous we let it vanish to reduce the equation down.)

]]>**1) **

**2) .**

**Show that .**

Let .

*The looks DAUNTING. Let’s try a small case with both the ‘s replaced with .*

*If large enough, , we will need because if , we will have which is bad. Therefore we inductively show that all large enough will have . Then we will choose large enough odd and , and thus and , which blows up the inequality 1) by . *

*Right. We have a rough idea now: letting infinitely many non-fixed points exist and then we show that there are A LOT of non-fixed points, in which set we can do stuff. Note that the above ‘very large’ only requires .*

Choose . We claim that .

We suppose it is not true, and let and , with .

We want to control and at the same time, thus we want to use Chinese Remainder Theorem to get a good residue modulo and . But CRT requires co-primality of the different modulos that we are putting together, thus , because .

Let ,

. (This can be done now by definition of .)

We claim that for such , .

Because if not, , and . The second equation is true because if , we have . If , we then have . We thus need which is not true as and .

*(This is why we defined to be that number !!!)*

Thus we see that there exists an infinite arithmetic progression whose large enough terms are all elements of . We will just write for the arithmetic progression.

*We now want to let () have prime factors that we push to infinity, whereas themselves must not be tainted with those factors. And then we will compare it with other stuff, with the ‘other stuff’ better be coprime with , with having enough common factor with to blow up the inequality. (Don’t say this word at airport.)*

Choose , where to be a collection of different primes which are all greater than and all greater than the common difference in . This can be done as there are infinitely many primes. *(Yes, infinitude of primes is a GREAT theorem. It gives you a nice collection of modulos to Chinese Remainder on. And Chinese Remainder theorem is used almost always to only show the existence of a certain desired solution taking the form of an A.P., while not accentuating on the numerial value of the solution.)*

Let

We need to show that this can be done. All we need to show is that all the ‘s are coprime to the common difference of , which is true since we have chosen them to be large enough.

*(This step is why we took all the pain to show that elements of not only exist, but exist IN AN ARITHMETIC PROGRESSION, and then we can select the desired subsequence we want from it by Chinese Remaindering the arithmetic progression with other prime modulos. Only assuming that has infinitely many elements restricts what we can do on it.)*

*Note that what we are doing now DEVIATES from what we did with the simple case, because now it is rather difficult to control over what happens to as now we only have instead of what it actually is. Thus I wish to restrict all the values of and then compare it with OTHER positive integer’s function values instead of integers in the arithmetic progression that defined .*

Fix the , and suppose , where .

Now let

for each .

and .

Again, we need to show that this can be done, and it suffices to check that . Assume this is not true, and . Recall that .

The above inequality is impossible due to the fact that . Thus the Chinese Remainder construction of is indeed possible.* (Note that the requires the full power of .)*

*OMG SO MANY CHINESE REMAINDERS… We use it to control the remainders of the variables dividing each other so as to manipulate the gcd’s.*

*Finally, we have an with all its possible function values being star-crossed, and a , not necessarily in , with all its function values holding such an enmity with function values of . This will blow up the inequality.*

Substitute and into the inequality, . The RHS as . The LHS , where , and since , LHS . Contradiction.

**Comment**: while the idea is easy, it takes great care to babysit all the details in the Chinese Remainder Constructions, as one easily runs into modulos not being coprime or remainder actually which must be evaded if we wish to estimate the gcd. All the first half is done to generate an A.P. of ‘s elements. I admit it can be annoying to see such problem posed in an exam, but we now see that all the crazy and daunting ‘s are indeed red herrings (its status of red herring should be apparent on first glance), the factor that really matters here is the FINITENESS of all the indices, for which a Chinese Remainder will always work because we only need finitely many congruence equations to hold simultaneously–and we can use A LOT of difference congruence equations to make our life easy as there’s not taxation on considering many congruence equations at the same time.

**Suppose satisfies:**

**1) if and only if **

**2) **

**Show that prime, implies .**

This problem seems to be the precursor of the one in 2014, and is left to the reader as an exercise.

]]>**Show that **

*All right. This looks like a spin-off of Cauchy equation, which, when fixed on rational numbers, admits only trivial solutions, and the deduction is done by first doing it on integers and then extending it to rational numbers.*

*The codomain being hints that it is not advisable to iterate the function. We might want to perform preliminary investigation on it first.*

Let the condition 1 be , condition 2 be .

, or, . Thus, , since .

. This can be used to inductively prove that .

. Since where , and

Let Thus is strictly monotically increasing.

*That was the (trivial) preliminary investigation. We have obtained some upper bounds on the function’s values at integers, but further investigation seems beyond reach. This is because we cannot really find any upper bound on the function values that does not involve further reference to function values. *

*(In fact, at this point, if the fixed point condition is ignored, fits into all the above equations. I did not realise this when I attempted the problem, but it does, in hindsight, point at the necessity of utilising the fixed point condition.)*

. This can be used inductively to prove that .

*YES! I has it! Upper bound! From now on we will just try to compare the function value with the powers of in order to apply a sandwich squeezing type of idea to fix the because we were worried about growing too rapidly.*

**Claim**: .

Fix . We define , and . We thus have .

*(Here we are defining extensions of floor and ceiling functions.)*

Let be , and be . Let , as since , we have . We will let be large enough such that . Thus , from which .

as and .

*At this point we are still extending the property of to obtain more points with . But the next inequality involves the idea of approximating the “ polynomials” with “multiples” of . The approximation idea is why we defined that generalisation of floor and ceiling function in the first place: we cannot use ‘s polynomials to attack straightaway, we then do it approximately and then analysis will take care of the error.*

On the other hand,

as (by the definition of and , and our choice of being large enough), and the fact that .

Dividing (*) by (**) (as both inequalities have their both sides positive), we get

.

*The equation (!) is our main breakthrough. We will very soon let both sides tend to their limit as , thus the inequality will practically look like . *

Thus it only remains to show that both and . The first one: , thus . Now let shows that .

*If we consider the worse case scenario, . But if the worse case does not happen, substituting into the functional inequality is illegitimate.*

If , . (The last inequality is true by the monotonicity of !) If , .

In either case, is inevitable.

This is the approximation idea at work. We have successfully controlled the error uniformly. We only need to show that as to establish the claim.

But this is obvious since (As , and . And now as , our claim is proven.

We already have , and thus we automatically have .

. Thus . This combined with the claim, which requires , gives .

Q.E.D.

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From this post on I will try to adopt a more casual tone in expounding the solution, with explanatory text mixing with the solution.

First, we see that the equation does not easily lead itself to non-trivial solution (the term is quite weird), therefore we decide to punish it by repeatedly substituting different values of variables until we can compare the information about the values to find a contradiction.

Setting , and cancelling gives .

Let be any integer, substituting will give us Since the LHS vanishes, we thus have or . Thus we only have to rule out the latter case and the proof is complete.

Notice that the functional equation, though strange in , behaves benignly when only the parity is concerned: in fact, setting , we have a ‘solution’ to it . Thus we want to consider parity. Setting , we have The RHS, incidentally, is always an even number, whereas the LHS is the function value of any even number. Thus maps even numbers to even numbers. (Note that the same cannot be said of odd numbers as is a solution to the equation.)

We aim to utilise the assumption that We let , for convenience. Let in the original equation, we obtain . Now we implant our initial strategy of trashing the function with repeated substitutions.

Let , we find Setting , we thus have thus . We thus know that and are all contained in the image of .

Setting , we have .

Setting , we have . Setting in the place of .

Setting , we have .

Subtracting the two equations above, we get

(This is why we used instead of when , as that would permit us to subtract and reduce one term.)

Upto this point, we haven’t used any property of (we haven’t in fact used the parity condition). But in an Olympiad problem, we should aim to use all the given conditions, as the problems are almost always tailored such that the conditions just suffice for a solution using elementary method. Now we observe that the LHS of the equation (*) is the difference between two terms with distance apart, while the RHS has a factor that is also a difference between two terms with distance apart, with another factor whose absolute value is at least if we can make sure it is not zero. We want it to be . This can be done by inequality.

If , then is a constant on even number (with that value being thus as ) and a constant on odd number (with that value thus being as otherwise would not be in the range of .) Now, we have shown that is also in the image of the function, thus this is a contradiction.

Thus . Consider the set , with its minimal element being attained at . Substitute into (*), and take absolute value.

.

By the minimality condition, , and because it does not vanish (as the LHS is non-zero)

Multiplying them, we have equality in the face of two inequalities. Thus the two inequalities must have equality holding at the same time, thus . This is impossible since we have shown above that

Q.E.D.

Comment: The problem is not very difficult, and many alternative methods exist, as the function itself is quite bizarre and repeated substitution per se gives more and more information from which any contradiction would finish the problem off.

The problem is proposed by Qiang Shen, and used in a practice exam at Renhui Jiuzhang MO Tuition Centre (Beijing, China) in 2015.

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Let .

Consider set . We claim that this set is bounded, thus finite.

It is bounded below by by the strict monotonicity. If it was not bounded from above, we have . Now, since and that is strictly increasing, Thus This is impossible because Thus is bounded, and thus finite.

We claim that . If , for large enough, we must have , which contradicts the monotonicity of . If , we then have which is unbounded for . But we know that which is bounded. This is a contradiction. Thus .

Apply to both sides of , we have

Consider the minimal element of , let it be , attained at . Let and in , we thus have $f(a\eta +a+b)=af(\eta+1)+b=af(\eta)+a\mu+b.$

Because is the minimal element of , it is thus necessary that In particular, the equality is attained if and only if Let Thus equality IS attained and we must have .

Recall that , letting in the problem’s condition gives and Since that implies

Let the maximal element of be , and repeat the same argument above, we would also have . Thus is a single-element set, or, the difference is a constant for all . This proves the statement.

Q.E.D.

**Comment**: Superficial investigation can only lead one to and , and it becomes difficult to move on from here as one has to disentangle the iteration of the function. The problem only gives very limited information about the values attained by , and the information accumulate near the range of the iteration. We see that this range is in general not the same as , and thus we have to find an argument to integrate the scattered local information. The definition of and plays this role by interacting directly with the function values elsewhere (considering s values and differences between two consecutive terms, where the ‘equality of the global inequality implies all local equalities’ is really the fruit of the global consideration in defining ).

After knowing that implies , iterating the linear function to try to give a full description of the set of positive integers on which the consecutive difference is minimalised is a natural idea — but turns out uneconomical: it is not clear how to force the subset that minimalise the consecutive difference and the subset that maximises the difference to intersect in that manner. In fact if that could be done the second problem condition becomes unused in deriving the solution–uncommon for Olympiad problems. Thus we seek to show that the subset of on which the consecutive difference is minimised (maximised) intersects the image of instead, which is already done.

I reckon this problem is quite nice (except its original formulation in terms of sequences which uses sub-sub-scripts which is painful to read), as one does not have access to any function value at any point (thus avoiding some ad hoc phenomena) but instead requires conceptual consideration. It is impossible to gauge the consideration of the function with any non-trivial solution, as there isn’t any.

When the functional equation acts on integer sets, it is almost always (practically, the ‘almost’ is redundant in Olympiads) important to consider properties characteristic of integers (extremal elements, number theory, discreteness, etc.) as if a solution without these considerations is possible the problem proposer would just define it on a wider domain. Because direct comparison of values maybe impossible, it might also be useful to consider asymptotic behaviours of certain auxiliary functions to gain information about parameters (here the proof that .)

The problem is proposed by Gabriel Carroll, USA, and appeared in the IMO 2009 as Q3.

]]>**1), if and only if .**

**2), .**

Answer: .

Proof: does not fit into the criteria because it requires . Fix .

**Bertrand’s Postulate (a.k.a. Tchebyshev’s theorem):** , with prime.

Now we aim to find a such that .

For , take . For odd, , apply Tchebyshev’s theorem to , there exists a prime . But now is even and thus is not prime, and thus we have , which shows that is the desired prime number as . For even, apply Tchebyshev’s theorem to to yield a prime which gives . The has been chosen for all .

Let , . Let .

(E.g. let , then .

Define an equivalence relation on : for , if and only if . It is easy to verify that this is indeed an equivalence relation. Partition into equivalence classes , by this relation. Let be indexed by the set .

**Axiom of Choice:** If is a family of sets indexed by the set , then the Cartesian product is non-empty.

Apply Axiom of Choice to the equivalence classes of , and we get a family of elements , with each corresponding to exactly one .

Let be the -adic evaluation of . (That is, reduce to simplest form , if does not divide , and if .

Let . For , suppose . Define . (E.g. if .Similarly, . This gives .) We set

Indeed, we show first show that the function satisfies the conditions.

First, we see that does not vanish for non-zero .

We claim that if . This is obvious by substituting the definitions of and , and then LHSRHS.

Now, we verify the conditions. Condition 1 is already checked. If , . If , let . For with , we have , and thus . Since , there are such s. .

Thus . We have found the desired function, for each of .

Q.E.D.

**Comment**: to be written later.

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First, set gives us inequality for all .

Let the inequality be denoted by .

gives us for all . Similarly, we have , by switching and .

Adding the two inequalities up, we get . Let in this inequality, we see that .

If , we see that from , we get either or . We now try to rule out the latter case, by observing that replace in the inequality with and we get . This demands to be non-positive.

Now, since is positive, is non-positive, we have for that certain , which contradicts our initial inequality . Thus this cannot happen and .

It remains to show . Now, we re-use the inequality by letting be any negative number, which gives off and thus .

gives . Assume now that , and let , which requires since is positive, and thus we see for large enough, , which is impossible as now is negative and . Contradiction. Thus has been wrong which combined with gives .

Q.E.D.

**Comment**: The functional inequality does not lead to simple non-trivial solution, and we thus hope to exploit the “badly-behavedness” of the condition by reducing it to something simple yet non-trivial. (In this problem’s case, the inequaity .) Since the condition is an inequality, it is not easy to derive an *equality *out of it. In this solution we established equality from both sides, to thus force the equality to be true.

The motivation for the substitution is to yield a term on one side, being symmetric to the term on the other side. Since on the other side only appears once and appears outside , we can afford to substitute a relatively complicated form of . Once we reach , substituting is simply giving up the strength of that inequality while preserving its non-triviality. It should be noted that substituting leads to weaker inequality .

The difficulty of the problem lies in the difficulty in dealing with the iterated application of , given that it is not easy to find a non-trivial solution to this equation–under the pressure of IMO and the anxiety of the windmill problem. (I personally do not know of any elementary non-trivial solution to the inequality, which nonetheless makes it a great problem.) Also, functional inequality is much less common than functional equations. I reckon that if it appeared on the IMO as Q2 instead with the windmill problem being Q3, it would be better done–people would likely spend more time on it. And it proves itself not as challenging as the windmill problem after all!

This solution is probably very close to the official solution.

The problem is proposed by Igor Voronovich, Belarus, and appeared in IMO 2011 as Q3.

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