A very interesting polynomial problem that was longlisted for the IMO!

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[@Krekkertje]

1. Notice that it works for p(x)=x 2. Have a gut feeling that it will also work for powers of x. 3. Check for an arbitrary n that it indeed works for p(x)=x^n (not very hard) 4. Decide that you’re just gonna assume that there are no other solutions and that even if there are any, you can’t be bothered

[@anshumanagrawal346]

lol

[@adityajha4887]

Sigma Mathematics

[@mohamedshafeeq7265]

Ofcorse sir.. I ve done the same.. Plz be check my procedure..

[@sirlight-ljij]

That's the joy when sometimes in these functional equations some bizzare, unexpected and totally bonkers function satisfies it as well. Like the general solution to Cauchy's equation f(x+y)=f(x)+f(y)

[@OuroborosVengeance]

You sound like a true mathematician

[@goodplacetostop2973]

0:01 I dunno why but I’ve read the problem on the thumbnail like : (prime number)(x-1)(prime number)(x+1) = (prime number)(x2-1) 10:07 Hey Michael, I was watching Bouldering at the Olympics and I was wondering at what time you will compete 😂

[@wojteksocha2002]

The proper way to deal with this problem is to write p(x) = a*x^n + q(x) where deg(q) < n. Then you compare coefficients on both sides of the equation and you'll get that deg(q) = 0

[@juyifan7933]

What do you mean "proper"? Was that solution in the IMO compendium or something? That looks super messy. The video solution is incomplete because it is missing alpha=w-1, where w is a root of unity, but as a solution method it looks more aesthetic.

[@wojteksocha2002]

@@juyifan7933 It's proper because: 1) representing polynomial in that way is a very typical move 2) solution in the video is incomplete, what if alfa is a complex number? What's happening then?

[@ChuckLincoln]

That's how I did it too, but there's no one proper way to go about it as long as it works.

[@leif1075]

Why in the seven hells woukd anyone write alpha plus obe minus obe..that's just silly and random and infuriating because no one would ever think of that..

[@pahom2]

It would also require lots of words and notes. You can only directly compare coefficients if x is an arbitrary number not equal to 0 1 i -i etc... then the problem turns into the challenge to find all that constants.

[@santinodemaria2818]

1:47 restarting windows

[@anshumanagrawal346]

😂

[@AndreasHontzia]

That was really elegant. Thank you very much!

[@manucitomx]

What a nice problem. Thank you, professor!

[@pietrobaiardi5020]

Great explanation!

[@user-bx7rw1pt4p]

5:38 if you suppose that p has a root then alpha could be complex. Why you didn't study the special complex case?(like i-1 , -i-1)

[@user-ik2kd9mb5t]

I'd like to add that a polynomial can have no real roots like x^2+2x+2

[@typha]

To ask your question another way, (for those who aren't convinced that this is improper) when did he rule out the possibility that p(x)=x^2+2x+2 ?

[@LucasCAPS]

I thought of that, but we would still get those zero exponents in the end, so all fitting polynomials would still be 0 and the monomials of coefficient 1.

[@typha]

hmm. Well maybe we would be able to still do the second part (starting at 6:50) in the same way as he did, but we'd have infinitely many options for alpha. But I'm not sure if things will end up being so cut and dry... let's see... we'd have for some {n_k} non-negative integers, only finitely many being non-zero p(x)= (x+1) Π_{k=-1}^{inf} (x+1-e^(iπ/2^k))^n_k. And then we'd have to do the same substitution business... but we can use a little geometry to help... the zeros of p(x) are all located in the complex plane on a circle of radius 1 centered at -1 (the center, -1 itself). So the zeros of p(x-1) will be located on a circle around 0, plus 0, and the zeros of p(x+1) will be located at on a circle centered at -2 plus -2, and the zeros of p(x^2-1) will be located at 0 and on a strange oblate figure centered at 0 only intersecting the unit circle at the points -1,1,-i,i . Since p(x+1)p(x-1)=p(x^2-1) the union of the set of zeros of p(x+1) and p(x-1) has to be contained in the set of zeros of p(x^2-1). So the largest the set of zeros of p(x^2-1) could ever even conceivably be would be the intersection of the previously mentioned set with the union of the other two. in other words, p(x^2-1) could only have roots at {-1, 1, 0, i, -1}, the zeros of p(x+1) and p(x-1) would also have to come from that set. We now have a very finite number of options and we could reason as he does starting at 6:50, but with a few more options thrown in the mix. but that was all by no means a trivial :/

[@ecuasonic_7]

7:13 when you thought you did really well on the AP Bio exam but instead got a 4

[@quinnharris3283]

To be fair that is still quite well

[@peterquartararo3249]

brilliantly presented.

[@TheKudo555]

From 6:00 on, α = -1 + e^(2 i ​π θ) with θ rational will give a finite sequence of roots. You must take into account complex roots, as even the roots of real valued polynomials can be complex, and you are talking about any root of this polynomial. (More so, real-valued polynomials can have all imaginary roots) To resolve this issue, as mentioned in other commentaries (see Behnam Rohani), we can do the same trick to show that all (α - 1)^(2^k) -1 must also be roots. This means 1 and all α = +1 + e^(2 i ​π θ) with θ rational are the only numbers that yield a finite sequence. 0 is the only number satisfying both conditions.

[@adityaekbote8498]

Any tips on learning functional equations ? Like a list of types of functional equations(like they have for diff eq) or any book recommendations?

[@omenuawo]

1. Functional equations and how to solve them - C. Small 2. Functional equations: A problem solving approach - B. J. Venkatachala 3. Functional equations - T. Andreescu First one has a style of a textbook. Although it requires a little bit of calculus, you can probably go through it without it. Second book presents some methods of solving functional equations through problems. And third book is a typical problem book, although the problems are divided into sections, according to the method used to solve them. If you go through all of them, in the order i listed them, you will cover pretty much everything you need to know and will obtain some intuition and problem solving skills for these kind of problems. There are pdf versions available online. (if you are interested in polynomials specifically, I can recommend you materials too)

[@n8cantor]

I think I found a different (maybe simpler?) method: Consider all the roots of P(x): r1, r2, ... The roots of the LHS are obviously r1 ± 1, r2 ± 1, ... The roots of the RHS must be the same, and for P(x^2 - 1) to equal 0, x^2 - 1 must be a root of P(x) Consider the (possibly complex) root of P(x) with the largest magnitude, let's call it rn Choose the root of the LHS, rn ± 1, with the plus-or-minus having the same sign as the real part of rn. Since rn±1 is a root of the LHS, it is also a root of the RHS. Thus P([rn±1]^2 - 1) = 0 The only values of P(x) that equal zero are its roots, therefore [rn ± 1]^2 - 1 = rk for some root rk of P(x) [rn ± 1]^2 - 1 = rk rn^2 ± 2rn + 1 - 1= rk rn(rn ± 2) = rk And since magnitudes are multiplicative: |rn||rn ± 2| = |rk| By our definition for rn, |rn| ≥ |rk| for all rk ∈ {r1, r2,...} And we chose the plus-or-minus to be in the same direction as the real part of rn, which means |rn ± 2| ≥ 2 > 1 (I'll leave showing that as HW) Multiplying those inequalities, we get |rn||rn ± 2| > |rk|, which is a contradiction, unless |rn| = |rk| = 0 Because its root with the greatest magnitude is 0, all the roots of P(x) must be zero. Therefore, P(x) = Ax^n As in the video, we can show A = 1 and that the constant polynomials P(x) = 0 or 1 work as well. It's also easy to show that all natural numbers work for n since: P(x - 1)P(x + 1) = P(x^2 - 1) (x - 1)^n (x + 1)^n = (x^2 - 1)^n [(x - 1)(x + 1)]^n = (x^2 - 1)^n (x^2 - 1)^n = (x^2 - 1)^n

[@ivankaznacheyeu4798]

Let's m is maximal by absolute value root of P(x) (including complex) Then m^2+2m and m^2-2m are also roots of P(x) abs(m^2+2m) m=0 or abs(m+2)

[@clementdato6328]

For the complex case, the answer would be the same. We already know immediately from video that the solution set is contained in the (unit circle + zero)(denoted S). Do the same trick with p(x+1). For x to be a solution, (x-1)^2-1 must also be a solution. There are only three points in S which remain in S after such transformation: 0, z, and z* (z’s conjugate). This means our solution set is contained in these 3 points. However, (z+1)^2-1 and (z*+1)^2-1 are not one of these 3 points so they are not solutions either. Only 0 can be the root.

[@WriteRightMathNation]

I missed where in the video it is shown that S is contained in the unit disk translated one unit to the left. Can you point out where, please?

[@fedorlozben6344]

Impressive method!

[@michaelz2270]

Because you have to include complex numbers, you have to do it slightly differently... go in the opposite direction and say that if alpha is a root, so is (alpha + 1)^{1/2^n} - 1 for some complex 2^nth roots. The only way the sequence of such numbers can be finite is if alpha +1 is equal to 0 or 1, corresponding to roots of -1 or 0. You can then eliminate possible roots -1 as you did it there.

[@Wurfenkopf]

I'll take a guess that this problem did not make it to the IMO because it doesn't admit non-trivial polynomials as solutions. Actual IMO problems generally do

[@JM-us3fr]

Oh that’s interesting. I thought it was because students might be led astray if they have to consider complex roots, while students who don’t know anything about complex numbers wouldn’t even think to do that.

[@Wurfenkopf]

@@JM-us3fr That seems unlikely to me because in IMO you're supposed to use more advanced concepts than complex numbers. Nothing to do with calculus and derivatives, but other things like congruences and modules will do

[@sirgog]

@@JM-us3fr The IMO allows problems that require complex numbers. There's always a non-calculus solution though.

[@ImaginaryMdA]

Yeah, I think it might just be a little on the easy side. I mean, I could figure out the solution. IMO is usually too hard for me.

[@beeble2003]

@@JM-us3fr Anybody who's going anywhere near the IMO knows a ton about complex numbers.

[@dmitrymiloserdov911]

If first root is -1+e^(i*pi/2^k) for any k, then we get finite set of roots

[@tomatrix7525]

Great stuff Michael. Side note, have you been watching Olympic climbing or olympics in general?

[@ImaginaryMdA]

I did a similar thing, suppose you have a root x, then this implies roots x*(x+2) and x*(x-2). For any complex number either x+2 or x-2 has a modulus strictly larger than 1, showing that for each root there must exist another root with a strictly larger modulus, if the modulus of x is positive. So either there are infinitely many roots (p=0), no roots (p=constant), or the only root is 0 (p=ax^n). This reduces further to p=0, 1, x^n through standard algebra.

[@mzg147]

This is exactly the way I did it too. More elegant than the solution from the video I guess...

[@dproduzioni]

I'm a customary viewer of this channel but I got amazed every time this guy hits the blackboard.

[@howareyou4400]

There is a much simpler solution: Suppose the leading two terms of P(x) is ax^(n+k)+bx^n, then examine the expansion of P(x-1)P(x+1), we will have all the "diagonal terms" the same as P(x^2-1), like [a(x+1)^(n+k)][a(x-1)^(n+k)] = a(x^2-1)^(n+k). What's left over is the "cross term". Specifically we examine the cross term of the first two leading term: 2ab[(x+1)^k+(x-1)^k](x^2-1)^n. We can see that this will produce a term of x^(2n+k) which any other cross term is unable to produce. As a result we can concluded that 2ab must be 0. So P(x) is either a constant or a single term of X^n

[@howareyou4400]

2ab should just be ab

[@behnamrohani843]

Another way to accomplish the same result is to plug in \alpha -1 into the equation (assuming \alpha is a root of P(x) ), and according to the same reasoning as before, it tells us that (\alpha -1)^(2^k) -1 must be a root for P(x), but now, the only values of \alpha that don't make P(x) a polynomial with infinite terms are 0,1,2. Comparing it with the previous roots \alpha=0,-1,-2, they have only 0 in common, so P(x) only has a root of 0 (Why is that? Notice that any \alpha results in new roots (\alpha -1)^(2^k) -1 and (\alpha +1)^(2^k) -1, so the choices of \alpha=-1,-2 would create infinitely many roots using the first form, and \alpha=1,2 is also a problem considering the second form).

[@jorgecasanova8215]

If you consider complex roots, the reasoning is not right. The only thing you can say is that (alpha +1) and (alpha -1) is a unit root. This implies Re(alpha) >=0 and Re(alpha)

[@behnamrohani843]

​@@jorgecasanova8215 That's right! if \alpha +1 and \alpha -1 are respectively, let's say, nth and mth root of unity, then we (might??) get finitely many values as roots of P(x) so gotta check that (we're just "circling" around so we only have n distinct values for powers of \alpha+1 and m distinct values for powers of \alpha-1). Since \alpha+1 and \alpha-1 are unit roots, we must have -1

[@user-jc2lz6jb2e]

1:05 I expected you to say x^n for all natural n.

[@CauchyIntegralFormula]

If alpha+1 is a root of unity, then the list is also finite. I think a cleaner way of doing this is to let alpha be a root of greatest magnitude, so that all roots r of p satisfy |r|

[@HagenvonEitzen]

Or, once you have the two new roots (alpha+1)^2-1 and (alpha-1)^2-1, note that their *difference* is 4alpha, i..e, thier abslute difference is 4|alpha| and at least one of them must have absolute value at least 2|alpha|. This is bigger than |alpha| unless alpha=0. As alpha was of maximal magnitude, we must have alpha=0, i.e., p(x)=x^n

[@samuelmarger9031]

The problem is interesting indeed!

[@cmilkau]

if we assume that non-constant polynomials have at least one root, we must consider complex roots (note that even real polynomials can have complex roots). then, there are infinitely many choices for α making (α+1)^(2^k) stationary. Just start with say α+1 = -1 and keep taking square roots, α+1 = ±i, α+1 = (±1±i)/√2 and so on. All these will end in (α+1)^(2^k) = 1 for large enough k.

[@wavyblade6810]

What about the case when P(x) doesn't have real roots? Unless maybe I misunderstood and alpha can be complex

[@fevesvfr]

What about when alpha + 1 is a root 2^p of unity?

[@MrRyanroberson1]

4:30 what i find interesting is that the number of roots approximatetly doubles each application. from a, you get (a+1)^2 - 1 and (a-1)^2 - 1 from those you get (a+1)^4-1, ((a+1)^2-2)^2-1, (a-1)^4-1, ((a-1)^2-2)^2-1 The number of newfound roots and old roots differ by only one at every step. 6:40 Notice here that -2 does not provide finitely many roots, as (-2-1)^2 = 9, and 9-2=7, 7^2=49, we have 48 as a root. Similarly, -1 cannot be a finite root, we get 3 as a root, and therefore 7 and so on. 9:17 The algebraic comparison agrees with the finding that -2 and -1 are not valid roots

[@saadhaider9576]

What about complex roots?

[@Zaxx70]

If we plug x=alpha-1 into the equation we also get zero so (alpha-1)^2^k-1 is also a root, there should have been x-1 and x-2 terms in p(x) but ofc it doesn't matter since the exponents would be zero as well...

[@RadioactivePretzels]

Question: how to prove those solutions are the ONLY solutions? The given solution identifies only the set of p(x) solutions that have one or more real roots. What about possible polynomials that have NO real roots? (There AREN'T any that satisfy the functional equation, but how to prove that?)

[@lucassandleris4486]

If you let alpha=a root of unity -1 wouldn't that give finite options as well? I think they should specify that P has at least one real root.

[@niom9446]

elegant.

[@LittleLion72]

Elegant sol

[@cmilkau]

or, you could realize that the product of any two solutions (polynomials p satisfying the equation) is again a solution, so you only have to examine polynomials of degree 1 (or 2 if you don't know complex numbers) and are done in like 5 minutes.

[@kylecow1930]

essentially the idea should be: find constant solutions. assuming p isnt constant let x be a root, we find a sequence from constantly squaring which gives us infiite roots unless the process terminates, by splitting up the starting points you can find all the points that this sequence stops changing, we can then just check A(x-a1)^n(x-a2)^m... to see what powers work

[@bobzarnke1706]

A more rigorous justification for the "special" values of α is to argue that, for p(x) to have only a finite number of roots, some roots must be repeated, ie, (α+1)^2^k - 1 = (α+1)^2^m - 1 for some k>m>0. This implies that (α+1)^2^m((α+1)^(2^k-2^m) - 1) = 0, which implies that (α+1) = 0 or (α+1)^2 - 1 = 0, ie, α = -1, 0 or -2.

[@filipmunteanu2211]

It doesn't work because that alpha was supposed to be complex.

[@aradziv89]

What about complex roots? alpha=i-1 might also work...

[@juyifan7933]

Yeah, that was missing in the solution. The case i-1 can be eliminated because then i²-1=-2 would be a root, but as shown in the final step that cannot happen. But in genral alpha=w-1, where w is a root of 1 needs to be dealt with.

[@drsonaligupta75]

I was thinking the same

[@DrTaunu]

The big omission is that the polynomial has real coefficients. In that case, if you have an imaginary number as a root, its conjugate is also a root. One of those will always lead to an infinite number of roots.

[@juyifan7933]

@@DrTaunu Thats not true, if a complex is a root of unity, so is its conjugate, that "argument" leads nowhere.

[@soranuareane]

What does it mean for a problem to be longlisted?

[@bernieg5874]

Aren't there finitely many roots when alpha=(alpha+1)^(2^k)-1 for some k? And doesn't this introduce a new infinite class of solutions?

[@Chid417]

Great answer. But I am curious - what if p(x)=0 does not have roots? Then there is no alpha such that p(alpha)=0 and there can be another possible p(x) s.

[@andrewtugume1314]

After looking at and exhausting all the constant polynomial solutions, he then looks for non-constant solutions which by definition have a degree greater or equal to 1 therefore must have atleast one root by the fundamental theorem of algebra.

[@Chid417]

@@andrewtugume1314 Ah do you mean he also considered about that? I muted the video and just watched the blackboard so I could not get it if he explained just in his speaking, not writing in blackboard

[@Chid417]

@@andrewtugume1314 And like many other's opinion, alpha can be complex, and we should consider about all alphas such that magnitude of (alpha+1) is 1 (or 0 but alpha is unique by -1 in this case).

[@andrewtugume1314]

@@Chid417 He first took out the solutions p(x)=0 and p(x)=1 which are the only constant solutions, so if non-constant solutions exists, they have to have atleast one root so he searches for the non-constant solutions. And indeed there's need to consider possibility of \alpha being a complex root or show that alpha can only be real. Many have already showed how we can do this

[@Red-Brick-Dream]

Everyone: polynomials These guys: pOlYaLs

[@yoav613]

No HW today,go to the beach🏄‍♂️

[@NobleMushtak]

This is nice, but there is a solution which does not involve looking at the roots at all, which I think is easier to come up with on your own by trying some example p(x)s and seeing why they don't work, at least for me. First, we know p(x)=0 is a solution. Next, consider the case where p(x) is a nonzero polynomial. As explained in the video, p(x) must be monic, i.e. leading coefficient=1. Let p(x)=x^(n_1)+a_2*x^(n_2)+...+a_k*x^(n_k) be the polynomial, where n_1 > n_2 > ... > n_k >= 0 and a_2, ..., a_k are nonzero. There are two cases: (1) k=1, in which case p(x) is just x^(n_1), which, as discussed in the video, is a solution for all non-negative integers n_1=0 (2) k >= 2, in which case a_2 is nonzero. Using this assumption, we will show that p(x) is not a solution, which suffices to show the only solutions are p(x)=0 and p(x)=x^(n_1) Then, the first three terms of biggest degree in p(x-1)p(x+1) will be: (x-1)^(n_1)(x+1)^(n_1) + a_2(x-1)^(n_1)(x+1)^(n_2) + a_2(x-1)^(n_2)(x+1)^(n_1) + a_2^2*(x-1)^(n_2)(x+1)^(n_2) which simplifies to: (x^2-1)^(n_1) + a_2(x-1)^(n_1)(x+1)^(n_2) + a_2(x-1)^(n_2)(x+1)^(n_1) + a_2^2*(x^2-1)^(n_2) Next, the first two terms of biggest degree in p(x^2-1) will be: (x^2-1)^(n_1) + a_2(x^2-1)^(n_2) Subtract q(x)=(x^2-1)^(n_1) + a_2^2*(x^2-1)^(n_2) from both sides. Then, p(x-1)p(x+1) is left with: a_2(x-1)^(n_1)(x+1)^(n_2) + a_2(x-1)^(n_2)(x+1)^(n_1) plus several terms of degree less than 2n_2 and p(x^2-1) is left with: (a_2-a_2^2)(x^2-1)^(n_2) plus several terms of degree less than 2n_2. Since a_2 is nonzero, from inspection, we see that the degree of p(x-1)p(x+1)-q(x) is n_1+n_2 while the degree of p(x^2-1)-q(x) is at most 2n_2 (i say at most because it is possible that a_2-a_2^2=0, in which case p(x^2-1)-q(x) has degree less than 2n_2). Since n_1 > n_2, it follows that n_1+n_2 > 2n_2, so the degree of p(x-1)p(x+1)-q(x) is strictly greater than that of p(x^2-1)-q(x). Thus, p(x-1)p(x+1)-q(x) does not equal p(x^2-1)-q(x), so p(x-1)p(x+1) does not equal p(x^2-1), as was to be proven.

[@ivankaznacheyeu4798]

P(x)=C or P(x) is polynomial having finite number of roots (including complex)P(x) = C => C^2=C => C=0 or C=1Now consider the case of P(x) having finite number of rootsLet m is maximal by absolute value root of P(x) (including complex) Then m^2+2m and m^2-2m are also roots of P(x) abs(m^2+2m) m=0 or abs(m+2)

[@yoav613]

I did not like this problem,but your explanation is great!

[@playgroundgames3667]

p^2(x^2 -1)

[@beautyofmath6821]

Great

[@xoriun8638]

what about non-constant polynomials without any roots?

[@lilylikesmarkies]

Try and show that no such polynomial exists. Hint: intermediate value theorem

[@OscarCunningham]

@@lilylikesmarkies x^2 + 1

[@seanziewonzie]

I don't know if it was stated, but alpha is allowed to take on complex values. Therefore, no such polynomials exist.

[@OscarCunningham]

@@seanziewonzie Then we don't get that it can only be 0, -1 or -2. It could also be one less than any root of unity. (I think the neatest way to complete the proof is to use the analogous logic to show that α - 1 must also be 0 or a root of unity. Then those circles only touch at α = 0.)

[@seanziewonzie]

@@OscarCunningham Indeed I think the 0, -1, or -2 bit was unjustified

[@mohamedshafeeq7265]

Clearly 0,1,x and powers of x satisfy this equation. Now check possibility of kx^n, for nonzero k. Then we can see this is possible only for k =1 Now check the possibilty of x^n +x^m.. which will lead to a relation(x+1/x-1)^n-m = -1., but this is not true for real x.. x has to be a complex number... So as a conclusion Solution.. 0 and non negative integral powers of real variable x. We dont want negative powers as we need polynomial solutions. Plz..give me suggestions for correction, if any

[@alfreds1347]

6:45 Wait, why can't it be A*x^l*(x+1)^m*(x+2)^2*Q(x) where Q(x) has no roots?

[@Jay-ss6cc]

this polynomial does not have to have roots, right?

[@valatko]

Every polynomial has roots

[@Jay-ss6cc]

​@@valatko Did OP consider complex roots?

[@rbnn]

Let q(x)=p(x-1) so q(x)q(x+2)=q(x^2). If r is a root of q so is r^2. Hence r is 0 or on the unit circle. If r is a root then (r-2)^2 is a root, so 0 is not a root, and all roots are on the unit circle. The only r on the unit circle for which (r-2)^2 is on the unit circle is r=1. Hence the only nonconstant q are (x-1)^n, so p is constant or x^n. [extending another user’s comment].

[@nevokrien95]

Alpha has to be e^(2piim/k)+1

[@someuser257]

Well obviously if x≠+-1 p^2=p meaning p=0 or 1😎😎😎😎😎😎😎😎😎

[@LechuvPL]

You can simplify the solution and also easily include complex roots (wich many pointed out made the proof incomplete) by substituting q(x) = p(x-1), then the equation is: q(x) * q(x+2) = q(x^2) so if r is a root of q then r^2 also is this means that roots must be either on unit circle or 0, because otherwise magnitude of r becomes larger/smaller anytime you square it, generating infinite roots we similarly write q(x) = A * (x-r1)^n * (x-r2)^m * ... each root will give us factor of (x-r)^n * (x+2-r)^n on the left side of the equation (x^2 - r)^n on the right side All the roots on the right side are in the form of +/- sqrt(r) so if r is on unit circle they also are on unit circle and if r=0 the roots are also 0 this means every root of the right side is either 0 or on unit circle On the left side we have factor x+2-r so it has a root r-2 We need to find r so that r-2 is either 0 or on unit circle, so it can appear also on the right side but since r is also either zero or on unit circle we know that real value of r

[@rbnn]

Very clear

[@telnobynoyator_6183]

My attempt before watching the video If deg P = 0, P(x) = 1 or P(x) = 0 If deg P != 0 deg P(x^2 + 1) = deg P + 1 deg (P(x-1)P(x+1)) = 2 deg P 2 deg P = deg P + 1 deg P = 1 P(x) = ax + b such that a(x - 1) + b + a(x + 1) + b = a(x^2 - 1) + b If x = 1 we have 2b + 2a = b, b = -2a If x = -1 we have 2b - 2a = b, b = 2a -2a = 2a, a = 0 and so is b, which contradicts deg P = 1. Thus there is no solution P where deg P != 0

[@mehdimarashi1736]

hmm, what if p(x) has no real roots? if alpha is a complex number, (alpha+1)^(2^k)-1 can be equal to alpha for some value of k, and we won't necessarily have infinite roots. :/? Finding such values foe alpha is not that hard. alpha = (-3+i.sqrt(3))/2 is such a number, and alpha^4 = alpha. Needless to say, the quadratic x^2+3x+3 for which alpha is a root, does not have the property we are looking for. It's easy to show that the only linear p(x) = ax +b that has this property is p(x) = +-x and the only quadratic p(x) is p(x)= +-x^2. We can easily show that x^n and -x^n have this property, and c.x^n for c/= +-1 does not have this property. I can argue that if P(x) and Q(x) have this property, P+Q can never have it, unless P+Q=0. I was thinking that since x^n is a base for the vector space of polynomials and every other polynomial is a linear combination of x^n's, no other polynomial can have this property. But now I noticed that even that is not enough, because I have not shown if P or Q does not have this property what can happen. This problem wan not supposed to be this hard! I wonder if we can argue that if P does not have any real roots, P(x-1)P(x+1) can never be equal to P(x^2-1). I tried it, but I wasn't successful at all.

[@zadsar3406]

You never proved the sequence ak = (α + 1)^2^k - 1; a0 = α has finitely many distinct values iff α is in {0, -1, -2}. Clearly, it doesn't work for α > 0 (the sequence is strictly growing) or for α < -1, but I don't think it's as clear that for, say, α in you don't get ai = aj for some i ≠ j as the sequence appears to be bounded in .

[@wojteksocha2002]

Additionally, what if alfa is a complex number? Then things start to get even more complicated

[@zadsar3406]

@@wojteksocha2002 Absolutely true. This approach probably doesn't work without a lot more work.

[@jimskea224]

@@wojteksocha2002 Well the field over which he's working is never defined. Since the variable used is x not z you could argue that it's implicit he's working over the reals (though he does mention i once).

[@wojteksocha2002]

@@jimskea224 saying that the variable is x so that means it's a real number is very naive. There exist solution to this problem that's not using complex numbers, but Michael's attempt requires that.

[@Nnm26]

Correct me if I'm wrong but I thought this is pretty clear from the fact that c^x is bijective.

[@georgekh541]

The sulution is actully wrong because alpha can be a complex number

[@hornedowl293]

Polinom may have no roots

[@WriteRightMathNation]

I think the detailed induction is unnecessary. Once we know that the roots are closed under the unary operation that adds one, squares and then subtracts one, let's compare that to the input. Given a complex number z other than -2, we have |(z+1)^2-1|=|z^+2z|=|z| |z+2|>|z|. Consequently, the set S of all magnitudes of zeros of p other than 0, -1, or -2 is a set of nonnegative real numbers that satisfies this property: For each r in S, there is some t in S such that t>r. It's well-known that any such set of real numbers is either empty or infinite, but one can given an easy argument if you wish, using if you prefer, induction, or some other fundamental approach. In a any case, that's why the polynomial p cannot have any zeros other than 0, -1, -2. From that, showing that either p=0 or p=1 is as shown. It's a nice problem, Michael. Thanks for your videos.

[@rbnn]

Your inequality fails when z=I-1

[@rbnn]

i-1

[@writerightmathnation9481]

@@rbnn Good point. Thanks. So there is a special case to consider.

[@ZipplyZane]

NOTE: BAD MATH FOLLOWS: p^2(x^2-1) = p(x^2-1) p^2 = p p = {0, 1} That's what I got from the thumbnail, where I just treated p as a variable. But I knew it was too simple, so p(x) must be functional notation. And, yet, it does get the right answer. Seems like one where how you do it is more important than the actual answer, which might be why the problem didn't make it in.

[@emanuelvillanueva9240]

p is a functional notation? we can do p^2=p?

[@ZipplyZane]

@@emanuelvillanueva9240 No. I just didn't realize it was function notation at first. Yet my mistake still wound up getting the right answer. I thought that was funny.

[@natepolidoro4565]

cool

[@harshparashar9210]

Could you tell the year in which this appear🙄🙄

[@jimskea224]

It didn't. It was long listed, so didn't even make it to the shortlist.

[@harshparashar9210]

@@jimskea224 but man every time year changes In which year it was longlisted If you know you could tell tha

[@sonmohikan5100]

P(x)=x İts fitts too

[@LiangLao2]

here is a question, you said alpha is root of p(x) then so is (alpha+1)^{2^k}-1 , you seem like assmuming alpha is real nubmer, but you have to prove it .

[@shalvagang951]

Please try imo2021 q 2

[@xwtek3505]

Question unclear: polynomial over what ring?

[@nombreusering7979]

9:55 l can be any number, not just positive stuff

[@aradziv89]

If it's negative it's not a polinome. For example 1/x isn't a polinome

[@nombreusering7979]

@@aradziv89 Yeah got it now, forgot about the polynomial part of the question. Toda ya gever

[@CM63_France]

Hi, I found a solution in a second : f(x)=x !!! ok, .....

[@yurya.davydov8947]

At the stage when we determine alpha values to be 0, -1 or -2, don’t we have to prove that there are no more possibilities? I suppose that is a nitpick, but anyways

[@francescfloresgamez657]

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[@laretslucky2899]

actually thats a bad place to stop, cuz there may be complex roots

[@danielmilyutin9914]

i guess you're right. If set alpha = exp(i*2pi*m/2^l)-1 and we can have finite family of roots. This is only by MP reasoning. However, @Behnam Rohani pointed out that both (\alpha -1)^(2^k) -1 and (\alpha +1)^(2^k) -1 are roots of p. Thus, only root 0 is accepted.

[@stewartcopeland4950]

?