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

[goodplacetostop]

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 😂

[wojteksocha]

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

[TheKudo]

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.

[bxrwptp]

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)

[ncantor]

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

[clementdato]

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.

[ecuas_]

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

[michaelz]

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.

[AndreasHontzia]

That was really elegant. Thank you very much!

[howareyou]

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:

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

[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

[behnamrohani]

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).

[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.

[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.

[manucitomx]

What a nice problem.
Thank you, professor!

[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| <= |alpha|. (If there are no roots, then clearly p = 0 or p = 1; otherwise, there is necessarily a root of greatest magnitude.) Then, plugging in either alpha+1 or alpha-1 gives us 0 on the left-hand side, so it must be the case that p((alpha+1)^2-1) and p((alpha-1)^2-1) are both 0, that is, alpha^2 + 2alpha and alpha^2-2alpha are both roots of p. If we think about the geometry of the complex plane, alpha^2 makes an angle of pi/2 or more with at least one of 2alpha and -2alpha, because for any z and w, one of (z and w) or (z and -w) is a pair with angle pi/2 or more. For one pair that has such an angle, its magnitude is at least sqrt(|z|^2 + |w|^2) by the law of cosines. As such, between alpha^2+2alpha and alpha^2-2alpha, the one with greater magnitude has magnitude at least sqrt(|alpha|^4+4|alpha|^2) = |alpha| sqrt(|alpha|^2 + 4). Since this is a root of p and alpha had the greatest magnitude across all roots of p, |alpha| sqrt(|alpha|^2 + 4) <= |alpha|, which is only satisfied for non-negative |alpha| if |alpha| = 0. As such, p has no non-zero roots, which gives us that p = 0, p = 1, or p = x^n for some positive integer n. It's easy to check that all of these satisfy the given equation.

[f-th]

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

[dmitrymiloserdov]

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

[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 <= 1
this means that r = 1 because for any other value r-2 would have real value < -1
that means 1 is only possible root and gives us:
q(x) = (x-1)^n
checking equation, like in video every n is possible:
(x-1)^n * (x-1+2)^n = (x-1)^n * (x+1)^n = (x^2-1)^n
since p(x) = q(x+1) then p(x) = x^n