Putnam 2024, A2, A Polynomial Functional Equation

[@Itachi_Uchiha3112]

please cover the rest of the problems as well in the following videos(especially waiting for problem A5).
much appreciated.

[@Dvir226]

Wow nice problem!
Can you note what are the most important properties of polynomials that we need to use when confronting these questions?

[@DrEbrahimian]

Check out my book, Putnam Guide. I have a chapter on Polynomials. blog.umd.edu/ebrahimian/teaching/

[@abdulllllahhh]

can you try B6? I feel like it was one of the more doable B6 questions, but i didn't wanna risk wasting time trying it in the exam.

[@DrEbrahimian]

I haven’t gotten to session B yet. I’ll try!

[@АлександрОрлов-ж6д]

is that for all real x?

[@DrEbrahimian]

Yes!

[@ChristopherBitti]

Don't we know that p(x) - x divides p(p(x)) - x because p(x) - x = 0 (mod p(x) - x) => p(x) = x (mod (p(x) - x) => p(p(x)) - x = p(x) - x = 0 (mod p(x) - x) ???
I did something similar for the main portion of the problem. I noted that (p(x) - x)^2 = p(x)^2 - 2xp(x) + x^2 so p(x)^2 = 2xp(x) - x^2 (mod (p(x) - x)^2)
Now assuming p(x) has degree n >= 2
This allows us to reduce p(x)^k mod (p(x) - x)^2 to something of the form a(x)p(x) + b(x) with degree n + k - 1, this can be proven with induction
Then p(p(x)) - x = c_n * p(x)^n + ... + c_1 * p(x) + c_0 - x (mod (p(x) - x)^2) and this right hand side is congruent to a polynomial with degree 2n - 1 by the above observation which has nonzero degree smaller than (p(x) - x)^2 and so it can't be congruent to 0. Thus n < 2. Figuring out the rest is easy as you showed in the video.

[@DrEbrahimian]

Yes, this is similar to what I did. Thanks for sharing.

[@ChristopherBitti]

@@DrEbrahimian Yes, same idea, just slightly different execution. Your way definitely seems more slick.