IMO, a Very Nice Number Theory Exercise.

Today we take a look at Question 4 of the 1998 IMO International Mathematics Olympiad. It asks of us to find all ordered pairs a,b that have the property that divide ab^2+b+7 divides a^2b+a+b. Enjoy! :)

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[MK-]

2:00 not equivalent. a|b implies b>a, but the reverse implication is false

[JohnSmith-vqho]

This solution was extremely well explained. I fully agree with what you said about explaining each step. I hope you continue to post more IMO/high level competition problems on this channel!

[dorian]

3:27 >mfw c|a+b <=> dab, approximated by the human eye.

[VaradMahashabde]

Papa : -1 is not Zero if you didn't know already
Anime girl : WOW!!

[sumukhhegde]

Very nicely explained. Even I had this type of question in one of my university entrance exam. But couldn't solve 🙈

[MichaelRothwell]

I couldn't resist trying this problem when you called it "a nice exercise in number theory", but it took me a few days to crack. My solution was pretty much like yours, but I used synthetic division to divide out the polynomials, which is neater I think.

[josephmartos]

Your right about that guy explaining hard competition math exercices like every of us know a bunch of hard math stuff. Very good job explaining every single step u make :)

[Markful]

F in the chat for the analytic philosophers

[thephysicistcuber]

I think the meme is supposed to say "pure mathematicians are just analytic philosophers who are good at math".

[octaviotastico]

I love the IMO videos, they're my favourite <3 love u papi

[raghualluri]

Excellent! I really love IMO problems, they are just so fun to do even if the math needed is elementary concepts. You can actually start making a playlist of like Contest Problems.

[anubratasaha]

Contest Math turned me away from math, cos I wasn't smart enough. Please do more of these problems, Papa. btw Loved this one 😘.

[RCSmiths]

Man, thank you for the Birthday upload man. Another great and informative maths video man!

[허공]

another method
a^b + a + b = k*(ab^2 + b + 7) for some positive integer k.
=> ba^2 + (1 - kb^2)a + (b - kb - 7k) = 0 ---(1)
let α, β are roots of a in equation (1), α + β = -(1 - kb^2)/b = kb - 1/b
since a, b, k are integers, 1/b must be integer. => b = 1
=> a + 8 divides a^2 + a + 1
=> (a^2 + a + 1) / (a + 8) = a - 7 + 57/(a + 8) => a + 8 divides 57(=3*19)
=> a = { 11, 49 } => (a, b) = { (11, 1), (49, 1) }

[HAL-ojjb]

9:19 Papa Flammy throwing shade at other channels left and right

But seriously, I really like that you always go through all steps and explain them, and only skip stuff that you covered already in other videos (like complicated integrals and stuff). That way it feels really like an accomplishment at the end of the video, and not as if we cheated by skipping the hard parts ^^

[sebastientraglia]

At 9:10 you should have specified that all (a, b) = (7r^2, 7r), for r € N, are solutions only because b divides a in those cases (since you multiplied the numerator by b in the beginning).

[AdityaKumar-ijok]

imagine trying to do this as a teen under time constraints:
mindf"ck 💯

[pukvandepetteflet]

This is so damn hard,
I’m trying to get into the IMO but first I have to be with the best ~30 Dutch students to be eligible, then we’ll have a great course and the 6 students who get the highest scores go to the IMO, rn I am with the best ~120 Dutch students and we’ll have the next test in September, but I think this might help getting there.
The explanation was great, I understood the steps even though I’ve never seen a “|” to use when something devides.
You might be the reason I participate in the IMO one day! Thank you keep it up!

[williamperezhernandez]

15:20 Just divide the polynomials to get residue of 57/(a+8) quickly getting a=11 and 49.

[diddierhilarion]

14:00 it should be b²-7 < 0 if it is equal to 0 the former inequality doesn't hold