Almost an IMO Problem | IMO Shortlist 2019 N2

#MathOlympiad #IMO #NumberTheory

Here is the solution to IMO Shortlist 2019 N2!
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.

[mcwulf]

I always start these kinds of questions with a=b=c. In this case we get a = cube root of 3. As all three variables can't be higher than this, at least one is lower, which means c=1. Repeat and we get b=1 or 2. Try both to get (3, 2, 1) and its combos.

[tonyhaddad]

For an easy way if a=b
a^3(a-2)=1 then we do not have a positive integer solution for (a)
There for (a) can not equal (b)

[AllanKobelansky]

Well done. Well presented. You are in Michael Penn territory. The road to 10k subs looks easily attainable.

[moonlightcocktail]

My solution was similar up to 3:00.
The equation is symmetric. Wlog sps
a >= b >= c.
a + (b3 + c3)/a2 = b2c2.
this is an integer, so a2 | b3 + c3.
Since a >= b >= c, b3 + c3 <= 2a3
b2c2 = a + (b3 + c3)/a2 <= 3a
bc <= sqrt(3a)
b3 + c3 is maximized when bc = sqrt(3a) and b = c = sqrt(sqrt(3a)). Then b^3 + c^3 = 2 * (3a)^3/4.
For a >=1, which a satisfies as it is positive,
16*27a^3 = 408a^3 < 625a^3 < 625a^4
2*(3a)^3/4 < 5a.
So, b3 + c3 < 5a.
But a2 | b3 + c3 < 5a, so a = 1, 2, 3, 4.
When a = 1, no solution exists.
When a = 2, no solution exists.
When a = 3, we get the solution
(3, 2, 1), and no other solution.
When a = 4, no solution exists.
Thus, the only solution is (3, 2, 1), up to permutation.

[sao]

2:29

how do you conclude a2 <= b3 + c3
??

[adamzoltan]

Or (abc)**2 = 36 thus abc = 6 so a and b and and c is less than 6 and you can only divide 6 by 1, 2, 3, 6 you cam just check for them

[jaysy]

Why not just remove common factor in 7:44? We can divide them out since a is a positive integer and therefore the expression a^2-a+1 is never 0.

[mra]

If negative was possible then, a=1, b=-1 and c=1, can be a solution.

[Navodayaclass_]

By intuition i found abc triples -1, 1, 1 ordered

[mumtrz]

Here's a homework:
a^b + b^c + c^a = (abc)^abc

[MgMG-igqg]

I don’t get how we get that a^2<=b^3+c^3. I do get that a^2<=a^3+b^3+c^3, but a^3>0 no?

[smashliek]

3:27 i dont get it, why can RHS exceed 2 ?

[ericlorentzen]

How did you determine that (abc)^2 <= 3a^3? In the worst case scenerio of a = b = c wouldn't it evaulate to: a^6 which is not strictly less than 3a^3? Please tell me what I'm missing.

[anupamrawat]

Amazing solution

Hope your channel grows soon
Your problems are great

[Nublolllzxx]

Hi
Why is it at 5:07,

a² - b | a³ + 1 - a(a² - b) ?

Thank you

[rounaksambhwani]

man pls can you recomend me any book of maths which starts from basic to this level pls

[tvvt]

4:02 wait how did we Ho from c^4 b <=18 to c^4>=32??

[turtlez]

At 5:07, how do we know a < b^2? Isn't writing that a^3 + 1 - a(a^2 - b) is positive an assumption?

[dionisis]

Please make a platform which we can write our favourite problems

[prithujsarkar]

Can you check this problem out from 1992 AIME Problems/Problem 15 ?
The problem states that
Define a positive integer $n$ to be a factorial tail if there is some positive integer $m$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?
One of my fav AIME problems :)