Can you explain why all three variables cant be higher than cbrt(3)? I dont quite get the implication

@@typo691 Now I look again I see that it's not a given. If the RHS is a constant, yes. But actually we need to demonstrate that it's true (or not).

@@typo691 using the AM GM inequality

@@raytheboss4650Hi! I know it’s from a while ago, but how would you do this with AM-GM?

there will never be another michael penn. He is far too powerful.

​@@clarinetowl7554the one piece is real

b3+c3 isn't actually maximized if b=c=√√(3a)though. I see that bc=√(3a), but b=c is actually the minimum over all such solutions.

Yes, that’s exactly what I meant.

Look again at the question statement and the previous line ;)

@@DoReMeDesignAh. I got it. Thanks for the reply!

Because a^2 divides b^3 + c^3. To see why this is just look at the original equation to find that a^2 clearly divides a^3 + b^3 + c^3 then get rid of the a^3

we don’t know whether a,b,c are consecutive so this formula does not actually works here

--- a < b^2 --- We know that the inequality a < b^2 is true , because for an equation M-N , where M and N are natural numbers , If M-N is negative , N-M is positive and vice-versa. This also applies when one of the numbers is squared . You can try this with any two natural numbers that you want. --- a^3 + 1 - a(a^2 - b ) --- According to the inequality a^2 - b < 0 , If we multiply both sides by a , we get a ( a^2 - b ) < a ( 0 ) , which doesn't change what it holds . Coincidentally , a^2 - b | a^3 + 1 - a( a^2 - b ) has the same effect that a^2 - b | a^3 + 1 - ( a^2 - b ) has.

a(a^2-b)

@@cantcommute understand.

Where does it even say a < b^2 ? Or where does it come into play ?

I'm guessing that since c and b are integers, and c < b, c/b has to be less than 1 and hence bounded by 1

0 is not an inetiger brother

No youre not dumb, he just speaks really fast! The step follows because he earlyier showed that (a^2)-b divides a*b+1 and if an integer divides another integer, the dividing integer is less then or equal than the divided integer. Dont let your head hang down if u dont hear/understand one Thing!

Sometimes i also cant catch up in These yt Videos, because its way above the Things we do in class and we dont do any number theory in class as well

@@fix5072 thank you, i get it now. :::)

@@fix5072 and one more question: if c²b

@@katagagyi3523 if youre talking about (c^4)*b =c thus b>=2 and the entire lhs woul be atleast (2^4)*2 which is bigger than 18

Sure thing!

@@letsthinkcritically when c=1 a³+b³+1=(ab)² (a+b)(a²-ab+b²)=(ab-1)(ab+1) Two condition a+b=ab+1 and a²-ab+b²=ab-1 (no solutions because (a-b) ²=-1) a+b=ab-1 and a²-ab+b²=ab+1 (from this system we get b=2 and a=3) How to find c=1 is insane solutions 🙏

Because of the initial inequality a >= b >= c.

Just a line above, he got that a^2 is a divisor of b^3 + c^3 and since all of them are positive, a^2 has to be less or equal to b^3 + c^3

@@reeeeeplease1178 thanks. What does he mean when he says that a^2 “is divisor” of …. and why can he say it?

Applying it to the solution, we that 9 (a^2) isa divisor of 27 (a^3)+ 8 (b^3) +1 which is right, but how do we know? And knocking off a^3 how do we know the inequality still holds?

@@MgMG-ig4qg well from the original equation we know that a^2 is a factor of (abc)^2 and since (abc)^2 = a^3 + b^3 + c^3, we know that a^2 has to be a factor of a^3 + b^3 + c^3 This implies that a^2 is a factor of b^3 + c^3 (because a^2 divides a^3)

@@reeeeeplease1178 riiiiiiight :). Thank you so much….

He didn't find it himself. He just writes it down. Why do you think he doesn't explain anything at all except the most obvious things ?

i think because C > B hence it (c/b) will be less than 1

@@icebeargt5142*C

Feel free to let me know here! I will make videos on problems that you suggest.

Thank you!!

because from (abc)^2

Oh thanks lol@@Jan-vz5ge

Did you see my question?

Because we established a² - b | a³ + 1 and naturally a² - b | a(a² - b)

@@jm2270 Oh I see. Thank you This KZfaqr doesn’t reply at all

​@@jm2270what does a^2 - b | a^3 + 1 mean ? I don't know if they use different symbols in my country but I don't understand that operation at all ?

@@Emre67511 x | y means x divides y.

Thank you!!

I learn problem solving tricks by browsing content in Art of Problem Solving (AoPS: artofproblemsolving.com). I usually recommend online resources rather than books nowadays because online resources are better organised and you can find them very easily.

@@letsthinkcritically thanks !!!!

You can try pathfinder for prmo by prashant jain

@@krishgupta6518 That book just has a ton of problems without solutions. If somehow you get stuck on a problem, game over. Plus it does not have advanced level problems.

Depends on the field. Here are some good books: Number Theory: ->104 Problems in Number Theory by Titu Andreescu Geometry: ->EGMO by Evan Chen Combinatorics: -> Problem Solving Strategies by Engel (old, but has some good explanations of basic techniques) -> Olympiad Combinatorics by Pranav Sriram (though this is advanced already) Competitions to look at (in order of difficulty): -> Junior Balkan Math Olympiad -> Tournament of Towns (especially for combinatorics) -> JUSAMO -> European Girls Math Olympiad -> National Olympiads of countries like Russia, Bulgaria, Brazil, Vietnam, South Korea, China, USA

Never mind I’m stupid

@@ryanclothier6853 ayoo broo 🥺

😅

because a^2 -b | 1 + ab then 1 + ab >= a^2 -b from a >= b+1 follow that a^2 -b >= a * (b+1) - b (it's simple substitution of a with b in a^2) then 1+ab >= a(b+1)-b

He multiplied the previous result by 9.

do it urself

@@AyushKumar-ot3zr you think I'd give homework without knowing the answer myself? 🤣

How to solve it ? Give me a hint

This equation has only 3 solutions that [a; b; c]= [2; 1; 1], [1; 2; 1], [1; 1; 2]

​@@KaranSingh-qw7cn use equality that a=>b=>c

Yes i also saw that

Probably that’s why it wasn’t chosen to be in the real contest?

That’s my point, I was trying to say that this would lead to a contradiction.

@@letsthinkcritically thank u for your reply and yes now i get it 💓💓

Wrong. AM-GM gives 3abc

@@alanjoelj1591 But if (abc)^2 is >= 3abc then isn't it contradicts original solution?

@@HemantPandey123 Where is the contradiction?

@@alanjoelj1591 Solution assumes (abc)^2

(only) solutions to the equation are (1,2,3) and permutations. (abc)^2 is clearly 36 which is > 3.

Fr I feel like he just writes it down without explaining at all