Greek Mathematics Olympiad | 2008 Q2

[@leickrobinson5186]

“That’s going to be, like, 8 or something”? 😂🤣

[@wise_math]

😅 yeah that sort of thing can be confusing for students. But overall good presentation.

[@tomatrix7525]

@@wise_math Haha, I love his style, personally.. Classic!

[@tonyhaddad1394]

What I love about your channel is that you are making many video in 1 day 😍😍😍

[@goodplacetostop2973]

12:55 Αυτό είναι ένα καλό μέρος για να σταματήσουμε More than 80k subs, let’s go! Here’s the daily... Considering the function f : ℝ → ℝ with these properties : - f(a + b) = f(a) + f(b), a and b being any real numbers - f(1) = 1 - f(1/x) = (1/x²) • f(x), x being a non-zero real number Determine f and calculate f(1/√2020)

[@KuroboshiHadar]

f: R → R f(a+b) = f(a) + f(b) f(1) = 1 f(1/x) = (1/x²)f(x) for all x ≠ 0 Firstly, notice that f(a+0) = f(a) + f(0) = f(a) so f(0) = 0 If p and q are non negative integers, we have → f(p*a) = f(a + a + ... + a) = p*f(a) → q*f(a/q) = f(a/q + a/q + ... + a/q) = f(a) → f(a/q) = f(a)/q Also, f(a-a) = f(a) + f(-a) = f(0) = 0 so f(-a) = -f(a) Now we can take r = p/q to be any rational number, and it will satisfy → f(r*a) = r * f(a) If a = 1, f(r) = r * f(1) = r Assuming f is continuous, this extends to all x in R since the rationals are dense in R, so f(x*1) = x*f(1) = x for all x in R Also, it satisfies f(1/x) = 1/x = 1/x² * f(x) = 1/x² * x = 1/x So f(x) = x f(1/√2020) = 1/√2020 Did I do it right? I don't know if there are other functions that satisfy the conditions though...

[@IanXMiller]

Or more trivially: f(x) = f(1+1+1+1+...+1)=f(1)+f(1)+f(1)+f(1)+..+f(1)=1+1+1+1+..+1=x so the only function is the identity function.

[@mikelolis3750]

change it to "Αυτό είναι ένα καλό μέρος για να σταματήσουμε"

[@light9744]

First, rewrite the function with f(x+c)=f(x)+f(c) where c be a constant and x be a variable. Then differentiate both side becoming f'(x+c)=f'(x) , which shows f'(x)=d where d is a constant. With this situation, we can conclude f(x)=ex+f where e, f are constants. As f(a+b)=f(a)+f(b) which shows e(a+b)+f=e(a+b)+2f , which leads to the result f=0. Then using f(1)=1, we can conclude f(x)=x. Therefore, f(1/√2020) is itself. (However, i don't know how to proof the function f(x) is differentiable or not, so the solution is with flaw. I hope someone else can do the proofing for me!)

[@Pklrs]

@@KuroboshiHadar assuming that f is continuous JUST at x=0, we can prove that f is continuous everywhere and then the rest of your proof you ve written

[@andreivila7607]

Wow! I loved the problem! Nice use of the Sophie Germain identity. I would like to see more number theory problems on this channel! :)

[@professorpoke]

If you want good number theory problems, then this channel is a good to stop.

[@TechToppers]

Do you know? I saw thumbnail for 2 minutes and I solved the problem by intuition.(I got that Ms Sophie would help I mean)

[@gilangNoDrop]

Yeah I solved it too by Sophie germain, we can prove that (x^4+2(2^(2^(x-2)))^2 - 2(x^2)(2^2^(x-2)) > 1 for x≥2

[@TechToppers]

@@gilangNoDrop Yup!

[@HagenvonEitzen]

08:00 Without reference to irrationality: Do x=-1 "by hand" and then: If x

[@parmilakumari3146]

Wow you find such cool NT contest problems

[@ИбадатЖұмабек]

Nice problems and solutions

[@acentasecond3721]

great video, you gave me some inspiration for a future video!

[@demenion3521]

for proving that y-z>1, we can almost factor it: y-z = x^8+2*2^(2^(x-1))-2*x^2*2^(2^(x-2)) = 1/2 {[x^4-2*2^(2^(x-2))]²+x^4}. as x obviously needs to be odd, the squared term is strictly greater than 0 and moreover an odd number. therefore y-z>=1/2 (1+x^4)>1 for all odd x>1.

[@KaedennYT]

That orange chalk looks quite nice this time of year. You should use it more often! I find it easier to see than the white chalk for some reason, too.

[@Kettwiesel25]

x^4 = 1 mod 5 for all x by small Fermat. Thus the first summand is always 1 mod 5. The second one is 4 mod 5 though for x>1 since 2^2 is and 2^x is divisible by 4, thus 2^2^x =1 mod 5 as above. So the only interesting x is x=1 and x=0, which is where we get 1+16=17, a prime, and 0+2^3=8, a non-prime.

[@antoniopalacios8160]

8:00 I think, actually there is a mistake with the parentheses. We have that for x=-1, 2^2^(-1)=0.25. To get sqrt(2) we need to put the parentheses 2^(2^(-1)). And so on for x=-2,-3,etc. Thank you.

[@reeeeeplease1178]

No, if you write x^y^z that notation means x^(y^z) instead of (x^y)^z. So something like 2^2^2^2 is evaluated from the "top" down: 2^2^2^2 = 2^2^4 = 2^16 ~ 60,000 If you want to do from the bottom up, you put parantheses: 256 = 16^2 = (4^2)^2 = ((2^2)^2)^2

[@MrRyanroberson1]

interesting little homework at the end. turns out when you factor to get y^2-z^2 the way you did, you do so exactly such that y+z and y-z are perfect squares themselves

[@davidepierrat9072]

if p is prime, p and phi(phi(p)) are coprime so you can't get anything by crt so no need to even try looking at it mod stuff

[@jayvaghela9888]

Great work you deserve more likes, love from India 🇮🇳

[@wesleydeng71]

Made the same first attempt and then switched to factoring.😀

[@Tomyato]

That’s powerful 🔥

[@egillandersson1780]

Damned ! I got the correct factorization but not the conclusion ☹️. Nice probleme and nice solution ! Thank you

[@aldinofahreza]

I trust that's only 1.. and you prove it 😂😂😂

[@xchomphk.9788]

x = 1 holds, for x>=2, 2^x + 2 = 2 mod 4. Meaning that 2^(2^x+2) = 4 mod 5. Its easy to prove any x^8 is 1 mod 5 through trial and error, therefore for x>=2, the expression is divisible by 5 , therefore the only solution is x = 1. Honestly a really easy olympiad Q2 problem even for my standards

[@СВЭП-и4ф]

you don't cover case where x mod 5 = 0 and it is really the hard part in this solution, i was not able to prove this case.

[@xchomphk.9788]

@@СВЭП-и4ф oh shit im an idiot good catch

[@T6e6r6o]

Math and flag nerd here, love all your videos. Just a little nitpick. Just like the flag of the United States, the union (ie. the stars part) of which, when displayed vertically, must be on the upper-left corner from the observer's perspective, the Greek flag's canton (ie. the cross part), when displayed vertically, must also be on the upper-left corner from the observer's perspective.

[@justforfun2238]

nerd

[@caesar_cipher]

Sheldon Cooper presents Fun with Flags

[@mondolee]

that’s interesting... does it mean that these flags when vertically viewed have to be mirrored?

[@ErmisSouldatos]

Also the stripes are horizontal

[@ErmisSouldatos]

The flag is turned around 90°

[@fivestar5855]

My brain just burned out

[@TheQEDRoom]

can we simplify this by noting that p must be a sum of two squares so p must be of 1 mod 4? then it follows that x^8 is 1 mod 4 also.

[@holomurphy22]

When x is odd we have x^2 = 1 mod 8. And here x is obviously odd. So its kind of trivial that x^8 is 1 mod 4

[@jotaro6390]

I solved it!

[@giuseppebassi7406]

Wait wait. Since i have used many times the Sophie-germain identity, i have already thought about that when i saw the title. I did the same method, but i have been stuck to prove that y-z is bigger than 1 for all x>2. In fact, i tried that with induction but without results. How can this be so simple that he gave that to homework problem? 😟

[@michaelcampbell6922]

Set f(x)=y-z. Find f(3). Use f’(3) to show f(x) is increasing for x>=3.

[@sayansircar360]

Make a video on madhava- Leibniz series

[@yla3727]

9:55 it looks like you used the formula of a^2 + b^2 = ( a + b )^2, which isn't correct as I learnt at school. I had checked your term and get bad result

[@AnkhArcRod]

:) He used a^2 + b^2 = (a+b)^2 -2ab. In this case, the 2ab turns out to be a square as well.

[@yla3727]

@@AnkhArcRod thanks and excuse me

[@yashvardhan2093]

What I did was a little complicated I first showed that if x is negative then exponent of 2 is fractional and hence it is not an integer only. Then I got the solution for x=1. Then I proved that there exists no solution for x being even . Thus I proved x is odd But clearly if it’s odd it’s unit digit is 1,3,5,7,9 . By unit digit cyclicity if I showed that in case of 1,3,7,9 unit digit of x^8 is 1 and that of the other part is 1 always thus the unit digit is 5 and hence divisible by 5. But when unit digit is 5 This didn’t work but then Sophie Germain struck me when I saw that 4 and remembered about the factorization

[@yashvardhan2093]

There is mistype I showed that the unit digit of other part is 4

[@SB-wy2wx]

Could someone please expand on his reasoning for choosing mod 5, his "8 = 2×4, 4 = 5-1" is not making sense. Why cant u do 8 = 8×1, and 8 = 9-1, so choose mod 9? Sorry if question is stupid, just new to Number theory

[@AlephThree]

8 is not coprime to 2 so you can’t apply FLT

[@jkid1134]

Good excuse for the colored chalk, I agree

[@keyanaskar1981]

You could have used sophie germain i dentity too ig?

[@the-hustle-guy]

Plz do this question Cube root of (5-x)=5-x³.

[@IanXMiller]

You get three roots of a degree six polynomial, only one of which is real.

[@mehmeterciyas6844]

@@IanXMiller no shit.

[@erenozilgili4634]

@ゴゴ Joji Joestar ゴゴ I got to x^3 +x =5 part but how to solve it can you explain?

[@blazedinfernape886]

@ゴゴ Joji Joestar ゴゴ It can be approached intutively because if f and f^-1 are continuous in an interval then f and f^-1 are images of each other with respect to y=x and an image and object can equal only at that line itself.

[@blazedinfernape886]

@@erenozilgili4634 You can use newton method to approximate the solution because the roots of the polynomial are not very nice(one ugly real root and two complex uglier roots)

[@antoniussugianto7973]

Clearly Never prime for even x

[@wise_math]

Right. But also have to consider the odds, and this is included in x >= 2.

[@raqmet]

Good video

[@jkid1134]

Can I copy somebody's homework?

[@justforfun2238]

y

[@user-A168]

Good

[@jonathanjacobson7012]

Can anyone recommend me a concise book that goes through the topics mentioned in this video? e.g. Fermat's little theorem, factorization, etc.

[@stephenbeck7222]

For Fermat’s Little Theorem, I would look at an abstract algebra book. Plenty of good ones out there and you can google reviews. The factoring stuff is high school level computation plus a lot of intuition/cleverness which is mostly just Michael’s familiarity with contest problems.

[@Miguel-xd7xp]

Maybe The Justin steven's website could help you

[@jonathanjacobson7012]

@@Miguel-xd7xp not much material there, but thanks.

[@Miguel-xd7xp]

@@jonathanjacobson7012 Also the AoPS (art of problem solving) can help, there are more people who can recommend a book

[@marceliusmartirosianas6104]

8^x+2^x+2= [[[8x+4x]]= [[ 6x+6]]=[[ x =1 ]] marcelius mrtirosianas

[@dertechl6628]

So let's see .. x raised to the power of snowman?

[@tilek4417]

Sophie Germain

[@shreyathebest100]

the factorisation wasnt so hard after all

[@benjaminbrat3922]

Sophie Germain identity for the win =)

[@mathissupereasy]

What is it?

[@ahuman6546]

@@mathissupereasy x⁴+4y⁴=(x²+2xy+2y²)(x²-2xy+2y²)

[@km-pw2fi]

But how would you apply it? The second term is always to the power of x and not 4?

[@besusbb]

tootoo the tootoo the ex

[@arkitson]

7:24 "None *IS rational" ;)

[@kenichimori8533]

This problem solution is not even prime number.=0

[@jawaduddin4244]

x=1 is one solution without watching the vid. 1^8 + 2^2^1 + 2 = 1+4+2= 7 And 7 is prime