Greek Mathematics Olympiad | 2008 Q2
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- เผยแพร่เมื่อ 28 พ.ค. 2024
- We solve a nice number theory problem from the 2008 Greek math olympiad.
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“That’s going to be, like, 8 or something”? 😂🤣
😅 yeah that sort of thing can be confusing for students. But overall good presentation.
@@wise_math Haha, I love his style, personally.. Classic!
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)
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...
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.
change it to "Αυτό είναι ένα καλό μέρος για να σταματήσουμε"
@@KuroboshiHadar you are right. There is only one continuous function. I guess the last equality follows that f is continuous.
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!)
08:00 Without reference to irrationality:
Do x=-1 "by hand" and then:
If x
Wow! I loved the problem! Nice use of the Sophie Germain identity. I would like to see more number theory problems on this channel! :)
If you want good number theory problems, then this channel is a good to stop.
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)
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
@@gilangNoDrop
Yup!
What I love about your channel is that you are making many video in 1 day 😍😍😍
great video, you gave me some inspiration for a future video!
Wow you find such cool NT contest problems
Nice problems and solutions
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.
That’s powerful 🔥
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
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.
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
Made the same first attempt and then switched to factoring.😀
Damned ! I got the correct factorization but not the conclusion ☹️.
Nice probleme and nice solution ! Thank you
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
Superb
I trust that's only 1.. and you prove it 😂😂😂
I solved it!
Great work you deserve more likes, love from India 🇮🇳
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.
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
Good video
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.
nerd
Sheldon Cooper presents Fun with Flags
that’s interesting... does it mean that these flags when vertically viewed have to be mirrored?
Also the stripes are horizontal
The flag is turned around 90°
Make a video on madhava- Leibniz series
Good excuse for the colored chalk, I agree
My brain just burned out
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.
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
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
There is mistype I showed that the unit digit of other part is 4
You could have used sophie germain i dentity too ig?
Can anyone recommend me a concise book that goes through the topics mentioned in this video? e.g. Fermat's little theorem, factorization, etc.
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.
Maybe The Justin steven's website could help you
@@Miguel-xd7xp not much material there, but thanks.
@@jonathanjacobson7012 Also the AoPS (art of problem solving) can help, there are more people who can recommend a book
Good
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? 😟
Set f(x)=y-z. Find f(3). Use f’(3) to show f(x) is increasing for x>=3.
Clearly Never prime for even x
Right. But also have to consider the odds, and this is included in x >= 2.
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
8 is not coprime to 2 so you can’t apply FLT
8^x+2^x+2= [[[8x+4x]]= [[ 6x+6]]=[[ x =1 ]] marcelius mrtirosianas
Can I copy somebody's homework?
y
So let's see .. x raised to the power of snowman?
Plz do this question
Cube root of (5-x)=5-x³.
You get three roots of a degree six polynomial, only one of which is real.
@@IanXMiller no shit.
Notice that if f(x) = 5-x^3 then the inverse of f is cuberoot(5-x). Therefore the function equals its inverse. Notice that f is strictly decreasing.
This implies f(x) = x and so 5-x^3 = x.
This is a simple third degree polynomial to solve.
I'll leave it to you guys to prove that f=f^-1 implies f(x)=x if f is strictly decreasing.
@@hybmnzz2658 I got to x^3 +x =5 part but how to solve it can you explain?
@@hybmnzz2658 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.
Sophie Germain
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
:) He used a^2 + b^2 = (a+b)^2 -2ab. In this case, the 2ab turns out to be a square as well.
@@AnkhArcRod thanks and excuse me
the factorisation wasnt so hard after all
Sophie Germain identity for the win =)
What is it?
@@mathissupereasy x⁴+4y⁴=(x²+2xy+2y²)(x²-2xy+2y²)
But how would you apply it? The second term is always to the power of x and not 4?
tootoo the tootoo the ex
mini writing mistake when proving the case for x=
This problem solution is not even prime number.=0
x=1 is one solution without watching the vid. 1^8 + 2^2^1 + 2 = 1+4+2= 7
And 7 is prime
7:24 "None *IS rational" ;)