Turkish Mathematical Olympiad | 2009 Q1 

The funny thing happens if you brute-force in a programming language with insufficient decimal precision. As expected 2,7 and 11 are answers, but 19262791 also comes out. The reason: sqrt(p**3-4p+9) with p=19262791 is really close to an integer(84...x10**9 + 0.9999961...)

Hi Michael. I am professor in Spain. I enjoy very much all your imos and Putnam problems. I watch all them. Very, very good job. Thanks for your videos.

Your channel has made me love math again I hope that in the future you'll make videos on Fourier Transform because it's the only topic which is missing from your channel and there are so many interesting concepts about Fourier transform

I think the hardest part is always coming up with the hints giving the right direction. How do you identify what approach/hints to take?

Here is a shortcut slightly from 8:45, aside from p={2,3}, taking the square of both sides of an inequality ((p^2-2p-9)/9) =< n, We will get the desired bound for p, by dividing p^3 as in the lecture, hence directly concluded that p=

I am very happy now,please make more videos about Turkish Mathematical Olympiads. Also you can try this one: For all x,y which are real numbers, (i) f(f(x²)+y+f(y))=x²+2f(y) (ii) x≤y --› f(x)≤f(y) Find the all functions which is f:R--›R (Turkish Mathematical Olympiad 2012 Second Round Day 1 Question 3)

Never occurred to me to use divisors to set up bounds! Thank you for bestowing these techniques upon us for free :)

For everyone asking about comparing n against p^2/4 at 8:48 and the early bound of looking at cases p>11 at 1:53 the comparisons aren't required at all and can be proved without it. They are for simplifying the argument a little. The bound of p>11: For any prime p, n^2 = 9 (mod p) implies (n-3)(n+3) = 0 (mod p) hence p divides either n-3 or n+3. Dropping the assumption p>11 doesn't affect any of the later calculations. The comparison with p^2/4: Combining the bound (p^2-2p-9)/3 < n with the identity n^2 = p^3-4p+9 we obtain the inequality (p^2-2p-9)^2 < 9(p^3-4p+9). After expanding, simplifying and factorising we obtain p(p-2)(p^2-11p-36)

Always the last constant determine the roots and always less than last remainder theorem and it tells something about how the last constant is distributed along the function. With the highest power as one. 2 7 and 9+2.

if we set the equation to x^2 and rearrange we can factor it as p(p^2-4)=(x+3)(x-3). p is prime so must divide either (x+3) or (x-3). The difference between these two terms =6 therefore let ap =x+-6 and p^2-4= a^2p+-6a. Now gather all terms on the left-hand side: p^2-a^2p-4+-6a=0 and use the quadratic equation to solve for p: p=(a^2+-sqrt(a^4+16+-24a))/2. Now we look at the discriminant. To be a whole number a^4+16+-24a must be a perfect square but a^4 and 16 are both already perfect squares. This means that 24a must be the cross term and it must equal to 8a^2. Therefore a=3. We can now rewrite the equation as p=(9+-sqrt((9+-4)^2)/2 and this simplifies to p=(9+-9+-4)/2 =(18+-4)/2 or +-4/2 which means p=2, 7, 11 after we discount the answer of negative 2

If you look at the original expression mod 4, you get p(cubed) + 1 is congruent to zero or 1. This means that p = 2 (which checks out) or p is of the form 4k + 3. This reduces the work considerably.

I don't understand why we need to consider two cases as in 9:36 and further. I was able to skip that assumption that "was going out from nowhere" and solve more straighforward way. On that board we had already found that 3n≥p²-2p+3. Squaring that, we have 9n²≥p⁴-4p³-14p²+36p+81. But we search for p's where n²=p³-4p+9 (from the problem statement), so we just substitute that into this inequality and we get a polynomial inequality for p. It only looks quartic and hard, in fact is's easy to solve: the free term 81 cancels out, so after combining all the terms we have p⁴-13p³-14p²+72p≤0. Let's forget for a minute p is a prime, just let's factor it out and leave a cubic poly'al. It happens that 2 is a root of that cubic, so it factorizes to (p-2)(p²-11p-36)¸ with irrational roots, one of which is negative, anf the other is around 13.6. So we have all four roots or our quartic (around -2.6, 0, 2, around 13.6). Solving this inequality by intervals method we find useable intervals are [around -2.6, 0] and [2, around 13.6]. Now, remembering p was a prime, this translates into p≤13. Easy part, testing each of p = 2, 3, 5, 7, 11, 13 we have left with 2, 7 and 11, whose are the answer (corresponding n's are 3, 18, 36).

15:25

Greetings from Turkey

Cheers from Turkey! Good problem anyway. Turkey's questions are really good compared with most of the countries -especially the term 2005-2015 the brightest of Turkey National Team-. I hope you can offer more questions from Turkey:)

Another way to solve it: p^3-4p=(n-3)(n+3) either n+3 or n-3 must be divisible by p so p^2-4=a(n+-3)=a(pa+-3)=a^2+-3a. Using the quadratic equation p=(a^2+-sqrt(a^4+16+-24a)/2 however the square root term must also be a perfect square otherwise the answer cannot be an integer therefore 8a=+-24 a^2. Therefore a=+-3. Feeding this back into the original equation (unnecessary because we could have used the previous equation but it is what I did) we get two separate quadratic. Nothing else is possible p^2-9p-22=0 or p^2-9p+14=0 one solution is impossible (-2), the others are 2,7,11

Another possible answer is to prove that n is divisible by 3. To do this, transpose 9 to the LHS and factor the RHS. Now assume for a moment that p>3. This means that 3 does not divide p (since p is a prime) but 3 must divide at least one of (p-2) or (p+2). Now, this means that p^3-4p is divisible by 3, and so n^2 is also divisible by 3. But this means that 3|n. Now, let n=3m. We have m^2=(p)(p-2)(p+2)/9+1. Rearranging, we have (m+1)(m-1)=p(p+2)(p-2)/9. But since p is odd, so are (p-2) and (p+2). This means that (m-1)(m+1) are both odd.But this implies that gcf(m+1,m-1)=1. Notice also that the two factors differ by 2. This is an important hint Also, since p is odd prime, gcf(p,p-2)=1, gcf(p+2,p-2)=1, gcf(p,p-2)=1. This also implies that any of p-2 or p+2 is divisible by 9. Assume 9|(p-2). This means that the product of any two of the three factors p, (p+2), and (p-2)/9 must be either 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p. Now assume 9|(p+2). This means the product of any two of the three factors p, (p-2) and (p+2)/9 must be 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p. Now, p=2 will be a solution. Since we assume that p>3, what remains is to test at p=3.

Problem proposition - From Polish edition of Championnat des Jeux Mathématiques et Logiques final (2003, Question 18): Find all natural n such that : 3^n+4^n+5^n...+(n+2)^n = (n+3)^n.

Respect from one math teacher to another math teacher

Great shirt 👍🤗

there should have a resolution more short and more clear. i m sure of that

Instantly thought that primes must end in 1, 3, 7 or 9; and putting those digits through the formula needs to end up with a units digit of 1, 4, 5, 6, 9 or 0. (ie not 2, 3, 7 or 8). 1 minus 4 plus 9 is 6 (3*3*3) = 7 - 2 + 9 is 4 (7*7*7) = 3 - 8 + 9 is 4 (9*9*9) = 9 - 6 + 9 is 2 - so that's the only thing I've ruled out. The prime can't end in 9. Now let's see what I should have been doing...

Let p^3-4p+9=k^2 -> (p-2)p(p+2)=(k-3)(k+3), the product of three consecutive numbers with the difference 3 should be described into the product of two numbers with a difference of 6. If p-2 or p+2 is 9m (m=0, 1) (p-2)p(p+2) = m*3p*(3p+/-6), so satisfies conditions. Of course, I should strictly prove that m bigger than 1 cannot satisfies the condition, but ...

Who in the world would even think of doing this?? Why not move the 9 to tbe other side and factor it as p(p-2)(p+2)= (n-3)(n+3) and then plug in values to see what works??

Eqn factorises as (p-2) p (p+2) = (m-3)(m+3) For p>2 we have three consecutive odd numbers so one must be a multiple of 3. I'm fact a multiple of 9 as RHS is a multiple of 9 if it's a multiple of 3. RHS can be rewritten 9(k-1)(k+1). k=3m. Obvs p can't be 9 so either (p-2)=9 or (p+2)=9. In fact both give solutions with three consecutive odd numbers (k-1),(k+1),9 and 9,)k-1),(k+1). p=7 or 11.

There is a lot ways to solve, but try to think of this that way: p^3-4p=k^2-9 p^2(p-4)=(k+3)(k-3) Then we can think 'bout which of multipliers from k+3 and k-3 devides by p. For one reason it can' t be only k-3, because then it devides by p^2 and there is unequalety k-3>=p^2. Analogicly it can't be only k+3, so both of k+3 and k-3 devides by p, so 6 devides by p. And then we check p=2 or 3 by hands. And get that only 2 reaches

英語と数学の勉強が同時にできる一石二鳥

but... what was the rank of the elliptic curve?

So when you get to pm^2 +- 6m = p^2 -4, why can't we solve as a quadratic in p, and try to find m values that make p prime? When I try it doesn't work btw, cause I end up with the sqrt of m^4 +- 24m + 16 which can't be factored into squares afaik. Thx

Nice.

So when he got to P|+-6m-4, why did he say that, because P≧11, P divides half of +-6m-4? Does it have something to do with the -4?

Let = represent congruence , that is a = b mod p means a is congruent to b mod p, p is prime. 64 = 100 mod 3 then 8 = - 10 mod 3 but not 8 = 10 mod 3 100 = 9 mod 13 then 10 = -3 mod 13 but not 10 = 3 mod 13 144 = 64 mod 5 then 12 = - 8 mod 5 but not 12 = 8 mod 5 So n^2 = 9 (mod p) we only have (n = 3 mod p) OR n = -3 (mod p ) but not both. Many books also use + AND - instead of + OR - , wondering if this is trivial in the video. I frequently look for materials in your videos to train students for math contests. Thank you.

That's a hard one. As usual I lasted till about half way.

please answer this. you said on 5:03 that if it is not even it must divide the odd part of the right side but that doesn't hold for every divisor. For instance, 3 divides 21, therefore, it divides 19 + 2. Similarly, we could say that it should divide 19 something that is false.

Huh, I'm wondering about the +/- portions being added together. Can we do that? I suppose it's because they're both "+/-" rather than one of them being "-/+" ? I was thinking they may partially (or completely) cancel, otherwise. Thanks

The assumption early on that p>=11 seemed unmotivated. Hardest part of proving something difficult is to not use a truth that follows from already knowing the answer.

I enjoy watching these proofs, but they punctuate how weak I am this area (number theory, analysis). Can anyone recommend entry level book(s). (I have some math background - engineer, calculus tutor).

türkler için söylüyorum bu soru 2009 tübitak lise matematik olimpiyatları 2.aşama 1.gün 1.soru

why did u take n≤p²/4... why not p²/5 or something else? we would have got a quadratic equation in both cases..

Surf Arrakis 😄 Miller's World would be another prime location.

How about your mega favorite number??

The technique provided gives us new insights, but the solution is way too complicated. Just consider 2, get it out of the way and for odd primes, consider primes to be 3k+/-1. For both cases lhs is 0 mod 3. This implies it is 0 mod 9 too as rhs is a perfect square. Hence p(p^2-4)=9n^2-9=9n(n^2-1) .case wise bashing-p+2=9,p-2=9 yields p=7,11. Hence p=2,7,11. Other cases don't yield solutions. Very easy :-)

I tried making the ff. Soln, and got the same answers as you, but I have no idea how to justify one of the steps I made...: Express the RHS as a DOTS. Then, we have (n-3)(n+3) = (p-2)(p)(p+2). Obsv: p even, then p 2, but RHS 0, so n 9. That's ok. So we need p odd. But p odd means n even. So we have the solutio: p = 2 Let n be 2g, then the lhs is ( 2g+3 ) (2g - 3), so p divides one of these terms, let one be Ap, the other is Ap + 6 or Ap - 6. We have either, Ap(Ap + 6) = (p-2)(p)(p+2). Ap (Ap - 6) = (p-2)(p)(p+2). Then transpose around to get A = ((p^2)-4)/(Ap + 6) or ((p^2)-4)/(Ap - 6) Treat the denominator and numerator as polynomials of p and use synthetic division, to get A = (1/A)p + (6/A^2)(p^0) + ((36/(A^2))-4)/(Ap+6) or A = (1/A)p + (6/A^2)(p^0) + ((36/(A^2))-4)/(Ap-6) Here is the step I have trouble with. There is a weak argument here, since A is an integer, so the term ((36/(A^2))-4)/(Ap-6) must be 0. I don't know how to connect the two but this is what solvds the problem, anyways if this term is nonzero we have an ugly-looking fraction for all three terms, and... I just don't know how to justify it. Even if the (1/A)p + (6/A^2)(p^0) terms were non integers, it's still more nice and likely to be that the sum of those two is an integer (?) Without the " interference" from the third term. But, if it is zero, then A = +3 or -3. It's not really important in this solution if we chose to use 2A + 6 or -6 earlier... isn't that strange? Now we solve the equation for p: The -3 solution: A = (1/A)p + (6/A^2) + ((36/(A^2))-4)/(Ap-6) -3 = -p/3 + 2/3; 9+2 = 11 = p. The 3 solution: 3 = p/3 + 2/3; 9-2 = 7 =p In addition to the 2 from earlier we now have all solutions of p.

"notice how i've changed the sign but that's kind of ok as well" Yet he had to explain why p divides something written as p*(...). If you're gonna explain all the obvious parts then explain the not so obvious ones as well

I reached the equation (p-2)*p*(p+2)=(n-3)*(n+3), which means that p divides either (n-3) or (n+3), but couldn't proceed any further without your help. Thanks!

CAN SOMEONE TELL ME WHAT WE NOTICED THEN WE CHOOSED P^2/4 AS A BOUND FOR N IT STILL SEEMS ABSURD FOR ME

Unless I'm missing something, saying p|(3m+2) means p≤3m+2, isn't valid because of negative numbers. For example, p=13 and 3m+2=-52. Obviously this particular m and p doesn't satisfy the equation, but it still is a counter example to p|(3m+2)→p≤3m+2

I m morrocan students and Next years l will study in first years in university can participate international mathematical competition IMC ? if I can how to inscrire in this competition

Can someone explain the simplification with the signs at 5:14? I can't quite follow that

nerdesiniz beyler

Oh ! Yet another transatlantic misunderstanding ! I read Q instead of 9 : a bit more complicated, isn't it 😅

I would never be able to do that. Not with that method anyway. Maybe there's another....?

Subbing for the Dune shirt lol

2:12 => assumes p >= 11 14:50 => gives all prime numbers under 13 as solutions to try, and then so 2 and 7 are. Excuse me? Where did I get this wrong please

It’s a solution but not a very elegant one; could it be that the setters in the olympiad had in mind something simpler? E.g. transform the equation into (n-3)(n+3)=p(p-2)(p+2), or, because obviously both parts are divisible by 9, into 9(m-1)(m+1)=p(p-2)(p+2), and then prove that either p=2, m=1, or both sides represent a product of three consecutive odd numbers, one of which is 9, i.e. p=7 or p=11.

RU-vid has changed

It's 9! Wow. I thought what is this q means :)

2:20 Why he can take square on both sides? (n^2 ≡ 9 (mod p) ⇒ n ≡ ±3 (mod p)). Is this a theorem?

I got three different prime number which satisfied the conditions. Those primes are 3,7 and 11. I got these answer through brute force 😅😅

5:12 5:17 ...wow, I've never seen that in my life. Still not convinced it is true.

Valla helal

Why could we make the assumption that n is bigger than or equal to 11?

think i solved in a similar but (for me) easier way. After reasoning mod p we can say p|a+3 or p|a-3. We also know for FLT p³=p (mod 3) so 3|p³-4p+9=a² and 3|a because 3 it's prime. I write the first case (p|a+3), the second is similar. a²=k²p²-6kp+9=p³-4p+9 so p²-k²p+(6k-4)=0 delta=k^4-24k+16, p is an integer so delta must be a perfect square. It's easy to see that delta=(k²-1)²+2k²-24k+15. Because it's a square it can't be (k²-2)

Can you solve this problem? Find all primes of the form p^p+2 such that p is also a prime. I’ve asked lots of people and none of them could help.

Very hard

Okay, and that's a good place to stop.... ...you know.... at the end...

That was quite long....

Not a prime, but 646³ - 4∙646 + 9 = 16419²

Definetly a really hard problem, I tried this, couldn't solve it and then I watched the solution: impossible!

Your hairs in every video, are even like this, like that! 😀😉 ...

plzz try this cool problem : find all functions that satisfy the following condition , for all x,y real numbers , f(x^2)-f(y^2)《(f(x)+y)(x-f(y))

🇹🇷🇹🇷🇹🇷🇹🇷🇹🇷

Noice

😩😩😩

can you please try this question: find all natural numbers a, b and c such that a+2020=b+c 2020a+1=bc *if someone here in the comments know how to do pls help me too :)

🇹🇷🇹🇷🇹🇷🇹🇷

No offense, but if that's how math is taught in the US, I can see how people hate it