FUN Ecuadorian Math Olympiad Number Theory Problem 

He has room for backflips now

i love this very formal number theoretic proof of a lemma that i would've summarized as "well it's gotta be somewhere"

I love it when you solve this kind of problems and teach such nice tricks which can be applied in similar situations and I also appreciate the idea of the two blackboards. The channel is getting better and better, I'm sure you'll get to pi-thousand subscribers soon!

Working through the bonus problem where p² + p + 1 = n³, the steps are mostly the same but with an occasional sign flip. Suppose p(p+1) = n³ - 1 = (n-1)(n² + n + 1). Then either p|(n-1) or p|(n² + n + 1) Case 1 - Suppose p|(n-1), i.e. n-1 = px for some natural x. Then p ≤ n-1 which implies p+1 ≤ n But also p(p+1) = px(n² + n + 1) which implies p+1 = xn² + xn + x ≥ n² + n + 1 > n (since x and n are Natural numbers) So n ≥ p+1 > n, which is a contradiction. Therefore p does not divide (n-1) Case 2 - Suppose p | n² + n + 1, i.e. n² + n + 1 = px for some natural x Then p(p+1) = px(n-1), so p = xn - x - 1 Plugging p in, we get n² + n + 1 = nx² - x² - x Solving for x, we find that the discriminant f(x) = x⁴ - 6x² - 4x - 3 must be a perfect square. (Similar to the video except we have a -4x term instead of a +4x term.) Similar to the video, set g(x) = (x² - 4)² and h(x) = (x² - 3)² (so instead of 3 and 2 we're using 4 and 3 as the differences for the consecutive squares). f(x) - g(x) = 2x² - 4x - 19. Note that if you solve for x this is >0 if x>4 or x4 part. h(x) - f(x) = 4x + 12, which is always >0 for all Natural x. Therefore if x>4 we have h(x) > f(x) > g(x), which puts f(x) between two consecutive squares. So there can be no solutions where x>4 (just like the video, only with slightly different consecutive squares.) So the only possible solutions are x = 1, 2, 3, or 4and only if f(x) is itself a perfect square. Recall above f(x) = x⁴ - 6x² - 4x - 3 for reference f(1) = 1-6-4-3 = -12, not a perfect square f(2) = 16 - 24 - 8 - 3 = -19, not a perfect square f(3) = 81 - 54 - 12 - 3 = 12, also not a perfect square f(4) = 141, not a perfect square Therefore f(x) is never a perfect square for any natural number x, and thus there are no solutions.

I actually really like the new format! Feels a lot smoother

I never understood why this channel only has 251k, this is a gem!

So Stephanie is the name of our unsung heroine of the description short stories. Shoutout to her!

I think that this new format has some potential but there are a few problems. The lighting is a little bit off and some parts of the board are more illuminated than others. Other thing is that you have always had previous steps summarised somewhere in the corner and now despite the fact that you have more space some previous facts were erased and it was harder to recall what was already done. Like the equation for n in terms of x and the problem statement itself that was only on one of the boards

Love how you are choosing problems that tie into your MathMajor channel and the recent proof writing series. Do continue this trend of having your problems tie into your lectures. Love it.

I like the new format.

Thorough as ever. The first p|ab bit can be easily proven with prime factorisation.

It first appeared in St. Petersburg olympiad in 90's, then appeared in Balkan Olympiad

I would love a series from you doing more 'modern' multivariable calculus as to arrive to the generalized stokes theorem with differential forms

The studio looks awesome! I also like this problem a lot; Using trial and error on p, 19 is quite large (even for a prime, there's seven smaller ones) so if one tested seven or fewer of them, one would like to conclude that there is no solution, however in such a situation, one would be wrong.

production quality is insane

Really nice problem and solution. Two blackboards look optimal.

Hello everyone, p^2-p+1=n^3...(*), as you said there exists a natural x>0 s.t. px=n^2+n+1 ....(1) p-1=×(n-1) .......(2) we have from (1): px=(n-1)(n+2)+3 multiplying both sides by x, one obtains: px^2=x(n-1)(n+2)+3x . Using (2), one can write: px^2=(p-1)(n+2)+3x. This equation is equivalent to , p(x^2-n-2)=3x-n-2 (it is easy to prove the positivity of both sides). For x>3, we have : 3x^2-n-2>3x-n-2 and since p is a prime (greater than 1) the last equation does not have a solution. Now we need to check the last equation for the cases: x=1, x=2, and x=3. By easy calculation, the equation (*) has a solution for the case x=3 while there are not solutions for the two other cases. For x=3, one can easily determine p and n which they are : p=19 and n=7. The answer to the question is : there is only one prime which is 19. Thank you for this good problem.

amazing quality and production, keep up the good work!

Beautiful!

I love the look of the space!

I like your video and the format! Please keep going! Thank you!

I don't find any primes less than 10,000 for which p^2 + p + 1 works

15:31 this is a bit of minor nitpicking , but -3 is in fact a perfect square, if one works over the Eisenstein integers , specifically it's (1+2ω)^2

Beautiful problem and excelent problem solving!

Wow amazing video... Love the new format

What a setup man!

The following solution maybe of interest since it only uses basic facts about divisibility, and the simple observation: if (4i+r)(4k+s)=4l+t, where 0=2 and thus x>0. But then (n-1)(n-2)>= p(n-2)=p(px-1)>= p(p-1)=(n-1)m>m and thus 0>-n^2+3n-2+m=4n-1>=11, a contradiction. Hence m=px for some x>0 and thus p(p-1)=(n-1)px, i.e., p-1=(n-1)x. Hence m=((n-1)x+1)x. From 0==n(n+1)=(n-1)x^2+x-1==(n-1)x+x-1=nx-1 (mod 2) we get n=2k+1 for some k>0 & x=2y+1 for some y>=0, and thus m-1=n^2+n=4k^2+6k+2=2(2k+1)(k+1) & m-1=((n-1)x+1)x-1=2k(2y+1)^2+2y=8ky^2+(8k+2)y+2k, i.e., 2k(k+1)+1=y(4ky+4k+1). From 2 | k(k+1) we get y=4z+1 for some z>=0 and thus 2k^2+2k+1=2k(k+1)+1=(4z+1)(4k(4z+1)+4k+1)=4k(4z+1)^2+ 16kz+4z+4k+1, i.e., k^2=2k(4z+1)^2+8kz+2z+k. Hence 2z=kl for some l>=0 and thus k^2=2k(2kl+1)^2+(4k+1)kl+k, i.e., k=2(2kl+1)^2+(4k+1)l+1. Assume(!) that l>=1. Then k>=2(2k+1)^2+4k+1+1>=2*3^2+4k+2, a contradiction. Hence l=0 and thus k=3 & z=0 & y=1 & x=3. This gives n=2k+1=7 & p-1=(n-1)x=6*3=18.

Hey I'm Ecuadorian!

The colors are excellent by the way

I did really like this one! Thanks

When getting to p | n²+n+1, I tried seeing if there was any way to use Eisenstein integers to solve the problem. You can immediately deduce that p cannot be an Eisenstein prime, because if it were it would divide either n-j or n-j², meaning n-j = p(a+bj), so pb = -1. Not happening. Therefore, p = qq* for some Eisenstein prime q. From there, I was kind of stuck (I didn't try very hard though). If anyone comes up with anything coming from that, I'd be interested to hear about it.

37(n+n+n)=nnn for 0 < n < 10. Sin (666) + Cos (6*6*6) = -1.618... (golden ratio). Belphegor's prime is the palindromic prime number 1000000000000066600000000000001 (10^30 + 666 × 10^14 + 1).

Thank you, professor and congratulations on the new set-up. Now that there is room, are the backflips making a comeback?

Hey, Michael! The constant term of h(x)-f(x) should be positive seven, not negative seven.

Nice studio 👌💥❤️

Cool math coming from my home country, cool.

you are right

I wonder why we spent time prove this lemma, while it's obvious from the "Fundamental theorem of arithmetic" by the factorization of a and b which gives the factorization of ab.

That “W” alliteration though

If you wondering like I was, why can't p divide both n-1 and n²+n+1 its because this implies RHS has factor of p², but LHS has factor of only p which wouldn't work

Nice! Did you know that there is quite similar problem from Romanian Masters 2023? Solve x^3+y^3=p(xy+p) where p is prime, x, y are positive integers

the lemma proof is circular!

I like your videos I have learned a lot of formulas from your courses Thank you so much

Really good❤

Nice bound on squares. I came to equation through simillar reasoning that (a^2-3)^2+4(a-3)=(a+x)^2 from wich I suspektes only a=3 leading to x=3 is possible solution. Otherwise one could investigate integers u^2+v^2=z^2 maybe?

6:15 If we’re working with natural numbers, you should’ve dispatched (easily, I admit, but necessary nonetheless) with the case n=1 when n-1=0 and p|0 🥴

0:33 But you needn't prove that. That is the definition of "prime".

For the homework p^2+p+1 = n^3 I get: p(p+1)=(n-1)(n^2+n+1) case 1) p|n-1 => n+1 n=1/2 (x^2-1+/- sqrtD) with D=x^4-6x^2-4x-3 Whenever x>=5 we have (x^2-4)^2 < x^4-6x^2-4x-3 < (x^2-3)^2, so try x=1, D=-12 x=2, D=-19 x=3, D=12 x=4, D=256-96-16-3=141 so D is never a perfect square and there are no solutions

I love your videos! What is the best way to tackle writing proofs without getting any hints? Does it come with time? Where do most math people get these crazy intuitions to use certain lemmas? Thank you.

2 boards are good. But do you know what is better? . Make a 3rd board on the ceiling. I doubt that Michael can climb !!!

Hi everyone. There is a writing mistake: h(x)-f(x)=2x^2-4x+7 for x>=4

Still not correct, but now i have it: if (4i+r)(4k+s)=4l+1, where 0

very nice ❤

Sorry, the correct form is: if (4i+r)(4k+s)=4l+1, where 0

I came up with a brute force solution by mentally checking all primes less than 100. The only values I could find were p = 19, n = 7. I tried to solve this one algebraically, got nowhere.

The lemma is evident fact

A faster way to prove p1|ab ⇒ p1|a or p1|b: Let a = ∏p^α1 and b = ∏p^α2, where p represents prime factors. If p1|ab, then the exponent of p1 in the prime factorization of ab is positive. This means α1 + α2 > 0 for p1. Therefore, at least one of α1 or α2 must be positive for p1. So, p1|a or p1|b. Proving Prime Factor Divisibility Here's the corrected text with explanations: A faster way to prove p1|ab ⇒ p1|a or p1|b: Let a = ∏p^α1 and b = ∏p^α2, where p represents prime factors. If p1|ab, then the exponent of p1 in the prime factorization of ab is positive. This means α1 + α2 > 0 for p1. Therefore, at least one of α1 or α2 must be positive for p1. So, p1|a or p1|b. I Would you explain any part of this proof in more detail Certainly, I'd be happy to explain this proof in more detail. Let's break it down step by step: Prime factorization: We start by expressing a and b in terms of their prime factorizations: a = ∏p^α1 and b = ∏p^α2 This means a and b are the product of prime factors (p) raised to certain powers (α1 and α2 respectively). The meaning of p1|ab: When we say p1|ab, we mean p1 divides ab. In terms of prime factorization, this means p1 appears as a factor in the prime factorization of ab. Exponents in multiplication: When we multiply a and b, the exponents of like prime factors add. So in ab, the exponent of p1 would be α1 + α2. Positive exponent: For p1 to divide ab, its exponent in the prime factorization of ab must be positive. This means α1 + α2 > 0 for p1. Logical conclusion: If α1 + α2 > 0, then at least one of α1 or α2 must be positive (or both could be positive). It's impossible for both to be zero or negative and still sum to a positive number. Final deduction: If α1 > 0, then p1 appears in the prime factorization of a, so p1|a. If α2 > 0, then p1 appears in the prime factorization of b, so p1|b. Therefore, we can conclude that if p1|ab, then p1|a or p1|b (or possibly both). This proof uses the fundamental theorem of arithmetic (unique prime factorization) and basic properties of exponents and divisibility to establish the result quickly and elegantly. I will pulblished in an other moment. Now the answer is in a comment. I prefer this way its more safer. Mafia can nakes me made for a comment just for hate. Any way is not possible to admet all those crames. They want to introduce wrong knowledge to cover tge ways of manupilation that they use. They can create Earthquakes change meteo and sayi g that is not predictible. And they want to makes me made for My article published. Even Microsoft and Alphabet is implicated. I think Opening an investigation Is possible because we can find proofs of crimes

spot on!!

I think I saw this kind of format in a ug entrance exam in India called the CMI entrance. I'd suggest you to have a look at those papers! You'd like them.

Do Steiner math.

I know that p=3 is not a solution, but shouldn't we also check that case? Since p could divide both n-1 and n^2+n+1 and that happens when p=3.

@ 14:14 it should be (x^2-2)^2

Do by bound that will easy its a easy problem

The two consecutive squares come out of the blue. Why have you chosen those squares?

😮😮😮niiice problem

Were students expected to solve this problem from a standing start in an exam room?

رائع قبل المشاهدة

I misread the thumbnail at first and thought it was when is sqr(p) - p + 1 = a prime which is an interesting question, but I couldn't see an obvious pattern, I got to p has to be 6m+1 as for 6m -1 it is divisible by 3 but there appear to be an infinite number of solutions, thus I thought it seemed a little too difficult for an Olympiad question, so I went back to to check the video and noticed my mistake. 2 3 3 7 7 43 13 157 67 4423 79 6163 My Solution to the actual: so first note cube(n) = 1 (mod p) thus 3 divides p-1 (Fermat's little theorem) as n is not 1 and less than p thus p = 6m + 1 then cube(n) = 1 (mod 6) so n = 6q + 1 sqr( 6m + 1) - 6m = cube(6q + 1) 36sqr(m) + 6m + 1 = cube(6q)+ 3sqr(6q) + 3(6q) +1 some manipulation produces 6sqr(m) + m = 36cube(q) +18sqr(q) + 3q thus m = 3k and p = 18k +1 18sqr(k)+k = 12cube(q) +6sqr(q) + q pk = q(12sqr(q) +6q + 1) gcd(q, 12sqr(q) +6q + 1) =1 and gcd(p, k) =1 so q divides p or 12sqr(q) +6q + 1 divides p but p prime so if a number divides p then that number = 1 or p 12sqr(q) +6q + 1 can't be 1 so q = 1 so pk = 19 thus k =1 and p = 19 and n = 7 I saw a comment about a Test Problem: sqr(p) + p + 1 = a cube similarly 3 divides p-1 thus p = 6m + 1 then cube(n) = 3 (mod 6) so n = 6q + 3 thus cube(n) = 0 (mod 9) but sqr(p) = 1 + 12m (mod 9) then sqr(p) + p = 2 (mod 9) and so sqr(p) + p + 1 = 3 (mod 9) contradiction no such p and n exist

Great❤

h(x) - f(x) = 2x² - 4x + 7 ?

How about p^2+p+8=n^3? I already found a solution by guess and check. But are there others?

Couldn't that first lemma be proven in one step by just using the fundamental theorem of arithmetic?

I remember this math problem pretty easy for even 9th grade

Why isn’t the proof of lemma just… write out the prime factorizations of a and b. What is it with this guy’s love of lengthy, obfuscating proofs?

Is it possible to prove the same if P is not a prime, but any positive integer? It seems. that result should be the same. If it is possible, then it is a part of positive integer solution of the equation y^2-x^3=1 in real numbers.

Not exactly rigorous proof, but: Lemma: all math-olympiad problems that ask you to find all numbers having a certain property have a finite set of answers :) (or a formula, but there is no formula for prime numbers [that we know of :) ]. Proof by exhaustive search of all problems in all math olympiads :) Therefore, we start enumerating prime numbers and stop when 19 is reached. The proof that only one exists is left as an exercise to some other troll :)

The long shots of your new environment are quite dramatic!!!

What is the brand of your chalk?

Do you have to prove the first lemma in the real Olympiad or can you just state it?

Michael's previous solve of the same question can be found here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-mYCVZitGsFQ.html Personally, I think his newer solution is cleaner and more elegant.

guy's too smart for me ...

Case 1 you neglected n=1 case, in which case x is 0. It doesn't work, but you didn't rule it out.

Am I supposed to understand whats going on?

p=19, n=7 🤪

niceeeeeeeeeeeeee

How stupid is, in order to try factorizing the equation pp-p+1=ppp , subtracting p from both sides of the equation and then: pp-2p+1 = ppp-p = (p-1)^2 = p(pp-1) = (p-1)(p-1) = p(p-1)(p+1) So p=1 (but is that a prime?) or p-1=pp+p The latter solves the same as pp+1=0 so has no real solutions and hence no (extra) primes.

June emoji

You can't divide 57 by 3, 57 is a prime. Grothendieck told me so.

Basado

I boycott all products advertised on RU-vid #BoycottYTads

How would you get such a time consuming problem done during a math olympiad??

Going too fast. Got lost really quickly. Math experts make the worst teachers because they assume their audience already understands some fundamentals.

The new format is great

It should be mentioned that n 1 (which is easily checked) in order to conclude (in Case 1) that p | n -1 implies n -1 = px, where x is a natural number.

Awesome transition at 11:03.... I would like to see it "behind the scenes"

what an amazing studio!!!

343 came up in Domotro's camping/math live-stream last night ~@2:02:40, as a better alternative to 333, because 7*7*7, so that's a cool coincidence.

Why such a convoluted proof for the lemma? The prime factorization of ab contains p (because p is a prime), and this guarantees p is a factor of either a, or b, or both.