Polish Math Olympiad | 2017

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We look at a nice number theory problem from the 2017 Polish math olympiad.
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Комментарии : 196

@kuba-xz3rl 4 года назад

An alternative approach: Let k be a positive integer such that n^2+np=k^2 p^2=(2n+p-2k)(2n+p+2k). Both factors are positive and different, therefore 2n+p-2k=1 and 2n+p+2k=p^2 => 4n+2p=p^2+1 n=((p-1)/2)^2, which obviously works.

@L4wLiP0p 4 года назад

Great solution!

@knode1993 4 года назад

oooh my god.............. wow

@horseman684 4 года назад

nice approach

@yaasdpalala4492 4 года назад

My god

@shouligatv 4 года назад

Awesome ! Just wanna add that n is indeed natural because p-1 is even ofc

@szymekf5663 4 года назад

Greetings from Poland!

@quantumgaming9180 7 месяцев назад

Greetings from Not Poland!

@MyzticF 4 года назад

Video: When is n^2 + np a perfect square? Me: ..... p = 0.

@shreyasadireddy87 4 года назад

@@bobnich not in natural numbers

@zouhairees-sqally3830 4 года назад

@@bobnich Yeah that's wrong zero is not a prime number

@m4riel 4 года назад

@@bobnich 0 has infinite divisors, where primes only have 2.

@subnow4862 4 года назад

@@bobnich 0 isn't prime lol

@kkgt6591 4 года назад

Me : what's a perfect square?

@jrkirby93 4 года назад

n^2+np? I thought it was p=np?

@darealpoopster 4 года назад

simple solution, n=1, smh

@hrishikeshkulkarni2320 4 года назад

Nicely explained.... At the last step, since p is prime, the two possible factorizations of p^2 should be (1, p^2) or (p, p). Not (1,p) and (p, p)....

@jeremycai5870 4 года назад

Very nicely explained, love your videos so much. +They are free!

@RexxSchneider 3 года назад

I'm a big fan of creating tables to get a handle on what's going on, but it's not necessary in this case. You don't need to fetch the "deux ex machina" solution where p=2k+1 and n=k^2 out of thin air.. Start by observing that n^2+np = n(n+p), so the perfect square has two unequal factors (since n < n+p). We can let n(n+p) = (ab)^2 where a,b are natural numbers. Then we can look for a solution where n = a^2 and (n+p) = b^2. But that immediately leads to a^2+p = b^2, which gives p = b^2-a^2 = (b-a)(b+a). Since the only factors of a prime number are itself and one, we can conclude that b-a = 1 and b+a = p. So b=a+1 and p=2a+1; where n=a^2. That is an analytic derivation of the solution you found from constructing a table.

@goodplacetostop2973 4 года назад

10:15

@goodplacetostart9099 4 года назад

Hey old friend Good place to start at 0:00

@ilyafoskin 4 года назад

As always, a good place to smash that thumbs up

@matthias7790 4 года назад

A polished lesson! Thank you

@pratikmaity4315 4 года назад

Hi Michael, I hope you are doing well. This is my request that can you pls make a video on INVARIANCE PRINCIPLE with some examples of contest type problems? Thanks.

@vikaskalsariya9425 4 года назад

MIcheal, you look a bit tired in this video. I think that you ought to take a break for like a week. As much as we all love your content to watch, we wouldn't like you to be forced to upload a video everyday. :3

@keepitplainsimple1466 4 года назад

Yeah I agree He should take a short break

@RodelIturalde 4 года назад

Is this a good place to stop? (For a bit).

@keepitplainsimple1466 4 года назад

@@RodelIturalde yes

@goodplacetostop2973 4 года назад

RodelIturalde I guess it is

@stephenbeck7222 4 года назад

From other videos we know he has at least one kid and we also know he has a full teaching schedule. I agree that he deserves from these contest problems/‘fun problems’ but maybe these problems are the only things that give him energy to keep at it!

@majkgmajkg2613 4 года назад

Greetings from Poland! :)

@factorial1059 4 года назад

Polska gurom!

@MichaelPennMath 4 года назад

I have a collaborator who is Polish and have tried to convince him to do a problem for the channel in Polish!! Like this comment if you want to see it!

@mateuszsroczyk5203 4 года назад

@@MichaelPennMath Great idea! Polska gurom!

@goodplacetostop2973 4 года назад

Michael Penn Ask him how to say a good place to stop in Polish 😂

@silnikpradustaego7213 4 года назад

@Oskarnik Można też np. „Fajrant”, „No i chuj”, „Co by baba chciała, tego nie ma” itd. Są jeszcze krótsze niż Pan wymienił. Dużo tego jest. Biało-czerwone to barwy niezwyciężone!

@B.0MJiR 7 месяцев назад

it can be solved easier: n^2+np=m^2, n^2+np+p^2/4=m^2+p^2/4, (2n+p)^2=4m^2+p^2 and that's a pythagorean triple. if you can derrive the formulas for them, then p=a^2-b^2 m=ab 2n+p=a^2+b^2, p=a+b, because it's prime and factors only 1*p, therefore a-b=1 b=a-1, m=a(a-1), 2n+p=2n+a^2-(a-1)^2=a^2+(a-1)^2, n=(a-1)^2. p=2a-1 a=(p+1)/2. m,n depends on a, but a depends on p, therefore there is the only one n corresponding p. #andthatsagoodplacetostop.

@noname-jv6hx Год назад

2 years ago i struggled against this problem and couldnt solve it , but idk what happened today i solved it easily......... 😂😂😂 imo or ipho where should i compete in??????

@shafikbarah9273 Год назад

I believe I have amuch simler method: Please make me know if I missed anything n(n+p)=m² gcd(n,n+p)=gcd(n,p)=d and since p is prime we have just two cases for d d=1 d=p For d=1: each of n and n+p must be a perfect square and with a bit of factoring we will come to a conclution that the only solutions for this case are p = 2a+1 n=a² Which prooves each of the condition of the existence and uniqunes d=p ==> n=kp So p²k(k+1)=m² ===> k(k+1)=c² Which is impossible So for every p = 2a+1 ( odd prime) we have a n=a² that makes n²+np a perfect square

@Hakuna_Mataha 4 года назад

Wow he solved P =NP

@alxjones 4 года назад

n^2 + np > n^2 > 1 for all valid inputs, so any perfect square solution has to look like (n+m)^2. Expanding and rearranging gives n^2 + n(2m + m^2/n). In order for 2m + m^2/n to be prime, we need m^2/n = 1, otherwise whatever's left shares a factor with 2m. So we have m^4 + m^2(2m + 1) = (m(m+1))^2, but this is only valid if 2m+1 is prime. So for each m such that 2m+1 is prime, it is the only prime that resolves to a perfect square. This is clearly reversible, so for each odd prime p = 2m+1 for some m, n = m^2 is the unique solution.

@Goku_is_my_idol 4 года назад

Awesome solution

@AlephThree 3 года назад

Brilliant!

@Arbmosal 4 года назад

To find the original solution, one can also use the fact that if you sum up the first N odd number, you will get N^2. In the quadratic formula we have p^2+4x^2 in the notation from the video. writing 4x^2=(2x)^2, the question becomes "for what m is p^2 the m-th odd number? " This amounts to realizing that if p^2=4x+1, then (2x)^2+2(2x)+1=(2x+1)^2. And since p is odd p^2 is 1 mod 4, so there always exists an x with p^2=4x+1, giving us the solution to the original equation

@noname-jv6hx Год назад

Proffessor complicating problem for no reason 😂😂😂

@skwbusaidi 4 года назад

small correction y1-2x1=1 and y1+2x1=p^2 not p

@keepitplainsimple1466 4 года назад

Pretty polished question not gonna lie 😂

@utsav8981 4 года назад

Sir, Kindly Do Indian RMO 2017 Problem 6(Inequality Question). It's a Really great Question, which will really help me in the topic of Inequality

@aviralsood8141 4 года назад

Official solutions are available and I think they are pretty good.

@Tekashiixine-yb5kh 4 года назад

Alternative approach: let n=k^2 n^2+np = n(n+p) = k^2(k^2+p) since n^2+np is a perfect square and k^2 is a perfect square, k^2+p must also be a perfect square if k^2+p is a perfect square, it must be in the form (k+d)^2=k^2+2dk+d^2 which means p = 2dk+d^2 since d divides 2dk+d^2, d must also divide p this is where the condition that p is an odd prime comes in handy since p is an odd prime the only divisors of it is 1 and p or itself, so d = 1 or d = p this creates 2 equations: p = 2k+1, k = (p-1)/2, k^2 = n = (p-1)^2/4 p = 2pk+p^2, 1 = 2k+p, k = (1-p)/2, k^2 = n =(1-p)^2/4 = (p-1)^2/4 in both cases n = (p-1)^2/4 and since p is an odd prime, p-1 is even and (p-1)^2/4 is a natural number therefore the only value of n is (p-1)^2/4 n^2+np = (p-1)^2(p+1)^2/16= (p^2-1)^2/16,

@drsonaligupta75 4 года назад

I have another solution Factorise as n(n+p) { (a,b) means gcd of a,b} Case1 (n,p)=1 So (n,n+p)=1 So both n and n+p are perfect squares So p is difference of squares say P=a^2 -b^2 =(a+b)(a-b) Since p is prime a+b=p & a-b=1 Therefore b= (p-1)/2 But b^2=n by defination of b So n=((p-1)/2 )^2is the only possibility in this case Case 2: p divides n Et n=kp, so Kp(kp+p) is perfect square this implies K(k+1) is perfect square But k^2

@tsawy6 3 года назад

Fun question! I solved it by noting that n^2 + np has factors n*(n+p). Setting n=a*s where a is square-free and s is square, we can pretty easily show that a must be equal to 1: else n+p would need to have a factor of a=> a|as+p => a|p [and a=p falls down as that would set n*(n+p) = as*(as+a) = a^2*s*(s+1) and as s is a square, s+1 cannot be]. Thus it's clear that n must be a square number, meaning n^2+np is a square if and only if n+p is a square; meaning p must be the size of a square gap. Indeed, for every odd number this happens at least one time, as the gap between n^2 and (n+1)^2 is 2n+1 (which obv lists all odd numbers). To show uniqueness it's necessary to show that no larger square gap (i.e the gap between 3 squares, or 4) equals p; this can be shown by noting that the difference between two squares with a k gap is (n+k)^2-n^2 = 2kn + k^2 which has a factor of k=> All square gaps are composite for cases where k>1!

@mithutamang3888 3 года назад

Where, n^3+np is a also perfect squares?

@jucom756 2 года назад

note that there isn't specifyed 0 can't be n, so there is actually always a second solution of n=0. which still is possible with the proof you just have to make the ofshoot of what if x1=0

@Lucaazade 2 года назад

0:45 positive means strictly > 0

@Whoeveriam226 4 года назад

Hey🖐️. How did you manage to find all those n values, so that you get a perfect square in the end? Seems like there's no other way besides sitting for eternity and just checking almost each natural number by hand to get that pattern of n values

@manuelrodriguezmontero3138 4 года назад

Please,the legend of question six

@Eyalkamitchi1 4 года назад

Think twice should do a visual proof of these!

@tamarpeer261 3 года назад

n^2+np is a square n|x^2 either n^2|np, or n=k^2 Case 1: n^2|np=>n|p=>n=1 or p for n=1, 1+p=x^2 p=(x+1)(x-1) x-1=1 x=2,p=3 for n=p 2p^2=x^2 2*p^2 can’t be a perfect square Case 2:n=k^2 k^4+k^2p=x^2 k^2|x^2, x=mk k^2+p=m^2 p=m^2-k^2 p=(m+k)(m-k) m-k=1 k=(p-1)/2,m=(p+1)/2 (I like this definition of floor with odd numbers better n=(p-1)^2/4,x=(p-1)^2(p+1)^2/16

@surem8319 4 года назад

An alternative approach (which also works great for any intenger p): Since we want n^2 + n*p = m^2 we must have that n*p is the difference between two square numbers. The general distance between two square numbers will be: (n+k)^2-n^2 = k^2+2nk and therefore we want: n*p = k^2+2nk n(p-2k) = k^2 We do not have a solution when p-2k is 0 or negative so k should always be smaller than p/2 (and rounded down if p is odd) and so one could go through every k up to floor(p/2)-1 to find all (positive) intenger solutions for a given p. k can also not be 0 unless we want the trivial solution with n and m equal to 0. Now we see that for odd numbers ( >1) we will always have a solution when (p-2k) = 1 which corresponds to k = ((p-1)/2) and therefore n = ((p-1)/2)^2 is always a solution for odd numbers p (and therefore also odd primes) with resulting square number m^2 = ((p^2-1)/4)^2. But using this method we see that numbers like 16 and 9 got two solutions (for n =2 and n = 9 and n = 3 and n = 16 respectively) so we also need to show that this solution is the only solution when p is prime. The last part was shown pretty much the same way as in the video.

@JeffPeterson1989 4 года назад

0. n^2+np=a^2 1. n(n+p)=a^2 2. if n=mp, then m(m+1)p^2=a^2, a contradiction 3. => n and p must be coprime 4. => n and n+p are coprime 5. => n and n+p must be squares 6. call n=b^2 and n+p=c^2 7. p=c^2-b^2=(c+b)(c-b) 8. primes only have trivial factorizations => c-b=1 and c+b=p 9. rearranging, c=b+1 and therefore p=2b+1 10. p=2b+1 has a unique b that makes it work, i.e. b=(p-1)/2 C. Thus, there is a unique n=[(p-1)/2]^2

@appleslover 4 года назад

The flag didn't work

@ccdsah 4 года назад

Nice problem. Technically you have to prove that x1=x2 implies n1=n2. You assumed that x1 must differ from x2. It is rather trivial, but for the completeness of the solution it should be done

@barryzeeberg3672 2 года назад

don't you need to use mathematical induction to _prove_ that the formula containing "k" is correct? tabulating a finite number of cases does not _prove_ that.

@GuiveChafai 2 года назад

honestly the initial part is kind of meh.... Math is not about GUESSING. you can reasonnably derive that n = B^2.p/(2B+1) which means that 2B+1 should divide B2 or P, but dividing B2 is not a possibility... therefore 2B+1 /P which means P=2B+1 and n=B^2

@raystinger6261 4 года назад

I've solved this one by myself, but I didn't express n and p in terms of k since not every k will result in p being a prime. So I calculated k=(p-1)/2, where p is an odd prime, and so n is the square of that. In the end, what we have is given an odd prime p we have n=(p-1)^2/4 such that n^2+np is a perfect square. This way, our result is always tied to an odd prime.

@qwang3118 Год назад

If p = odd prime, then there is unique q such that p^2 + q^2 = square. Indeed, q = (p+1)(p-1)/2 (an even number). p^2 + q^2 = (q+1)^2. To see this, let p^2 + q^2 = r^2. p^2 = r^2 - q^2 = (r-q)(r+q). so r-q = 1, r+q = p^2. r = q+1, r+q = 2q + 1 = p^2. q = (p^2-1)/2 = (p+1)(p-1)/2. Now for n^2 + np = m^2. 4n^2 + 4np = (2m)^2. 4n^2 + 4np + p^2 = p^2 + (2m)^2, (2n + p)^2 = p^2 + (2m)^2. Using the above, 2n + p = q + 1 = (p+1)(p-1)/2 + 1. 2n = (p+1)(p-1)/2 - (p-1) = (p-1)(p/2 + 1/2 - 1) = (p-1)^2/2. n = (p-1)^2/4.

@bttfish 4 года назад

Simple solution: note that gcd(n, n+p) = gcd(n, p) = 1 or p. if gcd(n, p) = p then let n = mp. we have n(n+p)=p^2m(m+1). clearly m(m+1) is not a perfect square, so is n(n+p). if gcd(n,p) = 1, then we must have n = s^2, n+p=t^2 for some s, t in N. then p = (t-s)(t+s). Since p is prime, so we must have t-s=1. Now p = 2s + 1. Clearly, s uniquely exists with respect to p. n = s is the unique solution and n(n+p) = s^2(s+1)^2.

@Goku_is_my_idol 4 года назад

Amazing solution

@a_llama 4 года назад

finally one i solved alone! setting n = ab, where a and b are coprime and subbing in, we get a=k-n and bp=k+n (as a case), bp=a(2b+1), but p is a prime and a and b are coprime, which means a = b, and n=a^2 and p=2a+1

@keepitplainsimple1466 4 года назад

Congratulations! !

@Goku_is_my_idol 4 года назад

Same I solved it like u (k-n)(k+n)=np so assume k+n=nb/a and k-n=pa/b So b= 1 or b=p Solving i got the same

@particleonazock2246 3 года назад

@@Goku_is_my_idol Why do you assume k+n=nb/a?

@IvanisIvan 2 года назад

not going to lie, I thought this was going to be an entire olympiad in polish notation...

@abdlazizlairgi9690 4 года назад

Someone tell me how can i increase my potentials in math to solve this kind of questions...

@aviralsood8141 4 года назад

Practise good problems

@alexandrebailly9363 2 года назад

n^2+np = n(n+p) = k^2 Since gcd(n,n+p)=1, n = a^2 and n+p = b^2 Thus a^2+p = b^2 p = b^2-a^2 p = (b-a)(b+a) One of the term must be p and the other must be 1 Thus b-a = 1 and b+a = p We have 2 equations and 2 unknown values => there is a single solution b = (p+1)/2 a = (p-1)/2 And n = a^2 = (p-1)^2/4

@alexandrebailly9363 2 года назад

Sorry: 1. I see that I have the same kind of answer as abc (see in the comments). 2. abc nicely discards the case gcd(n,n+p) = p (the only possibilities for the gcd are 1 and p).

@a45sankhadiptoghosh64 4 года назад

Proof that if a,b,c,d are integers then, (a-b)(a-c)(a-d)(b-c)(b-d)(c-d) is divisible by 12,,,PLEASE DO THIS

@aviralsood8141 4 года назад

If there are two among a,b,c,d that have the the same remainder on division by 4 then it will have a factor of 4. If all of them have different remainders of 0,1,2,3 which covers all possibilities, you can just factor out the expression to show a factor of 4 exists. Something similar can be done for showing divisibility by 3.

@martothelfriky 4 года назад

We must prove that this product is a multiple of 3 and also a multiple of 4. We will use the fact that if two integers have the same residue modulo n, their difference is a multiple of n. To prove it is a multiple of 3: By the pigeonhole principle at least two of those integers have the same residue mod 3. Thus, their difference is a multiple of 3. To prove it is a multiple of 4: We have two cases: Case 1: Three or more of those integers share parity, say a,b, and c, which would mean that (a-b)(a-c)(b-c) is a multiple of 8 and therefore also a multiple of 4. Case 2: The four integers are split into two parity pairs, say a,b and c,d. Then (a-b)(c-d) is a multiple of 4.

@malabikasaha2452 Месяц назад

All primes are odd except 2

@napoli57rat 4 года назад

Assume n^2 + nP = m^2 for some m>0. Solving for n we obtain n = (-p + sqrt(p^2 + 4m^2))/2. For n to be a natural number, it must be that p^2 + 4 m^2 = k^2 or p^2 = k^2 - 4 m^2. We conclude that p^2 = (k-2m) (k+2m). Because p is prime, then it must be that k-2m = 1 and k+2m=p^2 which implies p^2 = 4m+1. Substitute this in the equation for n and obtain n = (-p + (2m+1))/2 = m + (1-p)/2 or equivalently m = n + (p-1)/2. Now we can substitute this in the original equation and obtain n^2 + n p = n ^2 + ((1-p)/2)^2 - n (1-p) which simplifies to n = (1-p)^2 / 4. QED.

@edwardhuff4727 4 года назад

To prove: for each prime number p ∈ ℕ, there is exactly one r² ∈ ℕ such that r⁴ + r²p is a perfect square. We will show that if p is odd, then p = q(2r+1), with q ∈ ℕ and r ∈ ℕ. If p is also prime, then q = 1 and r = (p-1)/2, which is unique. Suppose p is odd. Then (p-1)/2 ∈ ℕ. Suppose r² ∈ ℕ is some number that makes r⁴ + r²p a perfect square. Let q be such that qr ∈ ℕ and such that r⁴ + r²p = (r² + qr)². Expanding, r⁴ + r²p = r⁴ + 2qr³ + q²r². Subtracting r⁴, r²p = 2qr³ + q²r². Dividing by r² ≠ 0, p = 2qr + q². Recall qr ∈ ℕ, r² ∈ ℕ, and p ∈ ℕ. Since p = 2qr + q², it follows q² ∈ ℕ. Since q²r² = (qr)², it follows q ∈ ℕ. Since p = q(2r + q), it follows that p is not prime unless q = 1.

@giuseppebassi7406 3 года назад

My solution: let n^2+pn=(n+m) ^2 for (it must be bigger than n). We get n^2+np=n^2+m^2+2nm and then n(p-2m) =m^2. Dividing by p-2m (which has to be >0) we get n=m^2/(p-2n). Now i'm Going to demonstrate that m=(p-1)/2 (and so n is fixed) in order to have a natural value of n. GCD(m^2, p-2m)=(GCD(m, p-2m)) ^2=(GCD(m, p))^2=1 (that's obvious and i left for you as an exercise). So, in order to have a natural n, the denominator has to be equal to 1. Therefore n must be equal to (p-1)^2/4

@aggelosgkekas3113 4 года назад

Nice, next video p vs np I guess

@wolframhuttermann7519 4 года назад

You think too difficult and the solution is much more easier. You have to consider n^2+np = n(n+p) and a:= gcd(n, n+p). If a=p, there is no solution and if a = 1, n and n+p are squares. if p := 2k+1, and p := b^2-a^2 with b > a. So b+a=p and b-a=1. If p = 2k+1, then a =k and n = k^2. By the way b = k+1.

@mcwulf25 2 года назад

I did this differently. If m^2 is the rhs then multiply by 4 and complete the square: 4n^2 + 4np + p^2 = 4m^2 + p^2 (2n + p)^2 = 4m^2 + p^2 p^2 = (2n + p - 2m)(2n + p + 2m) But the factors on the right can't be equal unless m=0 (no solution) so one must be greater than the other, meaning that the LHS must factorise as p^2 and 1. So (2n + p - 2m) = 1 2m = 2n + p -1 (2n + p + 2m) = p^2 4n + 2p -1 = p^2 p^2 - 2p + 1 = 4n (p+1)^2 = 4n Therefore n must be a perfect square and p = 1 + 2*sqrt(n). Both sides are increasing functions, so there is a 1:1 relationship between n and p.

@bilalabbad7954 2 года назад

Is 15 a prime number 🤔🤔🤔

@Cloud88Skywalker 3 года назад

Shouldn't the last system of equations have been solved in order to say the uniqueness is proven? Because I don't think it's trivial when there are 3 equations and 4 unknowns (or 4 equations and 5 unknowns). Also, the chart that inspires the solution is given here but it's not easy to get to it. For example, for p=9, you'd get to 3^2+3·9 = 6^2 way before thinking you have to look for a greater n (n=16) and be stuck quite some time trying to make a pattern out of that.

@maxturgeon89 4 года назад

I know this is a minor point, but the assumption x_1 eq x_2 is unnecessary. We don't need the contradiction (and the law of excluded middle), we only need to prove that all solutions are equal to the same. Which is the conclusion of the argument.

@jiyoungpark6233 3 года назад

when P is zero...?😅😊😊

@AgentMCCityDE 4 года назад

Does odd prime mean anything else than that 2 is excluded? Because otherwise that wouldn't make sense as every prime is obviously odd

@bwcbiz 4 года назад

Is there an example for odd, non-prime p that actually has 2 solutions n?

@MClilypad 4 года назад

P=15, gives n=1 (results in 4²) and n=49 (results in 56²)

@shashankambone6920 3 года назад

@@MClilypad any idea which ones give multiple answers? Like a test or numbers that satisfy so and so equation?

@thomaskember4628 4 года назад

Since I have been watching maths videos I have noticed that Americans use the term "square number" whereas Michael says "perfect square" which is the British usage. Could it be that Michael has studied in Britain?

@l.w.paradis2108 Год назад

Superb problem!

@malawigw 4 года назад

I first tried np = (k+n)(k-n) and then considered the two viable cases. First case p = k + n, n = k-n: gives k = 2n and p = 3n which is prime only for n = 1 so this does not work. Second case np = k+n, 1 = k-n: gives k = n + 1 which gives np = 2n+1 and n = 1/(p-2) i.e. not an integer. Are there any other cases that I have forgotten, or does this method simply not work? :(

@malawigw 4 года назад

Think I know why now. I can not do those cases since n can be composite

@ВасилийТёркин-к8х 4 года назад

I've came up with a solution without guessing. since n^2 + np is a perfect square: n^2 + np = x^2 multiply by 4 and add p^2 4n^2 + 4np + p^2= 4x^2 + p^2 (2x)^2 + p^2 = (2n + p) ^2 so 2x, p and 2n + p is a pythagorean triple and 2x is the even number thus such natural h, k and m exists that 2x = h *(2km) p = h * (m^2 - k^2) = h (m - k) (k + m) 2n + p = h * (m^2 + k^2) since p is prime the only option is h = m - k = 1 thus m = k+ 1 p = 2k + 1 2x = 2k(k + 1) 2n + p = k^2 + (k + 1)^2 2n = k^2 + (k+1)^2 - (2k+1) = 2k^2

@panadrame3928 4 года назад

What about n = 0 ? 0 is a square... so you should say that n is a natural number, not 0

@marklevin3236 3 года назад

What if p =2 ?

@aahaanchawla5393 3 года назад

1*p = p²

@horseman684 4 года назад

I wonder if Australia has olympiads

@adrianshum 4 года назад

Sorry I don’t get the last part: p^2=(y-2x)(y+2x) and you said that two terms are either (p,p), or (1,p). Shouldn’t it be (p,p) or (1,p^2) instead?

@adrianshum 4 года назад

Oops I saw you have overlaid a “2” on the film, never mind then lol

@odiljonrahmatullayev6579 3 года назад

p=0

@raphaelnej8387 4 года назад

the solution to P=NP is N=1 or P=0

@particleonazock2246 3 года назад

truee

@ramanakv3272 3 года назад

P=(k^2-1)n where k is positive integer

@reshmikuntichandra4535 3 года назад

Why does p have to be prime? The argument at 5:07 holds true in general for ANY ODD NATURAL NUMBER

@rubenbohorquez5673 3 года назад

Yup, that's right. The prime part comes into play at 8:38 when you factorize p^2, if it wasn't prime the equation could have more solutions than just p^2*1 and p*p, but if p is prime it has only one solution, and that's what the problem was asking to prove

@dominiquelaurain6427 4 года назад

@9:31 : first factor is 1 and second factor is p^2 (not p) in the double equation...and it doesn't matter for followup. Video edited : two "2" added :-)

@stvp68 4 года назад

Busy with other stuff: hiring committees?

@takyc7883 3 года назад

Just wanted to say, I love your content and have an amazing day you genius! take a break from yt if u want to ever, dont stress about it.:)

@pawetrojan5571 11 месяцев назад

why there is 9 sa a prime number

@B.0MJiR 7 месяцев назад

not only 9, also 15.

@shamilbabayev8405 4 года назад

odd prime? what about 9 and 15 in the table?

@rfzafar 3 года назад

That was beatiful contradiction

@sanjayav8602 4 года назад

Sir, please try to solve JEE ADVANCED problems, one of most challenging competitive exams in india.

@easymathematik 4 года назад

I have one question: For which "age" is this exam? Who take the exam? I´ve taken a look at Jee. Some tasks are very nice! And most of them in the mathematical part is straight forward.

@braineaterzombie3981 4 года назад

@@easymathematik approx for 18 19 year old students

@easymathematik 4 года назад

I will prepare a JEE video. Not just solving it. I will exlain, how to get the right ideas. Most of the ideas are very obvious and natural. Hope to see you on my channel. :)

@easymathematik 4 года назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-iZnIVKHceAU.html

@Maths_3.1415 Год назад

JEE is nothing infront of olympiads 😂

@raphaelpoux934 4 года назад

For p=3n it works also.

@aweebthatlovesmath4220 2 года назад

P has to be prime and p=3n is prime only when n=1 so p=3 but it doesn't work for all n so contradiction.

@Maharshi_Riemann_RC 4 года назад

So we are given a non regular and non concave polygon,we have to show that there will be two perpendicular lines which will divide the area of that polygon into 4 equal areas.And if you don't make a particular video,can you help me with some idea, I've tried this using IVT

@luisisaurio Год назад

Brouwer Fixed Point theorem

@eldattackkrossa9886 4 года назад

michael penn solves the NP problem

@azerbenmed8597 4 года назад

I have a suggestion problem: Find all positive integers (p,q) such that gcd(p,q)=1 and : p+q^2 =(1+n^2).p^2 +1, for some positive integer n.

@azerbenmed8597 4 года назад

@Adam Romanov you have to express p and q in terms of n.

@TorstenAdair 4 года назад

You lost me on the third column. Where did you get those squares? 5, 4, 6^2?

@shashankambone6920 3 года назад

We had n and p in the 2 columns. The third is simply = n^2+np, so just substitute those values and he filled that Column like that.

@2kreskimatmy 4 года назад

prime numbers problems are cool

@youssefamen6872 4 года назад

Can u do Putnam 2005 B6 pls

@kuntalsengupta2598 4 года назад

p is only odd not odd prime, 9,15....are not prime numbers

@R0M4ur0 4 года назад

He used all odd numbers to find a solution. Once found, he proved the uniqueness of that solution for primes

@e-learningbymathex7947 4 года назад

Gd mng sir....

@kingfrozen4257 4 года назад

The magical square

@glitchslayer5630 4 года назад

Polska górą!

@RodelIturalde 4 года назад

Lovely added squares :)

@Saki630 4 года назад

this is a good one too

@kuldeepparashar7266 4 года назад

Thanks sir

@hagee7636 4 года назад

Nice cut

@Matieus 4 года назад

2:15 WAIT! 15 is not prime number ;)

@tom_szcz_org 4 года назад

Thanks for solving a problem from my home country 😁🇵🇱

@silnikpradustaego7213 4 года назад

@Tomasz Szczerba Fajnie, że są tu Polacy. Ja po obejrzeniu tego filmu nawet nie wiem, jaka była dokładna treść zadania.

@tom_szcz_org 4 года назад

@@silnikpradustaego7213 zawsze możemy sprawdzić na stronie OM 😜

@eneapane7594 4 года назад

When p=0🤣

@5271-q7m 4 года назад

Sans Granie

@ejsiisiwkwiwi3829hwjwjhdhf6w 4 года назад

pozdrawiam

@Parodiaseharlemshake 4 года назад

Michael, can you solve lim_{x -> 1} xsin(xpi)/ln(x^2 - x + 1) without L'Hospital's rule?

@shashankambone6920 3 года назад

Sin(xpi) = sin (pi - xpi). Then taylor series approximate both. (Essentially the same thing as L'Hopital.)

@Parodiaseharlemshake 3 года назад

@@shashankambone6920 I've solved. Thx

@shashankambone6920 3 года назад

@@Parodiaseharlemshake did you use some other method? If yes, i'd like to know it too.

@Parodiaseharlemshake 3 года назад

@@shashankambone6920 I've used the e definition, i'm Brazilian and the pdf that I made is in Portuguese. If you want to see anyway I can send for your email

@shashankambone6920 3 года назад

@@Parodiaseharlemshake oh. No thanks, i get what you're saying. Thanks anyways.