Polish Math Olympiad | 2017
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- เผยแพร่เมื่อ 14 ก.ย. 2020
- We look at a nice number theory problem from the 2017 Polish math olympiad.
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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.
Great solution!
oooh my god.............. wow
WoW
nice approach
My god
Greetings from Poland!
Greetings from Not Poland!
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)....
Video: When is n^2 + np a perfect square?
Me: .....
p = 0.
@@bobnich not in natural numbers
@@bobnich
Yeah that's wrong zero is not a prime number
@@bobnich 0 has infinite divisors, where primes only have 2.
@@bobnich 0 isn't prime lol
Me : what's a perfect square?
A polished lesson! Thank you
Very nicely explained, love your videos so much. +They are free!
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
Please,the legend of question six
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.:)
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.
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.
Awesome solution
Brilliant!
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.
Superb problem!
Greetings from Poland! :)
Polska gurom!
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!
@@MichaelPennMath Great idea! Polska gurom!
Michael Penn Ask him how to say a good place to stop in Polish 😂
@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!
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.
@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 :-)
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
0:45 positive means strictly > 0
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.
10:15
Hey old friend
Good place to start at 0:00
As always, a good place to smash that thumbs up
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
Proffessor complicating problem for no reason 😂😂😂
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!
n^2+np? I thought it was p=np?
simple solution, n=1, smh
Wow he solved P =NP
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,
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
Congratulations! !
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
@@Goku_is_my_idol Why do you assume k+n=nb/a?
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
Official solutions are available and I think they are pretty good.
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
small correction y1-2x1=1 and y1+2x1=p^2 not p
Thanks sir
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.
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.
michael penn solves the NP problem
Lovely added squares :)
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.
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.
@Adam Romanov you have to express p and q in terms of n.
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
Pretty polished question not gonna lie
😂
The magical square
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
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).
Think twice should do a visual proof of these!
this is a good one too
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
Where, n^3+np is a also perfect squares?
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.
Amazing solution
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.
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??????
That was beatiful contradiction
P=(k^2-1)n where k is positive integer
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.
Thanks for solving a problem from my home country 😁🇵🇱
@Tomasz Szczerba Fajnie, że są tu Polacy. Ja po obejrzeniu tego filmu nawet nie wiem, jaka była dokładna treść zadania.
@@silnikpradustaego7213 zawsze możemy sprawdzić na stronie OM 😜
Nice cut
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.
pozdrawiam
Does odd prime mean anything else than that 2 is excluded? Because otherwise that wouldn't make sense as every prime is obviously odd
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
Brouwer Fixed Point theorem
prime numbers problems are cool
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.
Nice, next video p vs np I guess
We can also derive it in another interesting way! Since n^2 + np will be a square, n*(n + p) is also a square. It means (n + p) must be an odd power of n so that n * (n + p) will be a square. Therefore, n + p can be expressed as n^3, n^5, n^7, .... and so on. Solving in this way, say, n + p = n^3, we have p = (n -1) * (n) * (n+1). It means that the product of any three consecutive numbers (n >1) measures the value of p for that n in order to make a square. Similarly if n + p is set as n^5, we have p = (n -1) * (n) * (n+1) * (n^2 +1). This also satisfy the condition. In the same way, if n + p = n^7, we have the value of p = (n -1) * n * (n+1) * (n^2 - n +1) * (n^2 + n +1) and so on..
How do you express p as a multiple of 3 consecutive numbers when p is prime? Isn't your method basically implying that, for n^2+np to be a perfect square, p must be in the above forms (all of which are based kn products, which mean p can't be prime?
@@shashankambone6920 Yes, in this problem, p can not be a prime number and rather, it is always divisible by the number 6 (six) as it is a product of three consecutive numbers.
@@rcnayak_58 that's wrong then because clearly the problem in the video, states for p being prime and also, Michael has shown prime numbers satisfying those equations. If p=5 and n=4, then n^2+np= 36 =6^2. By your equations, this (and many other) solution are not taken into account. What value of n does your equation give, for p=5. Michael alao showed (as was the problem) that for prime p, there is only one n which makes n^2+np a perfect square
@@shashankambone6920 So nice. Perhaps, 5 is the only prime that holds good for the problem. Try for another prime number. Perhaps not feasible. Here we made an attempt to show that p (any natural number) is a function of n such that p = (n-1)*n*(n+1). In your example, if n = 4, p can be 60 to make it a square!
@@rcnayak_58 but we're not choosing n. We're choosing p to find an n for the perfect square thing... also, there will be infinitely many examples like p=5 because in the video Michael proved, that for any prime p, there is 1 unique n such that n^2+np is a perfect square.... and he also had the formula for n based on p, so, every prime (maybe excluding 2) has an n such that...
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
odd prime? what about 9 and 15 in the table?
not going to lie, I thought this was going to be an entire olympiad in polish notation...
For p=3n it works also.
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.
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
Why does p have to be prime? The argument at 5:07 holds true in general for ANY ODD NATURAL NUMBER
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
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
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
What if p =2 ?
Is there an example for odd, non-prime p that actually has 2 solutions n?
P=15, gives n=1 (results in 4²) and n=49 (results in 56²)
@@MClilypad any idea which ones give multiple answers? Like a test or numbers that satisfy so and so equation?
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?
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? :(
Think I know why now. I can not do those cases since n can be composite
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
Yeah
I agree
He should take a short break
Is this a good place to stop? (For a bit).
@@RodelIturalde yes
RodelIturalde I guess it is
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!
Sir, please try to solve JEE ADVANCED problems, one of most challenging competitive exams in india.
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.
@@easymathematik approx for 18 19 year old students
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. :)
th-cam.com/video/iZnIVKHceAU/w-d-xo.html
JEE is nothing infront of olympiads 😂
the solution to P=NP is N=1 or P=0
truee
The flag didn't work
Michael, can you solve lim_{x -> 1} xsin(xpi)/ln(x^2 - x + 1) without L'Hospital's rule?
Sin(xpi) = sin (pi - xpi). Then taylor series approximate both. (Essentially the same thing as L'Hopital.)
@@shashankambone6920 I've solved. Thx
@@Parodiaseharlemshake did you use some other method? If yes, i'd like to know it too.
@@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
@@Parodiaseharlemshake oh. No thanks, i get what you're saying. Thanks anyways.
1*p = p²
You lost me on the third column.
Where did you get those squares?
5, 4, 6^2?
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.
Someone tell me how can i increase my potentials in math to solve this kind of questions...
Practise good problems
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?
Oops I saw you have overlaid a “2” on the film, never mind then lol
Could you explain Euler's identify
@@angelmendez-rivera351 the intuition behind it
@@eulersfollower7140 It is like... you write all the series and notice odd and even will help you get series of *e* .
Is 15 a prime number 🤔🤔🤔
I wonder if Australia has olympiads
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
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.
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.
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.
Sans Granie
Can u do Putnam 2005 B6 pls
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.
What about n = 0 ? 0 is a square... so you should say that n is a natural number, not 0
why there is 9 sa a prime number
not only 9, also 15.
Polska górą!
2:15 WAIT! 15 is not prime number ;)
Busy with other stuff: hiring committees?
p=0
when P is zero...?😅😊😊
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
p is only odd not odd prime, 9,15....are not prime numbers
He used all odd numbers to find a solution. Once found, he proved the uniqueness of that solution for primes
Gd mng sir....
"By natural number I mean a positive integer" - 0 is NOT a Natural Number.