Відео

Shitikanth

Please bring limits questions also

What app u use ?

Love your work

Man, you should post more nice jee and putnam questions, love ur vids v much.

In the minutes 6:13, what is the name of the property you used to obtain the cases?

nice very cool

Excellent impressive

My favourite problem from this putnam was a6 and b5

Great advice in the end!

👏👏👏👏👏👏👏👏👏

Very clear explanation! Amazing, thanks!

I was looking for problems that can be solved with Jensen's inequality

Why the site link doesn't work ?

looking at other comments ( which are very long and boring to read ) here is a simple summary . 1. gcd ( x , y ) = gcd ( x + y , y ) note that in the problem if p ^ 2 = a ^ 2 + 2 * b ^ 2 then a < p . and so gcd ( p - a , p + a ) = gcd ( 2 * p , a + p ) = 2 gcd ( p , ( a + p ) / 2 ) = 2. since (a+p )/ 2 < p . and any number less than a prime is co prime to that prime . this was the main crux of the problem.

geometry when?

Interesting the population found it tough. Perhaps there was less exposure to questions about Riemann sums.

My proof was the same, but I want to talk about the method I discovered it. Earlier I tried to take the modulo of the equation mod some numbers, so at least I knew p and a were odd and b was even. I looked and found two cases for (a, b, p, c, d) for small p: (1, 2, 3, 1, 1) and (7, 6, 11, 3, 1), where (c, d) is the pair s.t. c^2 + 2d^2 = p. So something I noticed was that c = b/2 for these examples. I tried to expand p = c^2 + 2d^2 => c^4 + 4c^2d^2 + d^4 = p^2 = a^2 + 2b^2. Of course you shouldn't preassume that c, d already exist, but I did this to see if good forms of c and d in terms of a, b, existed. So first I substituted c = b/2, but there was not a good result there. Somehow I figured out that one of the b^2s is equal to the 4c^2d^2 term, so I found that we could get c^2 - d^2 = a 2cd = b as a system of equations of c and d. For such c and d, (c^2 + 2d^2)^2 = a^2 + 2b^2. In fact we can usually find c and d now from this system of equations by solving a quadratic, it just won't usually be an integer (but somehow, when p is prime, it is!) So I tried to solve for d and got d = sqrt((p + a))/2. Then I discovered that sqrt(p+a) had to be an integer, except that in the 11 example I discovered it clearly wasn't: 11 + 7 = 18??? This confused me for a while until I realized if (a, b) is a working solution, so is (-a, b). The quadratic from this would give sqrt(p - a)/2, and indeed sqrt(11 - 7)/2 is an integer. So now I had to prove that p - a or p + a was a square, and since (p - a)(p + a) = 2b^2, it is (by the means shown in the video). I also got a second helpful hint that d^2 | one of p - a or p + a | 2b2, so d | b and as c = b/2d, c is an integer too. So we've found a (c, d) pair generating p from the (a, b) making p^2. It's funny that this fact was the last I discovered when it's the first thing you should prove when proving the problem statement. In writing the proof I started with this fact instead, but for this scratch work I had the reverse order. I didn't have the olympiad intuition or knowledge of the trick to solve this at first, but I was eventually able to power through and discover the tactic I was supposed to use. I'm ashamed to say that some of my first attempts were taking mod p, which led nowhere and got me stuck for a while. It's probably not a good sign to overuse the strategies you know. In the end it was a good problem.

Nice method You could also define another function h(y)= integral from 0 to 1 (x^2y-xy^2)dx Then since we need Max we can equate h'(y)=0 So integral 0 to 1 ( x^2-2xy)=0 Therefore y=x/2

Nice problem and even nicer explanation!!

Can't thank you enough, really helpful. Great explanation as always. شكرا

Overcomplicated way of forming a polynomial that satisfies that equation. Kx^n -x^(n-1) -x^(n-2) - ... x^3-x^2-x+K works for K=n Basically, in the expansion, the terms with n in them are big enough to outweigh all the other stuff. (And if you want this to be trivial, just make K huge)

thoughts on the answer of 2n-2 obtained by ua-cam.com/video/AFzvrCk_HYM/v-deo.html ?

great narrative dude, really cool and fun explanation

i.kym-cdn.com/entries/icons/original/000/035/627/cover2.jpg

Amazing as always

The Return of The King

3:59 if 2a0a1 is negative, why must a0 and a1 be opposite parity? why can they not both be same parity but opposite signs?

That's the easiest problem . Just put n=1 or 2 and compare.

Hmm so I tried to use Cauchy-Schwarz inequality and it kinda works, but not so efficiently ^^; So take our quantity S = Sum a_i / (2--a_i) and mutiply it by the quantity P = Sum a_i(2--a_i). Cauchy-Schwarz inequality says that S*P >= ( Sum a_i )² = 1, so we could simply try and prove that 1 >= P * n/(2n-1), or equivalently 2 - 1/n >= P But we have P = Sum 2*a_i -- a_i² = 2 -- Sum a_i² So we have to show that Sum a_i² >= 1/n But again by Cauchy-Schwarz we have n( Sum a_i² ) = (Sum 1)( Sum a_i² ) >= Sum a_i = 1 so we are done.

love this channel, great explanations and great problem selections. Keep it up

cool problem! thanks for showing me how to do it!!!! but what would happen if x=5221 instead???

This can be solved by giving n=2..and conparin with Tn and Sn with value of pi/(3*rt3)..then v would get A and D ..in less than 3 mins

the beast Creative Math is back baby!

This has been advertised on other channels. Nice job with this problem.

"The Return of the King"

7 is also a prime factor of a_2015. To prove this,let us compute some of the terms modulo 7.We shall not compute the values of the terms but the remainders when they are divided by 7.These will form a pattern.Starting from a_1, the consecutive remainders (using theory of congruences and the given relation between consecutive terms) are as follows: 1,2,0,5,6,5,0,2,1,2,0 Now,observe that the first 3 remainders are the same as the last 3 in succession,which means that the sequence of remainders gets repeated.The remainders are 0 in case of a_3 ,a_7 and a_11(last remainder in the sequence).After that,the entire cycle of remainders keeps on repeating as we compute them in succession.Thus,remainder,when a_2015 is divided by 7 is equal to 0,which implies that a_2015 is a multiple of 7.

Great!

Fun fact is that I solved this que under 2 min and then I came to know that it is said to be the toughest que..

Your age? Your voice is like a teenager.

I suspect there is an error at 4:12. You write "x+y = lambda1 * x" and "x = lambda2 * y", but in my opinion it must be "x = lambda1 * y".

I will give you one extremely ridiculous challenging integral open challenge to you if you can solve it. If you wanna try kindly reply. I am also unable to solve it I find it on an Instagram page.

sir there is one request can take THEORY CLASS ON POLYNOMIAL { from INDIA}

Wow that’s very interesting, actually very classic but beautiful!

اسطورتي

There's a mistake in the solution. Please ignore this video. I'm not allowed to equate the factors at the end and there are actually more cases than the ones in the video. Apologies

Nice problem

How do you like drawing in Miro?

Orz

Nice solution! Orz