He didn't end with its usual quote "it's good place to stop"

Meanwhile me in an exam situation: Well, let's try 1. Nope, doesn't work. Let's try 2. Yup, got it.

3:26 + 20

I think the easiest way to see that 9n^2-10n+5 is positive for natural numbers is to write it as n(9n-10)+5. The first term is negative only when n=1 where we get a total of 4 .

8:26 he picks c=5 because c=3 and c=4 have c^2 terms less than 20, which will make the resulting expression negative for low values of n(0,1,2).

I like the first hint you gave out in the beginning of the video. Great job!

I did it a slightly different way, completing the square and showing that if there is a solution, it looks like a^2 + b^2 = c^2 and therefore we can use the equation for a Pythagorean triplet. Though my approach is somewhat less elegant!

I inspected the first few numbers and found n=2 worked then did synthetic division to find there was a remainder of 57/( n-2) and figured that meant there were no other solutions.

The multiplication by 4 is a very smart idea. Thank you.

Can you please feature this nice integral problem? Integrate the (fractional part of 1/x)^k, where k is some natural number, from limits x=0 to 1. The expression for the final answer is to be written in form of a sum, which can be computed for a few values of k. Ans: sum (zeta(j+1)-1)/(binom(k+j)(k)) from j=1 to \infty

this bounding was also useful on EGMO 6

Sir we could have looked a perfect square as sum of odd nos and expression suggest that difference of two square is 5

Very well explained!!😃😃😃😃

this solution is so hard you can easily solve it with some creative inequalities edit: as it was pointed out in the comments this proof is wrong I am gonna post a slightly different one in the comments below first you observe n^3>3n^2 +5 for every n>=4 (1) so you check n=1,2,3 for solutions and you find n=2 as a solution and then you make the the argument that n^4 -n^3+ 3n^2+5=4 ] which leads to 3n^3+4n+4n

also love the "tap" on the board @1:06 !!! : )

for n in range(1,100): for i in range(1,100): if((pow(n,4) - pow(n,3) + (3 * pow(n,2)) + 5) == pow(i,2)): print(n) *python code*

Since the problem was to find *all* natural values of n that would make the given quartic expression a perfect square, and it turns out there is only *one* such value, then a "guess" of n = 2 is a very quick solution - which I noticed in less than 5 seconds. Nevertheless, actually *proving* it was very impressive - another great video!

Is it possible to use induction?

At 3:37 hew wrote A = 4n^4-4n^3+12n+5 that 5 should be 20.

Absolutely fenomenal

Great job

I did it in a simpler way imo: The general approach to this kind of problem is to try and bound the expression between two consecutive squares. n^4 seems like a good upper bound because of the -n^3 attached to it, which should overpower all the other terms in the limit. Thus, we shall try to find k such that n≥k implies n^4>n^4-n^3+3n^2+5, or in other words, n^3-3n^2-5>0. The derivative is equal to 3n^2-6n, which is increasing for n≥2, and by inspection n=4 makes the expression positive, so n≥4 should also work. Therefore, k=4. The next step should be to try to bound the expression below by (n^2-1)^2. Similarly, we shall try to find m such that n≥m implies (n^2-1)^2>n^4-n^3+3n^2+5, ie n^3-5n^2-4>0. Deriving the expression, we get 3n^2-10n, which is increasing for n≥4, and n=6 satisfies the inequality by inspection, so m=6. Because of that, we can already conclude n≥6 will satisfy the inequality (n^2-1)^2

Why go back to the quartic when have reduced it to a quadratic to find n near the end?

Great job sir What about this : 7n^3-3=m^2 What we can do

At 11:23 - an alternative is to show that f(n) = 9 n^2 - 10 n + 5 = 0 has no real roots (which is true because 10^2 - 4*5*9 = 100 - 180 = - 80 is not the square of a real number). Since f(n)=0 has no real roots, and f(n) is continuous, it cannot change sign. And since f(0) is obviously 5, f(n) must be positive for all real numbers. Edited to add: there's also a transcription error at 10:05 or so: A - (2 n^2 - n + 2) = 3 n^2 + 4 n + 16, not 3 n^2 + 14 n + 16. The conclusion is still correct (A is always greater than (2 n^2 - n + 2)^2 for any natural number n).

nice, clean handwriting

Brilliant

I got correct, I started with 1 next got 2 as the solution and gone till 6 and not got any so stopped.

10:05 3n^2+4n+16. In any case, always greater than 0. Thanks.

There is a much cleaner and easier way to solve this using quadratic residue. n^4 - n^3 + 3n^2 + 5 = z^2. z^2 is comparable to 1 mod 8, and so is the long expression. Hence, there are 4 cases: n odd, z odd; n even, z odd ... Modulo arithmetic immediately reduces it to 2 n odd, z even and vice versa. Then you write n^4 as 4^k(8m+1). After some trivial checking the solution is clear.

Put n is equal to it gives 25 . is first perfect square for that

Can anyone tell me the solution of this functional equation or its source please :) The problem: Let α,β ∈ Q* . Find all functions f: R ->R Such that: f([x+y]/α)=[f(x)+f(y)]/β

How did these questions are even made?

For me, the key insight was to try to bound n^4-n^3+3n^2+5 *between* "consecutive" squares, rather than analyze the "squared-ness" of n^4-n^3+3n^2+5 directly. To that end, I started by finding values of a and b such that n^4-n^3+3n^2+5 = (n^2+an+b)^2 cancels as many terms as possible, which yields a = -1/2 and b = 11/8. The net result is that you can strictly bound sqrt(n^4-n^3+3n^2+5) between n^2-n/2+1 and n^2-n/2+2 for large enough n...in fact, for n > 2, so n doesn't even have to be that large. And for even n, n^2-n/2+1 and n^2-n/2+2 *are* integers, fortunately, so you're done. For odd n, there's just one more small step: verify that the unique integer between n^2-n/2+1 and n^2-n/2+2, which is n^2-n/2+3/2, can't square to give you your n^4-n^3+3n^2+5. Wrap it up by checking n=1 and n=2 explicitly.

At 11:40 another easy way to notice it is always greater than 0 is by noticing with n=1 it is equal to 4, and then n=2 is also positive, and then notice than the exponential n^2 is growing at a greater rate than -10n once the n becomes around 3ish etc, so the value of the expression will only become greater, thus always being positive, given n being a natural

Do we need to prove there are no other solutions if you guess a larger c? Or you have proved it?

for Z the solution is -1

There was no need for checking the first case as A is even and the squared thing was odd ...

3:25 *20 not 5

7:19 I don't understand this argument here. There is no reason to compare the coefficient of the dominating term here because n is not going to infinity.

The solution was not too intuitive. Can anyone suggest some intuitive methods

Great work...keep it up

He didn't end this video with his usual quote "it's good place to stop"

Man.... I really like your videos but I get lost so often, can you suggest me some good videos to start with. I really want to understand everything you do.

Easy: when the sides are equal length.

I have an easier approach for this problem. From question we let n^4 - n^3 +3n +5 = k^2 (k is non-negative integer) Rearrange the equation we get n^2 ( n^2 - n +3) = k^2 - 5 Note: 1) Both terms at LHS must be non-negative integers, RHS must also be non-negative integers. 2) The term n^2 is a perfect square 3) Factors of 5 are only 1 and 5. Therefore, the gcd of k^2 and 5 can only be either 1 (perfect square) or 5 Case 1 (if k^2 is not a multiple of 5) n^2 ( n^2 - n +3) = (1)( k^2 - 5) By comparison Sub case 1i) n^2 = 1 --> n = 1 Sub n =1 into following equation n^2 - n + 3 = k^2 -5 k = sqrt 8 (rejected) Sub case 1ii) n^2 - n +3 = 1 No real solution for n. In fact the LHS term is not a perfect square. Case 2 (k^2 is a multiple of 5) We can rewrite the equation as n^2 ( n^2 - n +3) =5( k^2 /5 - 1) Where k^2 /5 is still an integer Since 5 is not a perfect square n^2 - n + 3 = 5 n = -1 (rejected) or n=2 Sub n=2 n^2 = k^2 /5 -1 4 = k^2 /5 - 1 k = 5 Thus the only solution obtained is when n = 2 Checking the solution by substitute n = 2 into original and u will obtain 25 which is a perfect square.

12:06 Do naturals not include 0? Although if you specialcase n=0 the inequality will still hold.

Soy el primero! 😅🎉

Haha I remember this last year

Loving the videos, but your audio is terribly quiet. Perhaps turn up the gain on the mic or video software?

Number 2 will satisfy your fomula end up as 25 and will be a square of 5.

14:38

Let n^4-n^3+3n^2+5=m^2 , where m ∊ N …. Eq1. If m = 1 or 2, then Eq1 has no solutions such that n ∊ N. We can see this as the LHS of Eq1 is ever increasing for n ∊ N, with a minimum value of 8, so m = 1 or 2 are not possible. So, for the rest of the solution we will assume m ≥ 3. This being the case, we note that 5 - m^2 < 0, where 5 - m^2 ∊ Z. Eq1 can be rewritten as n^4-n^3+3n^2+(5 -m^2)=0 … Eq3. Applying Strum’s theorem to Eq3, we find that the equation has two real roots, where one is negative and the other is positive for a given value of 5 -m^2, where m ≥ 3. Of course if Sturm's theorem is an over kill for you, then you can just find the critical points of n^4 - n^3 + 3n^2 + 5 and it will be discovered there there is only one, and it occurs when n = 0, and it is a minimum by the second derivative test. This being the case and the realisation that as n goes to infinite or negative infinite, then n^4 - n^3 + 3n^2 + 5 goes to infinite. Now with the minimum at n = 0, then the minimum is at 5 - m^2, which is less than zero and hence there is one negative real root and one positive real root, and so the remaining two roots are imaginary. Now let k = m^2-5, and place it in Eq3. ⇒n^4-n^3+3n^2-k=0 … Eq4. Now let’s assume that n = d is a positive real value solution to Eq4, where d ∊ N, and place this in Eq4. ⇒d^4-d^3+3d^2-k=0 … Eq5. ⇒d^2 (d^2-d+3)=k ⇒d^2 divides k. So, let k = td^2 ⇒d^2 (d^2-d+3)=td^2 ⇒ (d^2-d+3)=t, as d ≠ 0 ⇒ d^2-d+3-t=0 … Eq6. Let’s factorise Eq6 and see if we can find b and d ∊ N that satisfies the equation. Let (d+b)(d+(3-t))=0 ⇒ d^2+d(3-t)+bd+b(3-t)=0 ⇒ d^2+d(3-t+b)+b(3-t)=0 … Eq7. Comparing Eq6 and Eq7, we observe that -1=3-t+b and 3-t=b(3-t) ⇒ b=1, and hence t= 5. Placing t= 5 into Eq6 ⇒ d^2-d-2 =0 ⇒(d+1)(d-2)=0 ⇒ d =-1, or d= 2. Now as d ∊ N, then d= 2 is the only possible solution. Now with d= 2 and t= 5, and as k = td^2, then k = 20. As k = m^2-5, then m^2=25, thus m = 5. Placing m = 5 into Eq3, we find indeed that n = d = 2 is a solution, and it is the only solution for any m as our analysis has shown, as only when t = 5. (I.e. when m^2=25) is an integer solution possible to Eq3.

The 16-th Hungary-Israel Binational Mathematical Competition 2005 (imomath.com/othercomp/Hi/Himc05.pdf ) had strange problem number 3 (find all sequences of distinct positive integers x1,x2,...,xn such that 1/2=1/x1^2+1/x^2+...+1/xn^2). There seems to be no full solution nowhere to be found on the internet and all related posts seem to imply it is exceptionally hard, yet it was on the competition (unless its presence on the competition was "trolling" the attendees...). Still I think it's interesting to look at.

How much time is given per question in the Olympiad?

Please a humble request that make some videos teaching geometry of circumcircle and cyclic quadrilateral and pigeonhole principle a humble request by your Indian fan and subscriber

i made a little Python program to check for other solutions. for now, it found a second one: n = 33554535. i checked a couple of times and it looks like it satisfies given conditions. can someone verify it? maybe i made a little mistake somewhere

So, n=-2/5 is a natural number. Which means that this expressions are perfect square. OK!!! 😁😁👍👍

Lol, I competed in that and got that question wrong XD

I am not totally convinced by your argument. You have found a solution in the sequence of squares u(n, r) = (2n² - n +r)², but there are 3(4n² -2n + 8) positive integers between your upper and lower limit for 4n⁴ - 4n³ + 12n² + 20, and you haven’t shown that none of these other than the one you found are perfect squares. So in effect you have failed to demonstrate the uniqueness of the solution you found.

I tried to solve it from the thumbnail: I thought a "perfect square" meant a quadratic in the form (an+ b)^2 or a^2.n^2 + 2abn + b^2 And going from there I got n = something crazy Turns out "perfect square" just meant "a square number" So... I approached this completely wrong :/

First three terms take n out and equate.

I started off trying to square (ax^2 + bx + c) and compare it to the original quartic, and got stumped.

Plz do more problems form indian national mathematical olympiad or Regional mathematical olympiad from india 🙏🙏 it's a request

@14:38 mood

I hated it when you start guessing

3:36-+20, not +5

14:40 😂😂😂mental calculation can sure be a bitch.

Great solution. But, the only one anwser is ridiculous.

15:53 I mean obviously there are no solutions: you imposed that an even number is the square of an odd number.

he has been working out lol

n^4 - n^3 + 3n^2 + 5 = k^2 n^4 - n^3 + 2n^2 - n^2 + 5 = k^2 n^2(n^2 - n + 2) - n^2 + 5 = k^2 n^2(n - 2)(n + 1) - n^2 + 5 = k^2 This suggests solutions of -1, 0, and 2. The first two are not natural numbers, 2 is *a* solution. But I don't know how to prove there aren't more solutions.

Anyone who can prove or disprove this .... If x^p = y^p (mod n ) then x=y (mod n)... Its converse is true... Thanks..

what is that deep breathe