you gotta find *all* positive integers

I too. Same

Arnav Singh did I say it was, euler?

@@heh2393 it says all non-negative, which includes 0 too

Lol

And "okay, great" is the new implies sign(=>).

TAGPTS

LMAO

Quod erat demonstandum

@@danielalp6871 WoW

2009 ends with 9, that's the hint

@@allasar For much larger numbers the way BeeBee does it will be faster - take the number 734,413. The next perfect square that is bigger than this number is 857*857 = 734,449, Notice that 734,449 - 734,413 = 36 which is also a perfect square and so the factors of my number are 857-6 and 857+6 = 851 and 863. 851 * 863 equals my number 734,413. 863 itself is prime but 851 is not. Using the same method for 851 we see that 851 is 49 away from 900 (perfect square of 30*30) so its factors are 30+7 and 30-7, so 23 and 37. If one factors the original number starting from 1 and using all primes you will not get a 'hit' until you reach 23. There is no luck in using the perfect squares method - it will work or it will not. Heck just use the smaller example of 851. By the perfect squares method you know right away that 851 is not prime because 30*30 is 900 and 900-851 = 49 which is also a perfect square. You will know the factors in a few seconds.

@@allasar If you know the squares from 1 to 100 like the kid said then for some numbers like 2021 it would be faster to get the factors using the perfect squares method because you would not get a hit the other way until you reached 43. Of course with your quadrillion number the perfect squares method would not work and your first divisor found would be 7 . Then there are other numbers like if you add 2 to your quadrillion number- that number would take a LONG time to factor since the factors of that number are two large prime numbers themselves.

notice that 2009 = 2025-16 = 45^2-4^2 so (45+4)(45-4)=(49)(41)

For the number 2009 either prime factorization method OR the perfect squares method can be applied and the prime factors found relatively quickly . For a number like 2021 the perfect squares method would be faster since you would have to test several prime factors using the other method before getting a’hit’ with the prime number of 43. Of course there are numbers where the perfect squares method would not work especially if the number is very large but the same argument can be made for the other method too if the large number does not jive with the known divisibility rules

On second thought, I could totally see this as a highschool math contest problem.

This is definitely one of the easier questions. But a nice one!

@@hassanakhtar7874 This looks like one of the harder (but not hardest) Euclid problems.

It's very easy for a BMO Round 2 problem

it is easy to solve but hard to see. unless you encountered with similar problems before

math isn't a plural noun dude

Not always, though maths is more common. But I am British and was taught math not maths.

I speak Englishs

It is mathematics, not mathematic. So it's maths not math. Similarly, physics is a singular word but no one tries to drop the s. After all that who cares! The math is wonderful however we spell or say It!

Have to laugh at these Americans telling me, a brit with a maths degree, that I'm wrong about my own language in my own country!

Definitely agree there. I would probably have continued maths in university if there was this level of content. At the time there was only Khan academy (not to insult them it just wasn't in depth enough back then).

And how does (41, 1476) fit in your 28×35 triangle? 😮

@@ciberiada01 √41 is the hypotenuse of a 4, 5, √41 right triangle, while √1476 is the hypotenuse of a 24, 30, √1476 right triangle. Both have slope 35/28=30/24=5/4. Finally 4+24=28 and 5+30=35, so it checks out.

@@valeriobertoncello1809 Oh, what an elegant solution! Thank you, Valerio! 👏👍 I just didn't understand it at first. So, you take the right part: √2009 = √(7²41) 41 is obviously a prime, but because it's 4k + 1 prime, the *only* way to represent it by two perfect squares is: 41 = 4² + 5² {1} And why do you need perfect squares and not just *any* numbers? Because in this way, you can represent √41 as the right triangle's hypothenuse (apply the Pytagorean theorem). The same is valid for √(7²41) : √(7²41) = √(7²(4² + 5²)) = √(28² + 35²) And this is the hypothenuse of our "wrapping" triangle. Its sides are 28 and 35. ❕With {1}, we are sure it exists only one such triangle. Now, you do the same for √a and √b So, √a represents another right triangle's hypothenuse: √a = √(m² + n²) Same goes for √b : √b = √(p² + q²) , where m, n, p, q are the sides of these 2 smaller right triangles. So we have: √(m² + n²) + √(p² + q²) = √(28² + 35²) I imagined that if you align the 2 triangles, so that their hypothenuses √a and √b follow the same line, you get: √a + √b = √2009 (hypothenuses) m + p = 28 = 4×7 (first sides) n + q = 35 = 5×7 (second sides), But not any m and p between 0 and 28 will do! Because all sides must be integer and we must keep the same 5/4 slope, m and p must be multiples of 4, as well as n and q must be multiples of 5. Thus, there are exactly 8 pairs that satisfy this: m, p, n, q, √a 0 28 0 35 0 4 24 5 30 √41 8 20 10 25 √164 12 16 15 20 √369 16 12 20 15 √656 20 8 25 10 √1025 24 4 30 5 √1476 28 0 35 0 √2009

@@ciberiada01 Yes, exactly! I was inspired by 3b1b's video on π/4 = 1 + 1/3 - 1/5 + 1/7 ... that talks about Gaussian Integers and complex factoring of Natural numbers. Really good stuff! Here's the link: kzfaq.info/get/bejne/hMd8kqaTmZi7qY0.html

@@valeriobertoncello1809 👍 Really interesting topic!

Thanks bro

I almost missed it. Thanks novelty YT account!

You really commited to that joke

@@Lukasek_Grubasek and he got his reward

I made it to 300 ♪

Yea I see what you mean. Just because 2 times the root has to be an integer doesn't mean that the root has to be one, it could be a fraction where the 2 cancels the denominator.

Wait in hind sight I think I understand it now. I think there is a proof that states that a square root is either an integer or an irrational number. Because you can't multiply an irrational number with an integer to get anything other than another irrational number, the root has to be an integer. I wish he would have said that for even 5 seconds, it would have cleared up a lot of misconceptions.

ahahah

DUDE😭😂😂😂😂😂😂😂😂😂😂😂

This is exactly what I was thinking!

Easier test is to test 2009 for divisibility of 2, 3, 5, 7, ... Divisibility is easily tested by adding modulo each digit by "weight". Weights for 2 is 0, 0, 0, 1, for 3 is 1, 1, 1, 1, for 5 is 0, 0, 0, 1, for 7 is -1, 2, 3, 1, 11 is -1, 1, -1, 1, for 13 is -1, -4, -3, 1.

I can't believe the Mentats wouldn't be sponsoring this content! (Although CHOAM does admittedly have a lot of money.)

Good point!

Wdym squarefree part?

@@randomdude9135 argument inside the square root is not divisible (coprime with) (prime number)^2, i.e. you can have as many distinct prime numbers in the prime factorisation as you want but their powers must all be exactly 1

Same question.

Great question! Lets look at the board at 7:23. Notice that even if we used the negative counterparts of x and y we are going to take the square root of (x^2) and (y^2). And sqrt(x^2) = |x| because we use the positive answer by convention (the question itself is phrased using principal roots). Ultimately its just a choice to take positive integers because the important idea is that a,b are 41 times a perfect square.

They don't 'have' to be non-negative, they are defined to be non-negative.

@Андрей Босой You are joking right?

I was confused by that too - we know "2*sqrt(2009a)" is an integer, which allows that "sqrt(2009a)" could be a half-fraction, right?

TheBrownMotie yes , but 2009a is integer, and if sqrt(2009a) = m/2 (m integer) => sqrt(n) is rational (n integer, and not square)

We have the equation b = 2009 - 2sqrt(2009a) + a. We know that b is an integer so then it follows, because both 2009 and a aren't rationals, that 2sqrt(2009a) is an integer divisible by 2 ----> sqrt(2009a) is an integer.

a and b are integers. So, sqrt(2009a) has to be an integer. What am I missing? If 2sqrt(2009a) is an integer but sqrt(2009a) is not, then "a" cannot be an integer, right?

​@@andreybyl So sqrt(2009a) is rational, but not necessarily an integer? That's where I'm confused

Then I'm the man who loves to watch how others work. How about you?

Well, he is actually already a genius, maybe not to all mathematicians, but certainly to the average person. He is an Olympiad winner. So this means he doesn’t really work, he is like a boat on the river of creativity. He pushes the door open with his feet and finds short cuts. He doesn’t do unnecessary calculations, in fact, he is the artist of minimizing the amount of calculations in a problem…

So a hypotenuse of 2009.. metres? That would be a... great value.

@@siralanturing9103 I mean it could be 2009mm, or cm, don’t see what you’re getting at

@@siralanturing9103 well, this is a negligible problem. Just think of 2009 units. You don’t need to care if that’s feet, nm, miles or mikrometer. Just the relation between the number counts, not what the exact kind of unit Someone might use. The described relationships are invariant with respect to all units…

@@howmathematicianscreatemat9226 Yeah, I know. What I meant was imagine if we had a hypotenuse of 2009 m and we were told to find the sides in cm. That would've been something, no?

Yeah I had the same thought.

We have sum of integer plus square root equal integer number. It means that square root is integer number really. For example, 3+5+x=17 obviously can't be truth if x isn't integer number. So '2009a' is a square number (because our square root is integer). 2009=41×49=7²×41. So 41a is a square number and it's true only if a=41x² (x is integer).

You mean a = 369, b = 656

RIght, so \sqrt(x) = n/2, so x=n^2/4, but x=2009a is an integer, so n is even.