British Math Olympiad | 2009 Round 2 Question 1

[@jayarani31]

The first thing that came into my mind was a=0 , b=2009 lol

[@GammaFZ]

you gotta find *all* positive integers

[@thedevil_boy5934]

I too. Same

[@GammaFZ]

Arnav Singh did I say it was, euler?

[@kshitizpokhrel7482]

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

[@syeduzairuddin7518]

Lol

[@paulooliveira4349]

"That's a good place to stop" is the new Q.E.D

[@kingkartabyo6206]

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

[@emanuellandeholm5657]

TAGPTS

[@sahilbaori9052]

LMAO

[@nuzlock4481]

Quod erat demonstandum

[@glitched_matrix44]

@@danielalp6871 WoW

[@omgopet]

Ah, yes, 49*41, of course...

[@Han-tc7wx]

2009 ends with 9, that's the hint

[@johnsmarra4902]

@@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.

[@jmslaforzadeldestino4943]

@@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.

[@Hope16449]

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

[@jmslaforzadeldestino4943]

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

[@cycklist]

British 'Math' Olympiad? No way mate, it's 'maths' here.

[@natepolidoro4565]

math isn't a plural noun dude

[@ngc-fo5te]

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

[@HeyKevinYT]

I speak Englishs

[@liamgeorgemusic1085]

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!

[@cycklist]

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!

[@mnazaal]

Im currently a grad student but i remember 4-5 years ago I was searching for good quality content just like yours, but to no avail. Im really glad you're filling up that void!

[@Jacquobite]

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).

[@valeriobertoncello1809]

Another nice way to solve this: Notice that 41 is a 4k +1 prime, and can be expressed uniquely as the sum of two integers squared, 4² and 5². Thus √2009 = √(7² × 41) = √(7²4² + 7²5²) We can conclude that √2009 is the length of the hypotenuse of a right triangle with sides 28 = 4×7 and 35 = 5×7. Therefore, all the possible solutions of √a + √b = √2009 correspond to all possible combinations of integer sided* right triangles that fit inside the 28 × 35 triangle. *EDIT: by "integer sided" I mean that every side of the triangles BUT the hypotenuse must be an integer, obviously (because the hypotenuses have to add up to √2009).

[@ciberiada01]

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

[@valeriobertoncello1809]

@@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.

[@ciberiada01]

@@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

[@valeriobertoncello1809]

@@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: kzbin.info/www/bejne/hJKvkHaYaZeKr7s

[@ciberiada01]

@@valeriobertoncello1809 👍 Really interesting topic!

[@itzmrinyy7484]

I loved solving this problem on my own! Especially as I'm reading The Art & Craft of Problem Solving, it was really fun to play around with the equation and create various cases (like a=b), figuring out why they don't work, squaring it and getting deeper insight into the conditions the numbers have to meet, and finally going back and looking at my factors and realizing sqrt(2009) could be rewritten as 7*sqrt(41).

[@jbtechcon7434]

When he says, "That's a good place to stop" by God he means it.

[@goodplacetostop2973]

8:24

[@Manuel-pd9kf]

Thanks bro

[@emanuellandeholm5657]

I almost missed it. Thanks novelty YT account!

[@Lukasek_Grubasek]

You really commited to that joke

[@glasgowblackchigowski6117]

@@Lukasek_Grubasek and he got his reward

[@oak8376]

I made it to 300 ♪

[@maharanirani54]

I am so happy that I found this channel. Hopefully you won't stop post the new videos🤗

[@nickpatella1525]

ah yes the fundamental theorem of arithmetic, my favorite theorem

[@pixelchu]

KZbin recommended brought me here. This is some interesting stuff I’ll say.

[@ramaprasadghosh717]

dividing both the sides by √(41) one gets √(a/41)+ √(b/41)= 7 the unordered pair (a,b) € { (0, 41*49 },(41,, 41*36 }, (41*4, 41*25 },(41*9 41*16 }}

[@nahrafe]

"Reduce to smaller problem" Informatics guy: interesting

[@ramazanperdebai9979]

ahahah

[@moskthinks9801]

Indeed, if sqrt(a)+sqrt(b)=sqrt(c), then a, b, and c have the same squarefree part. (provided these numbers are naturals ofc)

[@ThePharphis]

Good point!

[@randomdude9135]

Wdym squarefree part?

[@CaradhrasAiguo49]

@@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

[@_AadiDevSharma]

Your method is somewhat good but it can be expanced to higher level problems.. for example if the question was:sqrt(a)+sqrt(b)+sqrt(c)+sqrt(d)=7 then we can't do it such a way done by you.. So here is almost same method but with different approach》 Write sqrt(2009) as 7.sqrt(41) As done by u x+y=7 ..we can extend it to my q and can say x+y+z+w=7 Now using pnc no. Of non integral solution are :(7+4-1)C(4-1) i.e 10C3. Similarly the ans of your's q must be (7+2-1)C(2-1) i.e 8C1 i.e 8

[@jim42078]

I actually solved one of these problems for myself for a change. So glad I found this channel.

[@dejans001]

Conclusion that sqrt(2009a) is an integer is wrong. You can only say that it is 1/2*integer or that 2*sqrt(2009a) is an integer Also the same mistake is made with 7sqrt(41a). My opinion that this solution will be enough for 5/7 points

[@Powerracer251]

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.

[@Powerracer251]

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.

[@bridgeon7502]

2009a is always an integer as a is an integer. So that means √2009a is either irrational or an integer

[@Maxmuetze]

Useful fact to remember for 4- or 5- digit number factoring: 1001 = 7*11*13 (in problems that are set up nicely with small-ish prime factors). To test 2009, just test 2009 - 2*1001 = 7 to see 7 is a factor. Even for, say, 15877 it's still pretty good: just check 15877 - 15015 = 862 to get 7,11,13 out of the way in one go. Or even better 16016-15877 = 139

[@TonyChurch112]

This is exactly what I was thinking!

[@bpark10001]

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.

[@trrn13p]

Why does x and y have to be non-negative if you square them?

[@refreshingmint9663]

Same question.

[@pedroteran5885]

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

[@harris5140]

I watched the problem solved by different people but I didnt get it. Your explanation is much more clearer.

[@Gameboygenius]

Michael always knows when it's a good place to stop.

[@Patapom3]

Would that be a failed solution if we only gave the (a,b) that are not equal to (b,a) also? I mean, half the solutions are mirrors of each others...

[@maxjackson6616]

you should do BMO2 2017 problem 2, it fits your style of videos very nicely. it was also in an IMO shortlist I can't remember the year though.

[@henkhu100]

after 2:40 the conclusion is that sqrt(2009a) is an integer, but is the correct conclusion not that 2xsqrt(2009a) is an integer because the term in the equation is 2 times the sqrt. And of course after that it is simpel to prove that the sqrt itself is an integer.

[@hansisbrucker813]

Why is that so?

[@henkhu100]

@@hansisbrucker813 2sqrt(2009a) is integer, so it is even or odd. If 2sqrt(2009a) is even then sqrt(2009a) is an integer and we can follow Michael. If 2sqrt(2009a) is odd then we have 2sqrt(2009a)=2p+1 with a and p integer. This gives: 8036a=4p^2+4p+1 with a and p integer. But that means even = odd. So we can continue with sqrt(2009a) is an integer. etc.

[@hansisbrucker813]

@@henkhu100 Why is it obvious it is an integer?

[@henkhu100]

@@hansisbrucker813 from what we see at 2.40 it follows that 2sqrt(2009a) = a-b-2009 and because a and b are integers the right hand side is an integer and so is 2sqrt(2009a) From my earlier answer you can then conclude that sqrt(2009a) has to be an integer as well. Because from 2sqrt(2009a) is an integer it followed that it has to be an even integer so sqrt(2009a) is an integer as well.

[@hansisbrucker813]

@@henkhu100 Ooooh I totally missed that.

[@bahbahbah8460]

2:31 why isn’t it “2 times the square root is an integer”? and as such the square root term is a multiple of half

[@ipudisciple]

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

[@verbumtech]

Way more efficient than my solution. I spent two notebook pages to figure out that a = 41*q^2

[@RealMadrid-lj6vm]

This same kind of problem came in PRMO india ( 1st round of 4 national rounds) in recent years

[@wise_math]

If a is not necessary integer then sqrt(2009*a) could be in the form X + 0.5, where X is positive integer, so that 2*sqrt(2009*a) is integer. But a is an integer so sqrt(2009*a) can't be of the form X + 0.5, so sqrt(2009*a) must be integer.

[@CosylandWhispers]

Came here to learn maths, left wanting that Dune shirt.

[@DarthGeddy]

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

[@stephenhousman6975]

Why do x and y have to be non negative? I thought only a and b need to be non negative. Based on how you defined a and b a negative x or y will still give you a non negative a or b respectfully. I think the reason should have been realizing that sqrt(a) and sqrt(b) will always be positive based on your definitions of a and b. So that would mean that x,y>7 would cause a,b>2009 which is not the case.

[@djvalentedochp]

If you take a look at the original problem, the maximum value of a and b are 2009. Since they are in the form of 41x^2, you may notice that x^2 must be less or equal to 49. You probably think, well, x can still be negative. Let's see it: if x = -7, then y = 14 but 14^2 > 49 If x = -1, then y = 8 and 8^2 is also > 49. So that's why x and y must be in [0, 7]

[@cantfindagoodchannelname7359]

@@djvalentedochp thank you, pls post this as a comment, I was breaking my head trying to figure this out

[@passerby6695]

@@djvalentedochp x and y can still be negative, the equation x + y = 7 is written assuming x and y are positive. The correct equation should be |x| + |y| = 7, which allows for negative x and y. So the main point is sqrt(x^2) = |x| and not just x.

[@andreybyl]

You can say “it follows that sqrt(2009a) is an integer” if only you have proof “sqrt(n) not rational, for any integers n when n is not perfect square” Without it u can only say “ 2sqrt(2009a) is an integer”

[@irockyou1337]

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?

[@andreybyl]

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

[@dclrk8331]

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.

[@phanibhushantholeti9446]

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?

[@irockyou1337]

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

[@somasahu1234]

Always thankful for the hints !

[@polevaulter2668]

Hi, I don't understand how it follows naturally from the equation that sqrt(2009a) should be an integer.. It can also be a decimal like 1.5 so that 2*sqrt(2009a) is an integer

[@shivanshsanoria4053]

That is because sqrt(2009a) can either be a natural number ( if 2009a is a perfect square) or an irrational number ( if 2009a is not a perfect square) but sqrt(2009a) can never be a terminating decimal like 1.5

[@polevaulter2668]

Shivansh sanoria, yes that makes sense. In the equation, it has to be a rational number. Thanks!

[@markmontgomery2171]

I'm late to the game, but this video just popped up as a suggestion. Why would sqrt(2009a) be an integer? Couldn't it be some integer + 0.5. I think you mean 2*sqrt(2009a) is an integer. Similarly, 7*sqrt(41a) rather than sqrt(41a).

[@jonp3674]

Yeah I had the same thought.

[@satyapalsingh4429]

Your method of solving the problem is interesting !!! I enjoyed a lot !

[@michaeleissenberg637]

6:10 i don't understand why you can cancel all square root of 41

[@xtal7632]

Distributive property

[@michaeleissenberg637]

@@xtal7632I still don’t get it sorry.. 😅

[@archanasawant3104]

@@michaeleissenberg637 factor √41 on LHS to get √41(√x²+√y²)=7√41 Divide both sides by √41 √x²+√y²=7 Cancel square and square root x+y=7

[@michaeleissenberg637]

@@archanasawant3104 ah thank you so much!

[@the_magisterate]

2:52-2:53, KING CRIMSON HAS ERASED 2 SECONDS

[@wailof1]

DUDE😭😂😂😂😂😂😂😂😂😂😂😂

[@ivansprogrammingworld4231]

the ad was timed perfectly with the snap

[@ganzir9697]

Can you explain the step at 4:05? I do not understand why sqrt(41a) being an integer implies that a is of the form 41x^2.

[@ganzir9697]

And how do you know that there needs to be a perfect square in there?

[@fangorntreebeard7346]

Around 4:20 min, I agree that sqrt(41*a) must be an integer, say x. Then x-sqr must be equal to 41*a as opposed to a = 41*x-sqr. What do you think?

[@mijmijrm]

can't we deduce that, because sqrt(2009) is the sum of 2 sqrts, then 2009 **must** have a square factor to allow the split into 2 sqrt parts. Therefore sqrt(a) + sqrt(b) = sqrt(2009) = p.sqrt(N) + q.sqrt(N) | where 2009 = N*(p+q)`2. So we look for the square factor of 2009 - and bob's your uncle. / finding that square factor could be a bit of a trek tho .. it's anywhere in the squares of odd nos 3..43

[@arthursafira3375]

Does the equation imply that square root of 2009*a is a HALF integer? Since it is immediately multiplied by two…

[@智障-k1s]

Why is the sqrt in fact a integer? Can someone explain?

[@kier-jd5ih]

me knowing that the thumbnail question is hard also me: Thats a good place to stop

[@denisluke7191]

2:40 How do you know, that a and b are integers?

[@AlecBenzer]

That's part of the problem statement.

[@funkychromosome6990]

Fundamental theorem of Arithmetic

[@arindamkashyap9420]

We are to find the solutions for a and b as non negative integers, so we took them as non negative integers

[@ayxanalifov2791]

Generating 41*49 = 2009 is very meaningless. Is there any way for competitors can calculate this by hand in competition? I think the only thing that makes the question complex is just knowing 41*49 = 2009 this equality. If I am wrong please correct me.

[@prabhudasmandal6429]

Understood the solution from your clear explanatIon .Thanks.

[@2mk4tom11]

The explanation of this problem is just ingenious.

[@2mk4tom11]

@jqbtube Why not?

[@manamritsingh969]

@@2mk4tom11 because some people are salty that they couldn't think this up themselves XD they don't appreciate the beauty of math rather concern themselves to emotionless problem solving alone

[@2mk4tom11]

@@manamritsingh969 Yeah exactly, in order to solve these higher level problems, you need to be more creative and think about all these things... it’s so amazing people can come up with this stuff and I’m able to witness it. Math is really something special, I agree.

[@howmathematicianscreatemat9226]

@@2mk4tom11 thanks man for this flowery words :) Yeah, man. The true geniuses are those who THINK UP those problems. This is often more poetic than solving it. A good logic mind can almost solve this but nowhere near could think up such problems, much less the proof problems.

[@kleinerkaiser7050]

so the answer is 41(n^2) w/ 0

[@hnnagarathna7286]

I tried it for around 10 mins and came back here I got near but it ended up in wierd variable 😭😭😭😭 he solved it in a better way

[@kagang8959]

I took some beer to the olympiad, but was told drinking and deriving was not allowed.

[@SasukeUchiha-ps4yn]

02:20 How did he conclude that it is an integer?

[@helloitsme7553]

3:02 it follow that -2sqrt{2009a} is integer. But yeah from that it follows sqrt{2009a} is integer because √n is either irrational or integer if n is integer is a thm

[@shawniscoolerthanyou]

You're not the guy that sings "Walter Reed", but I like these videos too.

[@adiramrakhani]

6:10 could someone explain how please?

[@olivercondliffe5993]

He cancels the root 41 from both sides so u basically just erase anything with root 41 from either side. This leaves root x squared and root y squared, the root of x squared is simply x, this is also the same for the y. This leaves x + y = 7

[@adiramrakhani]

@@olivercondliffe5993 ohh makes sense, thanks

[@laskobreeze4035]

I don't understand that fundamental theorem of arithmetic part with a = 41x^2 . How do we get "a ="?

[@blezki]

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).

[@antiinequality1907]

A man who loves to work. Go on like this, tighten your hands, and tell you well, my brother

[@allykid4720]

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

[@howmathematicianscreatemat9226]

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…

[@alvkarthik2018]

From 3:51 I couldn't understand your steps. Can anyone explain. Or suggest some books for improving my concepts.

[@84y87]

For root(41a) to be an integer, the term inside i.e “41a” needs to be a perfect square. And therefore, we can say from surety that “a” ought to be of the form 41x^2

[@AlecBenzer]

One nit I have with a lot of these videos is that they spend so much time laboriously talking through minor things like set notation and being super explicit, but then just kinda hand-wave over more interesting pieces of the proof like (3:47) sqrt(41a) integral => a = 41x^2 This is because any perfect square must have all even powers in its prime factorization, because for any n = p_1^(e_1)...p_k^(e_k), n^2 = p_1^(2e_1)...p_n^(2e_n). So in order for 41a to be a perfect square, a must contain at least one factor of 41 to make 41's exponent even, and then all the rest of a's prime exponents must be even as well (which implies that the rest of a's primes taken alone are themselves a perfect square).

[@tonyha8888]

Thanks for the lecture. Please do more Britsh Math Olympiad round 2 problems, Thanks in advance.

[@JB-ym4up]

It follows that x is a positive integer since a is. No actually negative values for x produce the same values for x² as positive values do.

[@Creativemathlearning]

You have many interesting lectures on solving math problems for students who are good at math. Thanks. You can add lectures on inequalities.

[@EqueenYT]

i dont know math language but all i know is you are extremely smart

[@QMJUN]

I'm Korean Student and I have a similar but easier for coming out my brain. because setting a as 41*x^2 is not thinkable way for most people. a = 2009 + b - 2*sqrt(2009*b) -> a should be integer, so 2009*b must be a square form of an integer k. 2009*b = k^2. 2009 is 7^2*41, in order to satisfy the condition, b should be 0, 41, 41*2^2, 41*3^2 , ... , 41*7^2 which is 2009.

[@benjaminparzy4835]

Congratulation for your valuable and useful work.

[@Kornsk]

Solution is straightforward. 41 being a prime, solves everything fast. Directly: a=41x^2, and similarly b. You don't need all that jargon.

[@rebelgames2546]

6:10 how can he cancel squares

[@maxjackson6616]

yay I managed to solve it, largely due to watching your videos!

[@dvi2948]

Why does it follow that 2 times square root of 2009a is an integer? Just because it's not? Doesn't make sense to me.

[@mayattv4986]

The first solution that I thought of was pythagorean theorem. 🤦‍♂️

[@siralanturing9103]

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

[@JustAzab]

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

[@howmathematicianscreatemat9226]

@@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…

[@siralanturing9103]

@@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?

[@zoroalmanji4311]

At 3:53 I didn't understand why he put a=41x^2. And I don't know what is the relation between that and the fundamental theorem of arithmetic. Can you or someone reply and answer my question plz??🥺

[@abcdefghabcdefgh350]

41 is prime, sqrt(41a) is an integer. Thus 41a can't be prime so it's composite, that implies a=41x^2 so that sqrt(41a) can be an integer (we need 41^2*x^2 in order for sqrt(41a) to be an integer)

[@finlayhutchinson7370]

He explains it in a way that makes it look easy

[@urnoob5528]

what about a=b=502.25? thats an answer too, wouldnt this mean you miss one more answer and thus wrong

[@darreljones8645]

Or, to do the multiplication and give the actual numbers, four of the solutions are a=0, b=2009; a=41, b=1476; a=164, b=1025; and a=369, b=496. The other four solutions are found by swapping the values of a and b in these four pairs.

[@ciberiada01]

You mean a = 369, b = 656

[@henri-leonlebesgue5471]

I clicked to verify my solutions and I was right. Thanks for making me feel smart ^^

[@jamirimaj6880]

why is x and y nonnegative? since they are both squared, should we be using the the negative sign as well? so like x= 8 and y=-1 is a possibility?

[@jonathanbaxter4611]

They are nonnegative because thats what the question stated

[@jamirimaj6880]

@@jonathanbaxter4611 I'm not talking about a and b, I'm talkimg about x and y.

[@muhammadsyafiq8882]

How did you get a=41x^2? That part confuses me

[@pedroteran5885]

Let N=41a. The square root of N is an integer, which is to say N is a perfect square. Since 41 is prime, it cannot be broken into the product of smaller numbers. For N to be a square, it must come from squaring something which already contains 41, that is, N is the square of [41 times something]. Letting x be that something, from N=41a and N=(41x)^2 we deduce a=41x^2.

[@tasosgiot7038]

PLEASE can someone explain why cant x, y be negative integers ? Even if x,y are negative, a and b are still non negative integers because they have a factor of x squared and y squared correspondingly E.g.: if x = -1 then a = 41 which is an non negative integer

[@Vitor-ph1oh]

Technically It Just doesn't matter because Will give the same solutions of the positive counterpart so..

[@eduardoteixeira869]

Thank you, but I came the solution simply factoring 2009 which given 7.√(41) and then it is not difficult to figure out that √(a)+√(b) should be of the following format x√(41)+y√(41) with x+y=7and you have the possibilities as described in your video. Thanks again

[@user-my7ki4it3s]

I solved it in exactly the same way 😁

[@ravirajamadan]

I solved it in very short and simple way : Sqrt A + Sqrt B = 7sqrt 41 => Sqrt A/41 + sqrt B/41 = 7. Since 7 is integer therefore a = 41k^2 and b=41m^2 => k + m = 7. Hence k,m = (0,7) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) and (7,0) which brings us to the final answer.

[@harshagali7269]

Another way of seeing this problem is that if you think about it, two square roots when added make one term only if they can be broken down and reduced to a form where the value inside the square root is the same in both terms. (a.k.a multiples of one variable) so you could reduce the given equation to y(c)^1/2+x(c)^1/2=(2009)^1/2 (where x and y are some variable and c is a factor of 2009 whose square root cannot be broken down or simplified) that reduces to (x^2+y^2)(c)=2009 and with the factors, it should be easy. Note I haven't actually double-checked (I am lazy) but I don't see anything wrong with this and if there is something wrong I would love to know about it.

[@mikebrau5354]

That's exactly what the video does.

[@vookhan]

Only thing I didn't understand is why x and y are nonnegative. If x=-1 is it like the same thing if x=1 so we will only have 7 solutions anyway?

[@aaykat6078]

I am already lost in second step, where did that -2/2009a came from?

[@_GOUTHAM]

sir please tell why can't we take a=2009*x*x,b=2009*y*y why prime factorization is required?

[@wyboo2019]

rewrite as: sqrt(a) = sqrt(2009) - sqrt(b) square both sides: a = 2009 + b - 2 sqrt(2009) sqrt(b) a = 2009 + b - 2 sqrt(2009 b) factor 2009: a = 2009 + b - 2 sqrt(7^2 * 41 b) in order for a to be an integer, the expression under the radical must be a perfect square. the expression under the radical is a perfect square when all the powers are even, so the full set of solutions for b is: b = 7^(2c) * 41^(2d+1) * 2^(2e) * 3^(2f) * 5^(2g) * ... where each of the bases are the prime numbers (7 and 41 do not appear again), and c, d, e, f, and etc. are any arbitrary nonnegative integers. the solutions are unique because prime factorizations are unique

[@ocloudx]

I am not sure if my method is complete I just set a=0 notice to have non zero solutions b must be in the form of 41x𠆢2

[@marcozhang2696]

Kinda confused, why does 2 root(2009a) have to be an integrr

[@TheLight2323]

Yeah this confused me as well. The root of any integer is either an integer itself or irrational rational * irrational = irrational

[@juanlorenzo315]

Because as b=2009+a-2*sqrt(2009a) and 2009, a and b are integers, then the last term must be an integer itself for the equation to prevail.

[@juanlorenzo315]

@@TheLight2323 but this is true, you can affirm that 2*sqrt(2009a) is an integer, but the sqrt can be irrational, not neccesarily an integer itself.

[@TheLight2323]

@@juanlorenzo315 huh?

[@hashirbashir]

​@@TheLight2323 it has to be integer so that when you subtract 2 root (2009a) from 2009 + a the right side becomes an integer as B is an integer.

[@jay-tbl]

I just found out that √2009 equals 7√41 and found all the combinations that make up that number(√41 + 6√41, 2√41 + 5√41, etc). And if √a is 2√41, then a is 4*41, and b is 25*41

[@subashkc7674]

Nice qsn wow love your way to solve qsn michel😊😊☺️

[@xl000]

Can you giv3 us your routine for upper body strength?

[@bishal5921]

Bro just do your maths on the blackboard, it enhances muscles as you are writing on an elevated surface unlike a copy. And hopefully you'll get guns and lats like him.

[@xl000]

@@bishal5921 so basically, he's lifting chalk ?

[@bishal5921]

@@xl000 exact-fuckin-ly dude

[@z.zsamad5603]

Thank you, I learnt something new today.

[@tatopolosp]

Not sure about this transition at 3:40 : 7*Sqrt(41*a) is an integer --> Sqrt(41*a) is an integer...

[@passerby6695]

It's because a is an integer.

[@martincooper2175]

How do we know sqrt(2009*a) is an integer? Considering we multiply this by 2 at the end, it could be equal to x.5, which would become an integer when multiplied by 2. Can anyone explain where I’m wrong?

[@MatheusAugustoGames]

"It follows that sqrt(2009 a) is an integer" Could it also be an integer plus a half? We're multiplying it by 2 on the equation

[@tanmayshukla5330]

Ya that's the same problem I noticed in this bro👍🏻👍🏻

[@tanmayshukla5330]

b = 2009 -2√(2009 a) + a So, a - b + 2009 = 2 × √(2009 a) So, we know that here L.H.S. is an integer, since a, b have to be an integer (in question:- non-negative integer). Now, dividing both the sides with 2 will give us √(2009 a) = ( a - b + 2009 ) / 2. So, this can be concluded that:- √(2009 a) is only an integer if ( a - b + 2009 ) is an even number.

[@melonenlord2723]

@@tanmayshukla5330 But only "a" has to be an integer, not √(2009 a). It could be something with an half if ( a - b + 2009 ) is an odd number. The point is, that "a" has to be an integer, so 2009a has to be one too. But the square root of an integer can't end with an .5, because for that the square of a .5 number has to be an integer. A .5 number squared will always end with .25, so it cant be integer.

[@blackholesun4942]

Understood and enjoyed this exercics

[@shawnkee8590]

I dont get how you can square both sides though.... I though you are only allowed to times for example square root b on both sides?

[@pianochannel100]

Why did they exclude negative solutions?

[@psioniC_MS]

After seeing √2009=7*√41 we know the sum √a + √b has to be of the form x*√41 + y*√41, since a and b are integers, we know x and y need to be integers too, and their sum x + y = 7.

[@psioniC_MS]

@Андрей Босой You can check that your answer does not work by either plugging in some values and using a calculator or by √(n²) + √(2009 - n)² = n + 2009 - n = 2009 != √2009.

[@amitagarwal2216]

I used all the hints and after prime factorization i got, 7√41, breaking it I got many possibilities, like a=3√41, b=4√41, adding them we get 7√41 which is equal to √2009