A Problem WolframAlpha Didn't Solve, But You Can (615 + x^2 = 2^y)

Рет қаралды 1,332,785

Күн бұрын

I didn't solve this problem myself, but I felt better when I learned WolframAlpha couldn't solve it either! But there is a way to solve it using careful mathematical reasoning. Thanks to Luka Khizambareli from Georgia for suggesting this and sending its solution! (*And you have to be really careful--I thank Aniket Gupta and Ryan Wilson for spotting a mistake in the first video I uploaded)
Update: As of November it seems WolframAlpha can solve it--I love to see how these tools just keep getting better!
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@MindYourDecisions 5 жыл бұрын

Apologies for the re-upload. In the previous version I only solved for positive integers. It is straightforward to then find the other solution. But I uploaded this corrected video presenting both because not everyone reads the comments. Also these videos are getting shared in a lot of places so I'm doing my best to present accurate mathematical results. I thank Aniket Gupta and Ryan Wilson for pointing out the oversight!

@PraneshPyaraShrestha 5 жыл бұрын

What is your email?

@MindYourDecisions 5 жыл бұрын

So where did I go wrong in the original video? I often get emails where people ask me to look for mistakes in their work, if they did not find the correct solution in one of my videos. One important skill in mathematics is reviewing your own work. It's not always fun, but it's really important to learn where you might make errors. So where did I go wrong? In my self-review, there are three places I could have realized my mistake. Place #1: I said the difference of factors is (2^n + x) - (2^n - x) = 2x. Then I solved 2x = 118 to get x = 59. The thing is the difference of factors can also be (2^n - x) - (2^n + x) = -2x. Setting that equal to 118 would have given x = -59. Place #2: When I solved y = 12, I could have then solved 615 + x^2 = 2^12 where I could have seen x = 59 or -59. Place #3: In checking my work, I should have seen there was an x^2 term, and it's a common mistake to overlook a negative solution. This is the kind of self-criticism I do for my math homework. I have noticed people don't like to admit if they are wrong. But I have found self-reviews and honesty are big strengths to improving math skills. So I thank all the legitimate criticism (if accompanied with legitimate honesty) for these videos--you're helping us all get better at math.

@PraneshPyaraShrestha 5 жыл бұрын

@@MindYourDecisions your email please. I have an important question that no one in Nepal has solved.

@Lqtan16 5 жыл бұрын

Pranesh Pyara Shrestha his email is at every end of his video

@PraneshPyaraShrestha 5 жыл бұрын

@@Lqtan16 oh I found it

@Rimtay 4 жыл бұрын

Can you imagine solving all the question and forgetting to include -59 to the equation get 0 because of it.

@sxz452 4 жыл бұрын

That's sooooo sad

@Goku_is_my_idol 4 жыл бұрын

Lol i did same😂

@namanlakhotia6393 3 жыл бұрын

@@Vegan_PhysicsEnthusiast u definitely dont have these questions

@funni111 3 жыл бұрын

@@namanlakhotia6393 if you do prmo training you will definitely get this type of questions

@namanlakhotia6393 3 жыл бұрын

@@funni111 prmo is not comp in schools though

@roshandon3157 4 жыл бұрын

*99% of the people who clicked the video underestimated this problem*

@radekvecerka1115 4 жыл бұрын

@BlazePlayz YT of course you can just blindly type in numbers, that´s easy, but what if the answer was smth like y=22 i think you would have some troubles then

@rishijai 4 жыл бұрын

@BlazePlayz YT Yep the key is to spot 3481 as a square

@Martin-zr2tb 4 жыл бұрын

Rishi Jai is there a certain trick to spotting a high number with a square root? Cus I did the blindly squaring numbers on my calculator around the sums + 615 being a power of 2 thing...

@nonidhgupta9405 4 жыл бұрын

I did but I solved it too

@ReenaKumari-vq2rk 4 жыл бұрын

@KZfaq Watcher i also did the same😁

@arnavraut9691 5 жыл бұрын

literally a week after every video criticizing wolfram alpha's capabilities is posted, they end up fixing it

@ashyourresidentenby5916 5 жыл бұрын

www.wolframalpha.com/input/?i=solve+615+%2B+x%5E2+%3D+2%5Ey+over+the+integers

@tgdhsuk3589 5 жыл бұрын

@@ashyourresidentenby5916 solve 604 + x^3 = 3^y over the integers lmao still cant solve these types of questions

@ashyourresidentenby5916 5 жыл бұрын

@@tgdhsuk3589 x is 5 and y is 6 HA solved it in like 15 seconds

@pdpgrgn 5 жыл бұрын

@@tgdhsuk3589 WolframAlpha solves this

@quantumsoul3495 4 жыл бұрын

Ethical hacking..

@antilogis6204 4 жыл бұрын

7:27 "All we did was use of principles of number theories." Right, very simple...

@lach993 3 жыл бұрын

WYSI

@rcnayak_58 5 жыл бұрын

It is easier to solve this problem with little logical way than to follow Presh`s complicated method. Look at the problem once. 2^y is always an even number irrespective of whether y is even or odd. Now we can write the equation as 615 =2^y - x^2. Here 615 is an odd number. Therefore, x^2 will have to be an odd number. Because, the difference between two even numbers is always even and the difference between an even number and an odd number is always odd. Since x^2 is an odd number, x must be an odd number too as the square of an odd number is always an odd number and the vice-versa is also true. Let x = 2n+1 for any value of n, odd or even. Now we have 2^y - (2n+1)^2 = 615. This can be written as (2^(y/2))^2 - (2n+1)^2 = 615. This now becomes the difference between two squares form. That means the left hand side will be the product of [(2^(y/2)) + (2n+1)][2^(y/2) - (2n+1)]. Putting, 2^(y/2) = a, and 2n+1 = b for simplification, we have (a+b)(a-b) = 615. Now 615 has to be the product of two numbers. The number 615 = 1 x 3 x 5 x 41. Therefore, we have four alternative pair of numbers whose product will be 615. They are (i) a+b = 615 and a - b = 1 or (ii) a + b = 205 and a - b = 3 or (iii) a + b =123 and a - b = 5 or (iv) a + b = 41 and a - b = 15. Solving (i) we have a = 308, b = 307 (ii) a =104, b = 101 (iii) a = 64, b = 59 and (iv) a = 28, b = 13. Here note that 'a' = 2^(y/2) that is a number power of 2. If we check the values of 'a' in all thee alternatives, only (iii) which is 64,satisfies our condition and the others are not.Therefore 2^(y/2) = 64 = 2^6. That is y/2 =6, That is y = 12 and b = 2n+1 = x =59.

@xmarteo 5 жыл бұрын

You assumed a to be an integer without proving that y is even.

@rcnayak_58 5 жыл бұрын

@@xmarteo I have simplified it in a better way in my next explanation. Pl look at it again.

@raphaelmillion 5 жыл бұрын

@@rcnayak_58 you don't even have to set b = 2n+1 because you don't use it again. Also, xmarteo's comment raises a valid concern, because then you would have to check for more divisors (divisors on Z[sqrt(2)] ). Or you could show that y is even, resulting in the same proof as MindYourDecisions. Ultimately, your proof is not shorter than MindYourDecisions'.

@ChristopherNight 5 жыл бұрын

@@rcnayak_58 Try your logic with 23 + x^2 = 2^y. You'll get that (a+b)(a-b) = 23, which only has one integer solution: (a, b) = (12, 11), and since 12 is not a power of 2, you would conclude that there's no solution with integer x and y. But (x, y) = (3, 5) is a solution, corresponding to (a, b) = (4sqrt(2), 3).

@TejvirJogani 5 жыл бұрын

@@xmarteo 2^y is always even because it's a multiple of 2 and an even number is defined as multiple to, so by definition 2^y is even. The problem though, is that this solution is almost as complex as Presh's, because it practically requires the same methodology of thinking. While I agree it might be efficient, you've to keep in mind that he was send this solution as well.

@AYUSHGAMEROFFICIAL 4 жыл бұрын

He:this is presh talwalkar Captions:this is fresh lakewater

@shambosaha9727 4 жыл бұрын

pressure walker

@enavoid 4 жыл бұрын

dash fallwalker

@ahmadm7618 3 жыл бұрын

it says Presh TalWalker here

@adityarajsrivastava6580 3 жыл бұрын

It says pressure locker

@GigaPlaya 3 жыл бұрын

Not Luke Skywalker.

@notwildcard377 3 жыл бұрын

UPDATE : Wolfram Alpha can now solve this problem. I loved your approach.

@lukkash 5 жыл бұрын

You can solve it beautifully in Excel by solving (with parameters) an equation as 615+x^2 - 2^y = 0 Alternatively you can calculate y=log(615+x^2)/log(2) and then an integer value of y should be found. All to be solved in Excel :)

@btf_flotsam478 Жыл бұрын

I'd give you a cookie, but you only really deserve it if you use SQL. (By the way, good luck brute-forcing the proof that there are no other solutions.)

@ethanfrommer7772 5 жыл бұрын

Hey presh. Just wanted to thank you... we had a really difficult math test that was more about thinking than calculating. It was really hard but everytime i started a new question, I thought about what Presh will do and how he will solve that, and that helped solving most of the problems. Thanks for improving and sharpening the thinking for me, and for other hundreds of thousands people

@HueHanaejistla 4 жыл бұрын

ethan frommer ok boomer

@Seekingpatience19 3 жыл бұрын

@@sahaj9810 wait how did u get the 2^10 , 2^11, and 2^12 are the only possibilities? im a little confused.

@suryakalawasnik6204 3 жыл бұрын

@@sahaj9810 he wanted to say what about 2¹³ .. 2¹⁴..2^15 and so on

@Super_Smash_Dude 5 жыл бұрын

That's interesting! I tried taking the log base 2 on both sides in order to isolate y.

@bancodrut 5 жыл бұрын

The solution to this problem really blowed my mind. The structure of it is the most imaginative and trickery of what I've encountered so far 😱 Thanks for sharing

@jerrygundecker743 5 жыл бұрын

I was with you right up to "You're watching Mind Your Decisions"....After that I said, "Huh?"

@SameerKhan-fd2qe 5 жыл бұрын

Fresh tall walker

@MathIguess 5 жыл бұрын

@@SameerKhan-fd2qe Fresh Tail Wagger

@harshitagoyal7134 4 жыл бұрын

After spotting that n > 9, I just used trial and error with n = 10, 11... to check which values would give a perfect square for 2^n - 615. Eventhough, trial and error isnt the 'best' way to solve problems, I think this would be a quicker approach :) THOUGH I ABSOLUTELY LOVED THE WAY YOU DID IT, IT IS ALWAYS FUN TO KNOW HOW THE OTHER PERSON SOLVED IT. I LOVE MULTIPLE APPROACHES

@l.w.paradis2108 2 жыл бұрын

This is so cool. I've just been casually watching, and got up to the 3-minute mark on my own. I thought of the first steps without watching. So, these videos are putting me on the right path to figuring out cool problems. THANK YOU.

@knk0112 5 жыл бұрын

Thanks Presh for making such videos.. observing the way these solutions work out their way to the answer really helps in thinking laterally.. it sharpens the thinking skills for sure.. thanks a lot!!

@jasonterry1959 5 жыл бұрын

@mindyourdecisions I took a game theory class in college this past semester (spring 2018) and you would not believe the amount of times you were listed as a source.

@TheGeneralThings 5 жыл бұрын

Brilliant! This is the content I can't get enough of.

@srimatresri 5 жыл бұрын

2^y - 615 must be a square. Went on substituting y > 9

@bharatnotindia.6296 5 жыл бұрын

That's cool.

@oskarjung6738 4 жыл бұрын

One more analysis, that 2^y is always even so sum of of 615 and x^2 should also be even, which is possible when x^2 is odd. x^2 is odd when x is odd integer. Hence odd integral solutions of X is only possible.

@critisizerr245 4 жыл бұрын

I also wwnt on substituting y value but soon after I though that this will not help me for IIT JEE I am in 10 Going go attempt jee in 2022

@critisizerr245 4 жыл бұрын

But i got the answer by putting 12

@SuperLabelPerson 4 жыл бұрын

I did the same thing. Once he started looking at digit repetition, I was like "y tho?"

@curiousscientist6077 4 жыл бұрын

This channel is very amazing! I loved it! Knowing this fact about Wolfram is very motivating!

@franklyanogre00000 5 жыл бұрын

x=59, y=12, I did trial and error of adding odd squares to 615 and comparing it to powers of two, only took three tries. Surprised you didn't get it at first, Presh.

@mohdhussain46 5 жыл бұрын

Frank Anthony Overton Jr. I did the same 😂😂

@akankshasarda5216 5 жыл бұрын

Me too😂

@saurabhusane5780 5 жыл бұрын

Exactly ...!

@mr.moodle8836 5 жыл бұрын

He could've used trial and error, sure, but that would've made for a pretty boring video. Trial and error is how I "solved" it, but I wasn't able to guarantee that it was the ONLY answer, and saying you've solved an equation without ensuring that it's the only possible answer doesn't really count as "solving" it.

@bidyutchakraborty2433 5 жыл бұрын

Getting odd square of 3481 or to get number 59 and its square and then matching with power of 2, in three trials, I think, is a chance. There are much more odd integers in between. However, proceeding logically seem to be more elegant solution to me.

@GinoTC 5 жыл бұрын

Holy crap the captions got your name right. Well, there's a space between tal and walker but damn, I'm glad I've been checking every video haha

@tamirerez2547 3 жыл бұрын

The approach of checking the last digits, as all simple thinking, is genius!! Great way to solve such a problem.

@sjoerdwiesmeijer7231 5 жыл бұрын

Am I the only one who always gets depressed when he says:'Did you figure it out'? . 'cause I am already happy when I can follow his explanation.

@mirawenya 3 жыл бұрын

Yep, I’m with you. Don’t usually even understand the explanation it seems.

@ihopeicanchangethisnamelat7108 3 жыл бұрын

the first thing i did when i saw this problem was google ‘what does solve over the integers mean?’. i promise you’re not alone.

@willnewman9783 5 жыл бұрын

In this video, he looks mod 10 to get that y is even. But it is much easier to look mod 4 to get that fact. And also, a lot of people are saying they solved this by guessing and checking. While that does find a solution, it does not find a proof of the only solutions, which this video does.

@sapien153 5 жыл бұрын

Nope. The video didn't prove that this is the only solution

@freddiehand6551 5 жыл бұрын

@Bragadeesh S yes it did

@sapien153 5 жыл бұрын

My bad. Agreed

@Theo0x89 5 жыл бұрын

I'll see your mod 4 and raise you mod 3. In fact, modulo 3 you get 615 ≡ 0, x² ≡ 0 or 1 and 2^y ≡ (-1)^y, so y must be even.

@danmerget 5 жыл бұрын

Neat. I used the mod 10 solution myself, which seemed more obvious to me since I count in base 10 and tend to notice base-10 patterns before considering other bases. (OK, I lie. I did briefly look at mod 2 first, but that only got me as far as "x must be odd" before I abandoned that approach and jumped to base 10. Mod 3 and 4 never occurred to me.)

@Pepa14pig 3 жыл бұрын

I feel proud for solving this problem not only alone but also with different thinking :))

@kwcy92 3 жыл бұрын

Can you share your way?

@Pepa14pig 3 жыл бұрын

@@kwcy92 Of course! It was a guessing game with math background 😂😂😂 We can rewrite the problem as 615=2^y-x^2 I know x will be an odd number (because 615 is odd and 2^y is always even). Because we know powers of two, we want a “bigger” number for y because difference needs to be 615 From there on I played a bit with numbers to see what happens when I use bigger x, smaller y... I started with x=y=9 and moved up by 5-10 for x and one for y and got there relatively fast. I know it’s not the best way to solve, but it worked 😂😂😂

@tatane79 3 жыл бұрын

This is really a brilliant, simple and elegant affordable solution! Thanks!

@UltimatumNo5 5 жыл бұрын

I must say, your maths videos are brilliant and I love your simple solutions even if I sometimes don't understand them or feel intimidated by the many who not only understand but reach the same conclusion before the actual solving takes place. Could you do a video on how to approach a problem mathematically - it's one thing to be aware of mathematical rules, but is there a way to break down a problem so those rules become apparent?

@Acid31337 5 жыл бұрын

It's easy, wolframalpha cannot make a sandwitch, but I can!

@ffggddss 5 жыл бұрын

What, make a witch out of sand? Well, it is nearly Halloween, after all. So I guess all you need is to go to the beach!! Fred

@yerr234 5 жыл бұрын

but can you tho?

@silkyfirst3097 4 жыл бұрын

Dont provoke them wolfram alpha is listening

@TAT4guitar 4 жыл бұрын

Did you try "sudo make a sandwich"? That usually does the trick

@quahntasy 5 жыл бұрын

I always depend on Wolfram Alpha for solving Equations in Physics. But this new information blows my mind. Now I have to go back and do all those tedious exercises from Goldstein.

@folf 5 жыл бұрын

It works now 🧐

@spacefertilizer 5 жыл бұрын

For mathematics, only use wolfram for getting hints toward a solution. Often wolfram does not give you all possible solutions or even managed to find them (this has happened to me a lot in calculus and even combinatorics). Try to understand the problem and the theory behind it and solve it from there. There are lots of mathematics that are not implemented yet in wolfram either.

@kienha9036 5 жыл бұрын

615 is divisible by 3, hence 2^y-x^2 is divisible by 3. Easily checking out we could see that 2-x^2 is not divisible by 3, then 2^y-0 or 2^y-1 is divisible by 3. Since 2^y-0 isn't, we can conclude that 2^y-1 is, then y is even

@ozanoruc3796 5 жыл бұрын

If you look at the equation in mod 3 it is clear that y is even which means that you can move x^2 to right side of the equation and do difference of two squares. And then try for positive divisors of 615

@TamaraTkacova 4 жыл бұрын

I‘ve been taking some number theory classes in my free time and we‘ve just started with modular arithmetics, and I was so happy when I got the solution before watching the video haha :) I did it mod 3 tho

@TechToppers 3 жыл бұрын

Same here :D

@indy197905 3 жыл бұрын

I thought it was an algebra problem. Then I kept watching and my brain melted.

@kalliboymusic 3 жыл бұрын

Same

@user-ng4sb5nl2o 5 жыл бұрын

When I first started to solve it, i gave up. But then I decided to try it again and got it, using (x-1)^2 = x^2 - (2x-1)

@sarthakgupta4323 5 жыл бұрын

To chekc whether y is even or odd we can simply use 3 as x^2 gives remainder 0 or 1 when diveded by 3 and 2^y gives remainder 2 when y is odd and 1 when is even therefore y must be even

@nonesuch27 5 жыл бұрын

As always with these videos, a lot of people don't grasp the concept of PROOF.

@arolimarcellinus8541 5 жыл бұрын

Well how do you say? There is no any proper explanation about proof. It's so abstract and we never realized we need to use inequalities for this kind of problem. You know in HS we always use regular method, but not with this long and tiring explanation only to find y=2n. How come common people realized that? Me as a Master degree student even cannot realize this

@heronimousbrapson863 5 жыл бұрын

nonesuch27 Most people don't grasp the concept of proof. Proof is really only needed in the study of pure mathematics. For everyday applications of mathematics, you don't need proofs, just like you don't need a degree in automotive engineering to be an auto mechanic.

@qc1okay 5 жыл бұрын

What? What does proof have to do with anything? The problem as worded says nothing about proof. It asks you to find x and y, only allowing integers. Period.

@dariobarisic3502 5 жыл бұрын

@@qc1okay If it asks to SOLVE the problem then it means you have to find ALL the solutions. If you find only ONE solution, you have to prove its the only solution. Otherwise you didn't solve the problem. Many people just say "easy, i guesed it in 3 tries", while that worked in this case because there's only one solution, it won't always work.

@qc1okay 5 жыл бұрын

@@dariobarisic3502 No it doesn't. Just that simple. Solve does not mean all. This KZfaq channel isn't a formal math class, and even if it were, its instructions must be clear to its viewers. If "all" were wanted, "all" had to be stated. Incidentally, I went thru the full process Presh used, and it both leaves out steps and overcomplicates others.

@SamerShennar 5 жыл бұрын

Way simpler: x=sqrt(2^y-615) So 2^y has to be > 615 By trial and error starting from 1024 we can quickly reach x an integer (59) at 4096.

@t_kon 5 жыл бұрын

Ye prove it that that's the only answer

@tomashertl8895 9 күн бұрын

So beautiful, so nice and so clear! Thanks Presh.

@johnconway8070 5 жыл бұрын

One of the most enjoyable problems I have seen for quite a while . Finding the solution as explained by Presh is not easy but it *is* very satisfying . Me, I simply used the trial and error method! Noting that 2^y had the be *at least* 2^10, I only had to try 1024 minus 615 , 2048 minus 615 and 4096 minus 615 to find which of these three differences gives a perfect square . Not a whole lot of work !

@johnm6011 Жыл бұрын

Me too.

@michaelfredericks6970 5 жыл бұрын

I knew x had to be odd because odd^2=odd and odd +odd = even and that x^2 had to end in a 1,3, or 7 because these digits when added to 5 produce unit digits that correspond to unit digits produced by 2^y (I.e 2,4,8,6,2,4,... cycling). So basically that narrows the last digit in x to be only 1 or 9, both of which when squared and added to 5 make a units digit of 6. Thus because 6 occurs in the units digit of 2^y then y is a multiple of 4. So I started with 2^12 as an initial guess (2^8

@whiploadchannel2047 5 жыл бұрын

You can also find that y is even using mod 3. A little faster than mod 10

@whiploadchannel2047 5 жыл бұрын

Tom Domenge sure, (mod 3) 615 = 0, x^2 = 1 or 0, and 2^y = (-1)^y. So 615 + x^2 = 0 or 1. If y is odd, 2^y=(-1)^y = -1 which is impossible. So y is even

@sandrastrilakos9962 5 жыл бұрын

I solved it partly using logic and finished it using trial and error, but your method is so much better because it demonstrates the logic required to prove it is the only solution. I look forward to sharing this with my senior students.

@spacefertilizer 5 жыл бұрын

Great and imaginative solution! Lots of people commenting on these problems don’t understand mathematics though. They believe one only has to do trial and error to find one solution, but forget all about finding every possible solution while also showing that there can be no other. This is the core thinking that should be taught. Wonder what they actually teach people in school nowadays.

@thephysicistcuber175 5 жыл бұрын

3:52 it was even easier to establish this by looking at the equation mod 3

@TechToppers 3 жыл бұрын

That's bit advances. Presh Talwakar makes things intuitive.

@d4slaimless 3 жыл бұрын

@@TechToppers not realy that hard. Every square is 0 or 1 mod 3, every even power of 2 is 1 mod 3, but every odd power of 2 is 2 mod 3. So the y is even. You would need to prove the initial statements though (but it's very simple).

@TechToppers 3 жыл бұрын

@@d4slaimless I know it is simple. But if you can avoid technicalities, then why not!

@d4slaimless 3 жыл бұрын

@@TechToppers I am not arguing that the solution in the video is very demonstrative. Just saying that going for the reminder of a division is not that hard either.

@ericzhu6620 3 жыл бұрын

wow i found a cuber under a maths video!

@anyone6240 5 жыл бұрын

Great video as usual! Write as x^2 - 25 = 2^y - 640 Then (x+5)(x-5) = 2^6(2^(y-6)-10) Difference of factors on the left is 10 which implies y=12 and so x= +- (2^6 - 5) = +- 59

@xXJ4FARGAMERXx 2 жыл бұрын

Finally! A solution I can understand!

@yurenchu Жыл бұрын

It doesn't necessarily imply y=12 . There may also exist a different integer value of y and some integers a>1 , b>1 , and k>0 such that (2^(y-6) - 10) = a*b and (x+5) = (2^(k-6))*a (x-5) = (2^k)*b For example, consider the equation (x+3)(x-3) = 2² * (2^(y-2) - 6) The difference of factors on the left is 6 , which according to your logic implies that y=4 . However, there are actually two solutions: y = 4 : (1+3)(1-3) = 2² * (2^(4-2) - 6) = -8 y = 6 : (7+3)(7-3) = 2² * (2^(6-2) - 6) = 40 (It's even possible to construct examples with more than two solutions.)

@musicsubicandcebu1774 3 жыл бұрын

Yeah, that was amazing. Thanks for taking the time to post these.

@legendhero-eu1lc 5 жыл бұрын

Thank you for the video! All of you friends are super awesome!

@anubis_05 5 жыл бұрын

hello please look into integrals by the Feynman technique

@UniqueNCS 4 жыл бұрын

Literaly just had this in a test yesterday and couldnt answer, cmon universe

@ritusaxena2714 4 жыл бұрын

Not possible

@montsaintleondr7491 3 жыл бұрын

Holy cow, I am spaghettified by this elegant solution! Thank you!

@ralphmacchiato3761 4 жыл бұрын

Love the techniques. Got me to get a condensed highschool mathbook from the library so I can brush up the forgotten gold.

@nellvincervantes3223 5 жыл бұрын

Explain the Difference between Quasi-Static and Non Quasi-Static @Mindyourdecisions

@usualunusualkid7149 3 жыл бұрын

1) This isn't a physics channel 2) @s don't work anymore

@mrKreuzfeld 5 жыл бұрын

That was awesome!

@xaxuser5033 5 жыл бұрын

We can observe that x is odd to minimize the cases from the beginning

@TorBruheim 5 жыл бұрын

Who on earth came up with this awesome solution? This is more a way of thinking and a method instead of traditional math. I have learned a lot from your videos, and I have expanded the way of seeing math solutions. Especially those solutions involving geometry. Thank you very much.

@SpaghettiToaster 4 жыл бұрын

It is traditional math, a branch called number theory.

@user-uc1hf8zm6k 5 жыл бұрын

2^y , which means it could only be2,4,8,16,32... Therefore,just to find out which 2^y number can be square rooted after minus 615.There is 615, so we can start with 1024,2048,4096... then find out that the answer is x=59,y=12

@user-zb8tq5pr4x 4 жыл бұрын

Thats a numeric solution, not analitic. Prove that that is the only solution.

@madhavstalks3925 5 жыл бұрын

The unit digit method always work. Good solution..

@steveholman5978 5 жыл бұрын

I solved this in about 3 minutes by simply listing the powers of two, subtracting 615, and looking for an integer square root. The first one even possible had to be more than 615, so I only had to check three powers of two before finding one with an integer square root.

@elmatadordeangel5004 3 жыл бұрын

Interesting solution. For my part I took x^2 to the right side, applied the difference of squares formula, since we need integers then I looked for multiples of 615: 203, 5, 123, 41, and then applied each solution and checked if the outcome would be an integer. Got the same results

@PraneshPyaraShrestha 5 жыл бұрын

Your videos are awesome

@PraneshPyaraShrestha 5 жыл бұрын

@@iotaop9573 always always

@hsman6614 5 жыл бұрын

Why did WolframAlfha give up when it reach x = 3?

@martinpiekarski1512 3 жыл бұрын

AI won't take over the world so soon.

@user-bf2bp2ww9m 5 жыл бұрын

Pls tell me how to draw picture for the following x^2+(y-x^2/3)^2=1

@beautyofnature7361 5 жыл бұрын

Very good Sir, I like your every problem solving videos

@prerakcontractor6609 5 жыл бұрын

Be honest and confess; how many of you guys checked whether wolfram alpha could solve this or not? Well, I did

@Packerfan130 5 жыл бұрын

and can it?

@prerakcontractor6609 5 жыл бұрын

Nope

@dmitryronin6898 5 жыл бұрын

it actually gives you integer solutions if you type just the equation in it.

@KelfranGt 5 жыл бұрын

I don’t see why not, and it was able to

@academyofmathematicalscien7986 3 жыл бұрын

kzfaq.info/get/bejne/m6uphNya3bWroIk.html

@vladpetre5674 5 жыл бұрын

@Presh, had you constructed the proof the other way around, it would have been a lot closer to how people/ students think. Start with observing x^2 and that if y is even then you can do a^2 - b^2 = 615 which is easy to solve. Then prove that y cannot be odd, thus the only solutions are the ones where y is even (x = +/- 59, y = 12). The way you started it by observing y has to be even and then stating ("This will be a key in our finding the solution") makes you look insightful, but does not help people develop thinking patterns in math problem solving (especially Diophantine equations)

@akaRicoSanchez 5 жыл бұрын

Well... I solved the problem the exact same way and in the exact same order. It's not really about being insightful but about using the fact integers are a strong constraint on the solutions.

@vladpetre5674 5 жыл бұрын

@Dan It's the same with last digit analysis. How do you know it is we need in this specific question? You don't know, but i bet factoring leads to results more often than last digit analysis.

@merveilmeok2416 5 жыл бұрын

Sometimes you come across a map problem and you think “I get it” and you “solve” the problem under a couple of minutes only to realize later the problem was above your “pay grade” and a much more complicated problem that you anticipated. This has happened to me back to back today ;)

@jimvinci8295 3 жыл бұрын

I solved this directly by the substitution 2 to the yth power = a squared, which gets to the answer more quickly, although the remainder of the approach is similar to what was presented. The above approach leads immediately to (a-x)(a+x) = 615 and by substituting the four ways to factor 615 (41x 15, 123 x 5, 205 x 3, and 615 x 1), the only pair of factors that produce integer solutions is 5 and 123. a - x = 5 and a + x = 123 leads to a = 64 and x = 59, and since a squared = 2 to the yth power, 2 to the yth power must equal 4096, so y = 12.

@kasperjoonatan6014 5 жыл бұрын

This was a very fine problem! It was good that 615 is quite easy to divide into prime factors, otherwise it wouldn't have been so easy without a calculator :)

@Akash-qy1gf 5 жыл бұрын

I did trial and error method and solved it in 2mins.....2^9=512 & 2^10=1024, 2^y has toh be greater than 615, so consider 2^10 =1024, so 1024-615=409 which is not a Perfect Square, 2^11=2048, so 2048-615=1433 which is also not a Perfect Square, and then 2^12=4096, so 4096-615=3481 which is Square of 59. Thus x=59 and y=12

@pjalvarez6420 5 жыл бұрын

Rahul Chhabaria sameee hahaha i tried 1024,2048, then finally 4096 and it worked!!

@samsonpl1110 5 жыл бұрын

-59 is second matching X :D

@ceegers 5 жыл бұрын

This is what I did. No need to make it more complicated than it is :P

@Cohnan13 5 жыл бұрын

@@ceegers Actually, it is needed to prove that no other solutions exist

@speedsterh 4 жыл бұрын

@@Cohnan13 Some people here don't what "solving" means

@italixgaming915 3 жыл бұрын

We can eliminate all values of y

@teambellavsteamalice 2 жыл бұрын

Nice problem! I didn't think of looking at end digits so didn't find y was even. Clever find! Btw, Wolfram alpha can solve with substituting y=2n: solve 615+x^2=2^(2n) over the integers

@Tailspin80 3 жыл бұрын

I solved this in about 60 seconds from the thumbnail. The right hand side has to be 1024 ... 2048 ... 4096 etc. So just keep trying until you find one which has an integer square root. Bingo - square root of 4096 - 615 is 59 (and -59). A familiarity with well known binary numbers helps. Because I solved it before I opened the video the no calculator rule didn’t apply. 🙂

@danielreese6185 3 жыл бұрын

Cool bean

@mcovaleda 5 жыл бұрын

Wolframalpha: isolate y on 615+x^2=2^y solves the problem

@EduardvanKleef 2 жыл бұрын

This solution is so brilliant, I'm stunned...

@invincibleflesh4526 3 жыл бұрын

Came for the problem and found the solution on my own, actually watched the video to see if there were others.

@anuragmishra3282 3 жыл бұрын

I solved the equation in seconds using a chart containing square of numbers upto 100 It was easier than the method in the video

@xVitOSx 5 жыл бұрын

Thanks macOS for "Grapher"

@kanchanaramar 3 жыл бұрын

It was clear that x^2 should be odd since when added to 615 it should result in a power of 2. The smallest feasible value of y is 10, but this results in a non-integer solution. So does y=11. y=12 gives an integer solution.

@rudra-thandavam 3 жыл бұрын

Y has to be even. There is no other way to solve the problem. Only if this ruled out, then we can check for Y as odd. But most cases such problems will come out easily when Y is considered as even. But good insights from you to check the combination of factors. I admit that I learned something new today. Thanks.

@ronmadan8003 5 жыл бұрын

Who else used a different strategy, but found a much faster way to do it? My strategy was to try different powers of 2 (greater than 9). For example: 615 + x^2 = 2^11 (2048). This implies that x^2 = 1433. That isn't a square value (for a positive even integer), so I tried 615 + x^2 = 2^12 (4096). That meant x^2 was equal to 3481. Since 3481 is a square, I was able to solve for x and y.

@chaitanyasingh1565 5 жыл бұрын

Wish you could see my raised hand ✋ ✋

@thiantromp6607 4 жыл бұрын

Ron Madan but you never proved or showed that was the only solution. That is the challenge with these problems.

@zoetropo1 4 жыл бұрын

Try or not try, that is not the question. Use the force of logic, Luke!

@Oswald_Anthony 4 жыл бұрын

Take the logarithm base 2 of both sides: Answer: y = log(x^2 + 615)/log(2) + ((2 i) π n)/log(2) for n element Z x = ± 59, y = 12 (that's a Wolfram's answer)

@kookoo275 3 жыл бұрын

This is the first thing I thought, surprised I had to go so low to find this comment

@TitusObbayi 3 жыл бұрын

I have to admit, when I first saw this problem I completely did not think it would be very complicated. Then I tried it and got stuck but I was still convinced there some basic algebra trick that I'm missing. Then after watching this solution, I realize there was no chance I was solving this without the explanation

@KyriZee 3 жыл бұрын

We can still optimise the solution. We can deduce from the original equation that x has to be odd, hence, at the stage where we investigate the last digits of x, we would need to do half the work.

@Exachad 5 жыл бұрын

Wolfram Alpha can solve it, X= +/-59 and Y=12 for integer solutions!

@t_kon 5 жыл бұрын

Dont do trial and error it's bad. First apply mod 3. 615 mod 3 = 0, and 2 mod 3 = -1, however x^2 mod 3 = 1 for x relatively prime than 3. Hence y must be even. Apply y = 2k and factorize it (2^k +x)(2^k-x) = 615. This is how you prove the only integer solution is (59, 12), (-59, 12)

@snuffeldjuret 5 жыл бұрын

trial and error isn't bad, it is an extremely valuable tool in figuring things out. When applying math on the real world, trial and error is crucial.

@t_kon 5 жыл бұрын

@@snuffeldjuret not really. Trial and error is useful in trying to solve any problems, but is never good to apply it directly. Why? Because you dont know if this trial and error pattern will continue on or not. There is not enough conclusion from just trial and error. It can gives you some valuable information but is never the way to go directly.

@snuffeldjuret 5 жыл бұрын

@@t_kon I have no idea what you mean with "apply it directly" and "the way to go directly". I am merely pointing out that "Dont do trial and error it's bad." is a blanket statement that is not always true. You can argue that it is for this task, but it is not true for all possible tasks.

@albertmcchan 5 жыл бұрын

Is doing mod 3 also trial and error ?

@SaiKiran-fd3gq 5 жыл бұрын

Take mod 16 on both sides we end with x = +3 or -3 mod 16 .But 59 is neither .Where am i going wrong.

@jeanphillippes2196 4 жыл бұрын

In C++ or equivalent you'd just have to cycle upwards, one step at a time thru powers of 2 greater than 616, subtracting 615 from that power of 2. Just take the root of result and see if it's integer.

@hikaru-live 4 жыл бұрын

When I see a "solve over integer with 2^y" I default to "solve this use binary representation" since 2^y in binary is s lone one followed by y zeroes. This makes coding up a solver much simpler

@leo17921 5 жыл бұрын

7:36 so now you say this every time. ok.

@mattjw16 5 жыл бұрын

WolframAlpha can solve it now.

@marcoantonioreyessantos9977 3 жыл бұрын

I also started as Nayak's solution: x has to be odd, then x=2n+1. Then, we have that 4n(n+1)=2^y-616, or n(n+1)=2^(y-2)-154. Now I went with y-2>=8, and found y-2=10 and n=29, which give x=59 and y=12. Cheers, Marco.

@leeoldershaw956 4 жыл бұрын

You can solve fast by inspection like in a SAT test by trying numbers. The power Y of 2 has to be bigger than 615 or Y>9. You only have to try the next 3 values of Y before you find a value of X that has an integer square root, 3481. You don't even need to do the square root, The squares of 50 and 60 are 2500 and 3600 so look for a number between them whose square ends in 1. 9 does so trying 59 as X works and solves the problem when Y =12. If it were a SAT question there might be answer choices that include 59 and +/-59 so that might trip you up.

@vvvlekocajjj2011 5 жыл бұрын

You could’ve just kept adding 1 to y from 2^y=512 until the solution x^2 is an actual square

@ashkara8652 5 жыл бұрын

Brute force is for computers. Not very elegant, not very practical for humans either.

@PrincessEev 5 жыл бұрын

That wouldn't necessarily work. If there are infinitely many solutions, we need something a little more elegant to encapsulate that fact.

@giuseppepapari8870 5 жыл бұрын

actually an exponential and a parabula do not intersect in more than two points

@vvvlekocajjj2011 5 жыл бұрын

Giuseppe Papari exactly

@danmerget 5 жыл бұрын

@@giuseppepapari8870 - That's a non-sequiter in this case. There are only two intersections IF the parabola and exponential have the same x-axis. But in this case the parabola (x**2) and exponential (2**y) have different x-axes since x and y can be different numbers Another way to look at it is to note that the problem would have infinitely many solutions if either x or y were allowed to be non-integer real numbers. If your "at most two intersections" argument were valid, then there would be at most two solutions among the non-integer reals.

@harold351 5 жыл бұрын

If you do trial and improve it's easier I got it in less than 5 minutes

@jujujulost1232 5 жыл бұрын

Thing is finding a solution by trial and error doesn't prove there's no other solution

@sumit3735 3 жыл бұрын

Mind your decision is best Chanel for students

@bearloscuro 4 жыл бұрын

Here's how I solved this with paper and pencil. Rewrote expression as x^2 -2^y=-615. X^2 will always be a positive integer value, so 2^y must be some power of 2 greater than 615 where the difference between the two expressions has a (positive) integer square root. 2^10 = 1024 & square root of the difference (409) is not an integer, 2^11=2048 and square root of the difference (1433) is not an integer, 2^12=4096, subtracting 615 yields 3481. Since the last two digits of 3481 are 81, there is a good chance that 59 could be the square root. I tried that and it worked. Thus, x=59, y=12.Yes, I know intuition and trail and error aren't proper 'solutions.' Thanks for this channel!

@silverbladeii 4 жыл бұрын

Minha solução: Se "==" representa congruência, temos: 615+x²==x² (mod 3). 2ⁿ=/=0 (mod 3) →x²=/=0 (mod 3) →x²==1 (mod 3). 2ⁿ==1 (mod 3) →n=2a 615=2²ª-x²→ 615=(2ª+x)(2ª-x). Edit: ok, vamos testar todos os pouquíssimos casos: 615=3•5•41, além disso, 2ª+x>2ª-x. As únicas possibilidades são (supondo x≥0): 2ª+x=615 e 2ª-x=1→2•2ª=616. Abs 2ª+x=205, 2ª-x=3→2•2ª=208. Abs 2ª+x=41 e 2ª-x=15→ 2•2ª=56. Abs E 2ª+x=123 e 2ª-x=5 2ª=64, de onde segue que a única solução é a=6→n=y=12 e x=±59

@Gutagi 4 жыл бұрын

parabens cara

@Gutagi 4 жыл бұрын

Translation for english: My solution is: If "==" represents congruence then we have: .... Testing all the possible cases (which are just a few) , we have that the only solution is...

@silverbladeii 4 жыл бұрын

@@Gutagi macho, são bem poucos casos

@Gutagi 4 жыл бұрын

SilverBlade como assim ???

@silverbladeii 4 жыл бұрын

@@Gutagi taí. Resolvido completo (mas é tão trivial que eu nem devia ter perdido meu tempo editando)

@jameskelly745 5 жыл бұрын

I did it completely differentlymin my head in a third the amount of time it took to explain that solution

@trashtalking_bug 5 жыл бұрын

r/humblebrag

@SpaghettiToaster 4 жыл бұрын

Let me guess... you guessed it and now think you got a a proof.

@iamrepairmanman 5 жыл бұрын

If you move X^2 to the right and assume y=even, break it into (x+2^(y/2))(2^(y/2)-x). The prime factors of 615 are 1 3 5 and 41. 2^6 is 64. 41*3=123, 123-64=59, 64-5 is also 59, meaning we've found at least one answer, y=12 and x=+-59

@fredericofp Жыл бұрын

Bruteforced this one by just estimating values for Y and subtracting 615 from 2^y and checking to see if the result was a perfect square ( by calculating the closest perfect square i could do in my head ) and got to the awnser in a couple of minutes