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

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]

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]

What is your email?

[@MindYourDecisions]

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]

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

[@Lqtan16]

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

[@PraneshPyaraShrestha]

@@Lqtan16 oh I found it

[@Rimtay]

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

[@sxz452]

That's sooooo sad

[@Goku_is_my_idol]

Lol i did same😂

[@namanlakhotia6393]

@@Vegan_PhysicsEnthusiast u definitely dont have these questions

[@funni111]

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

[@namanlakhotia6393]

@@funni111 prmo is not comp in schools though

[@arnavraut9691]

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

[@ashyourresidentenby5916]

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

[@tgdhsuk3589]

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

[@ashyourresidentenby5916]

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

[@pdpgrgn]

@@tgdhsuk3589 WolframAlpha solves this

[@quantumsoul3495]

Ethical hacking..

[@roshandon3157]

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

[@radekvecerka1115]

@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]

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

[@Martin-zr2tb]

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]

I did but I solved it too

[@ReenaKumari-vq2rk]

@KZbin Watcher i also did the same😁

[@AYUSHGAMEROFFICIAL]

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

[@shambosaha9727]

pressure walker

[@enavoid]

dash fallwalker

[@ahmadm7618]

it says Presh TalWalker here

[@adityarajsrivastava6580]

It says pressure locker

[@GigaPlaya]

Not Luke Skywalker.

[@antilogis6204]

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

[@lach993]

WYSI

[@lukkash]

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]

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.)

[@notwildcard377]

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

[@rcnayak_58]

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]

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

[@rcnayak_58]

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

[@raphaelmillion]

@@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]

​@@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]

@@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.

[@ethanfrommer7772]

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]

ethan frommer ok boomer

[@Seekingpatience19]

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

[@suryakalawasnik6204]

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

[@knk0112]

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!!

[@bancodrut]

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]

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

[@SameerKhan-fd2qe]

Fresh tall walker

[@MathIguess]

@@SameerKhan-fd2qe Fresh Tail Wagger

[@Super_Smash_Dude]

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

[@l.w.paradis2108]

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.

[@curiousscientist6077]

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

[@TheGeneralThings]

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

[@harshitagoyal7134]

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

[@srimatresri]

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

[@bharatnotindia.6296]

That's cool.

[@oskarjung6738]

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]

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]

But i got the answer by putting 12

[@SuperLabelPerson]

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

[@sjoerdwiesmeijer7231]

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]

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

[@ihopeicanchangethisnamelat7108]

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.

[@GinoTC]

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

[@franklyanogre00000]

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]

Frank Anthony Overton Jr. I did the same 😂😂

[@akankshasarda5216]

Me too😂

[@saurabhusane5780]

Exactly ...!

[@mr.moodle8836]

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]

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.

[@kienha9036]

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

[@tamirerez2547]

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

[@jasonterry1959]

@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.

[@quahntasy]

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]

It works now 🧐

[@spacefertilizer]

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.

[@Pepa14pig]

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

[@kwcy92]

Can you share your way?

[@Pepa14pig]

@@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 😂😂😂

[@user-ng4sb5nl2o]

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)

[@Acid31337]

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

[@ffggddss]

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]

but can you tho?

[@silkyfirst3097]

Dont provoke them wolfram alpha is listening

[@TAT4guitar]

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

[@nonesuch27]

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

[@arolimarcellinus8541]

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]

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]

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]

@@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]

@@dariobarisic3502 No it doesn't. Just that simple. Solve does not mean all. This KZbin 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.

[@UltimatumNo5]

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?

[@indy197905]

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

[@kalliboymusic]

Same

[@willnewman9783]

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]

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

[@freddiehand6551]

@Bragadeesh S yes it did

[@sapien153]

My bad. Agreed

[@Theo0x89]

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]

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.)

[@SamerShennar]

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]

Ye prove it that that's the only answer

[@musicsubicandcebu1774]

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

[@sandrastrilakos9962]

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.

[@TamaraTkacova]

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]

Same here :D

[@UniqueNCS]

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

[@ritusaxena2714]

Not possible

[@tatane79]

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

[@spacefertilizer]

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.

[@anyone6240]

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]

Finally! A solution I can understand!

[@yurenchu]

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.)

[@user-uc1hf8zm6k]

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]

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

[@ralphmacchiato3761]

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

[@legendhero-eu1lc]

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

[@prerakcontractor6609]

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

[@Packerfan130]

and can it?

[@prerakcontractor6609]

Nope

[@dmitryronin6898]

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

[@KelfranGt]

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

[@academyofmathematicalscien7986]

kzbin.info/www/bejne/m3bcgqyfrbR6prc

[@mrKreuzfeld]

That was awesome!

[@ozanoruc3796]

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

[@merveilmeok2416]

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 ;)

[@Tailspin80]

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]

Cool bean

[@PraneshPyaraShrestha]

Your videos are awesome

[@PraneshPyaraShrestha]

@@iotaop9573 always always

[@sarthakgupta4323]

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

[@steveholman5978]

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.

[@anubis_05]

hello please look into integrals by the Feynman technique

[@thephysicistcuber175]

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

[@TechToppers]

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

[@d4slaimless]

@@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]

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

[@d4slaimless]

@@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]

wow i found a cuber under a maths video!

[@johnconway8070]

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]

Me too.

[@beautyofnature7361]

Very good Sir, I like your every problem solving videos

[@Akash-qy1gf]

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]

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

[@samsonpl1110]

-59 is second matching X :D

[@ceegers]

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

[@Cohnan13]

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

[@speedsterh]

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

[@whiploadchannel2047]

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

[@whiploadchannel2047]

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

[@invincibleflesh4526]

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

[@montsaintleondr7491]

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

[@hsman6614]

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

[@martinpiekarski1512]

AI won't take over the world so soon.

[@michaelfredericks6970]

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

[@elmatadordeangel5004]

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

[@tomashertl8895]

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

[@mcovaleda]

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

[@nellvincervantes3223]

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

[@usualunusualkid7149]

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

[@TorBruheim]

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]

It is traditional math, a branch called number theory.

[@JLvatron]

Wolfram & Hart couldn’t solve it, but Winifred Burkle could! Thanks for this genius solution.

[@kasperjoonatan6014]

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 :)

[@madhavstalks3925]

The unit digit method always work. Good solution..

[@marcoantonioreyessantos9977]

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.

[@TitusObbayi]

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

[@ronmadan8003]

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]

Wish you could see my raised hand ✋ ✋

[@thiantromp6607]

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

[@zoetropo1]

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

[@Oswald_Anthony]

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]

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

[@EduardvanKleef]

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

[@user-bf2bp2ww9m]

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

[@vladpetre5674]

@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]

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]

​@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.

[@anuragmishra3282]

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

[@BluetarES7]

Thank you. You are inspiring. Very nice solution

[@md.abdurrahmantarafder2256]

Excellent solution!

[@mattjw16]

WolframAlpha can solve it now.

[@xVitOSx]

Thanks macOS for "Grapher"

[@robert33232]

That is interesting, thank you!)))

[@IS-eb9lf]

Did it in 15 mins. I found for myself this method on Programming/Math olympic competition. 10x 4 video. :)

[@t_kon]

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]

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]

@@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]

@@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]

Is doing mod 3 also trial and error ?

[@SaiKiran-fd3gq]

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

[@silverbladeii]

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]

parabens cara

[@Gutagi]

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]

@@Gutagi macho, são bem poucos casos

[@Gutagi]

SilverBlade como assim ???

[@silverbladeii]

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

[@diedoktor]

I just did guess and check and found 59 as a solution for x. Obviously 2^y has to be greater than 615, so I started with 1024-615, which wasn't a perfect square, so I did 2048-615, then 4096-615. I found the square root to be 59. So -59 and 59 are solutions when y=12. I imagine there are infinite solutions. EDIT: The solution in the video is much more elegant, and also shows there is only one solution. Very cool.

[@twinksy90]

X^2=2^y-615. For x^2 to be positive, y is at minimum 9. Then increase y incrementally until 2^y-615 is the square of an integer.

[@Exachad]

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

[@vvvlekocajjj2011]

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

[@ashkara8652]

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

[@PrincessEev]

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

[@giuseppepapari8870]

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

[@vvvlekocajjj2011]

Giuseppe Papari exactly

[@danmerget]

@@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.

[@kanchanaramar]

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.

[@bearloscuro]

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!

[@jameskelly745]

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

[@trashtalking_bug]

r/humblebrag

[@SpaghettiToaster]

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

[@harold351]

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

[@jujujulost1232]

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

[@jimvinci8295]

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.

[@peterlobzhanidze1277]

Shout out from Georgia 🙌🏼❤️

[@TheTck90]

Don’t use calculator but count 2^9 and factor out 615 :D

[@minghaoliang4311]

That wasn't that hard, was it?

[@washizukanorico]

You need a calculator for the first few powers of two? Like really?

[@AvelinoTiago]

The worst students know 2 powers from 0 to 20

[@SpaghettiToaster]

@@AvelinoTiago No.

[@goodjob822]

WolframAlpha solve it for now

[@sumit3735]

Mind your decision is best Chanel for students

[@andysolano7847]

I solved all by myself in less than 5 minutes!! I used congruences and it worked out great!