YouTube Math Challenge

[PYTHAGORAS101]

nice try , but it is not continuous

it = 0 ,

but the next numbers are 2,3 ,4 ect (in your sequence)

they do not equal 0 according to the fomula a2+b=c2

[Shortmandesigner]

What about a = -1, b = 0, and c = 1?

[beerbelly777]

Try my math challenge!! check out my channel, solve it!

[KatyB34]

I made sure not to look at the comments before this. Could a=0, b=1/4, and c=1/2? I substituted a as b-x and c as b+x in the equation and solved for x (x=1/4). I just started plugging in numbers from there and it works. (0, 1/4, 2/4, 3/4, 1...)

[PYTHAGORAS101]

You Americans hate the decimal system ,i notce you would rather use fractions than use a number with a decimal point .Give yourself a pat on the back ,you got it .

[Drakeman8]

Ok here it is. There is in fact a real proof of the solution(maybe you already know it). The solution can be found under the assumption this is a geometric series such that (b-x)^2 +b = (b+x)^2. If we expand this we find b=4bx. Then we can divide and find x=1/4.

I realize that this solution requires the assumption that this is a geometric series. But someone trying to find a solution could surely work under different assumptions and ultimately find it in this way. Cheers.

[PYTHAGORAS101]

Thanks 4 your wise comments friend , its very apreciated .Its just a tast of whats to come .This is just a snippit of the big picture .

btw geometry / math are the same thing .

(proof is to come )

[Drakeman8]

Thanks. I know; I was referring to the fact that it's a geometric series (i.e. takes the form of some multiple times x where x is the number in the series). For example, 2, 4, 6, 8 is a geometric series, whereas 1,3,5,7 is not.

[PYTHAGORAS101]

7 is not (geometric ) ,but 1234568 is , they are constructable .(devisions in a circle )

[Drakeman8]

Sorry, you're right. The type of series I'm talking about is not a geometric one. The solution still holds, as a result of the fact that the arithmetic difference between consecutive numbers is constant. Therefore we can assume, as is done in my solution, that the former number is some number x less than the middle one, and that the latter number is that same number x more than the middle. That's what I was getting at, I just used incorrect terminology in my explanation.

[Drakeman8]

The reason it must be assumed in my solution is because b is assumed to be the middle number of any three consecutive numbers in the series. Under the assumption it's geometric, the number before would have to be some number x less than the middle and the latter number would have to be that number x greater than the middle. We then plug in the former number, a, as b-x, and the latter number c as b+x and solve for x.

[Drakeman8]

I found something interesting while pondering your series. There is actually a very simple solution to how to find it. We know its a series such that if b=n, a=n-1, and c=n+1. Thus if we say (n-1)^2 + b= (n+1)^2, then we find that b=4n, a=n-1, and c=n+1. This holds true only when n=1/4, which happens to be the geometric series of the problem. This may only be coincidence though, because n was supposed to represent the series number. However, it gives the geometric multiple instead.

[Drakeman8]

Very interesting series. How does it tie into the pythagorean series? I recognize somewhat what you are getting at. All the ratios can be written with square numerators and denominators, as is appropriate from the equation. Is it simply the fact that each gnomon (i.e. b) has to also be a square. The relation to this and the pythagorean theorem has been shown in the past. Do you have an interesting new explanation of it?

[Drakeman8]

Wow, fellow youtube commenters. It appears you are the ones proving yourselves to be abusive. Stop splitting hairs about whether you have seen the series or not and thank someone for finally bringing some interesting math onto youtube. It's much better than those who claim an indian with vedic math abilities to have mystical powers and better yet than those who commit zero-division fallacy and call it sorcery.

[112rhavin3581321]

I'm sorry but I'm Irish and recognize this from higher level secondary school maths....I know the solutions below but I'm sure I've seen it before

[PYTHAGORAS101]

lol ,i have a theory you are thinking of a theorem .

[112rhavin3581321]

You know what your probably right, seems like theorems are all we ever did!

Look forward to any more Challenges you come up with, hopefully I'll have a crack at em before its solved this time.

[PYTHAGORAS101]

i just posted another last night ,i'll give you a clue ,its easy if you saw the last one

[beerbelly777]

If youre looking for a new math challenge, try mine on my channel.

[PYTHAGORAS101]

check out challenge 2

[corbier111]

It's always funny to read: 'I'm the only one...'

what makes you think you're the only one?

This statement is utterly abusive , as if you knew everybody on earth..

[PYTHAGORAS101]

fair comment , but it is not listed .

[corbier111]

...or you forgot to finish your sentence. You'll probably agree that there's a huge difference between saying you're "the only one in your apartment" and "the only one in the world".

[PYTHAGORAS101]

i am the 1st person that has published the properties of this sequence .(in the world)

you got a problem with that???

[corbier111]

not really, but I do have a problem with the fact that you still think you're the first to have published it or even found it...

[PYTHAGORAS101]

it was known in ancient times ,but is not listed in websites which get 100s of sequences posted per day .google it , its not there OK

[beerbelly777]

Heres the solution to your challenge:

The sequence that satisfies the conditions you ask is an arithmetic sequence. the formula for the nth term is:

a_n = 1 + (n-1)*(1/4)

The first few terms of your sequence: {1, 1.25, 1.5, 1.75, 2, 2.25, 2.5, ...}

Every term in this sequence is rational, so it meets your first criterion. Any 3 consecutive terms satisfy a^2 + b =c^2 where a=a_n, b=a_n+1, c=a_n+2th terms in the sequence. So this satisfies the second criterion :)

[PYTHAGORAS101]

Congradulations my friend ,you found the sequence . There is one minor error in your answer . The first few terms are incorrect .

You got this far ,correct them .

[PYTHAGORAS101]

beerbelly who are you mate , you joined YT yestersday , you have only seen my channel , nothing else . What happened , you zoned in on me and had the answer .I am feeling perplexed , i don't know what to think or feel . I am astonished ,please tell me how worked it out and how you got on to me .

thanks

+

[beerbelly777]

Well, I checked the terms, they are correct; they satisfy your conditions. I guess by "incorrect", you mean you want them in fraction form opposed to decimal form so its a fraction of 2 integers, a clearly rational number? I like working with decimal form, but here you go: {1, 5/4, 3/2, 7/4, 2, 9/4, 5/2, ... }

[PYTHAGORAS101]

1 is not the start of the sequence (keep to decimals )

[beerbelly777]

well, the following sequence satisfy your criteria: i. each term is rational

 ii. any 3 consecutive terms satisfy a^2+b=c^2 iii. 1 is not the first term in the sequence. iv. no terms are negative valued

a_n= 1 + (n-5)*(1/4)

first few terms:

{0, .25, .5, .75, 1, 1.25, 1.5, 1.75, ...}

I figure 0 is the first term since ideally, if you dont want to include negatives, then you would want to include all non-negative terms, thus 0 would have to be in it. It should be the first term.

[PYTHAGORAS101]

Well done Sir ,i am impressed .You even included the zero . i was wondering if you would .You are the 92nd viewer of my challenge ,i thought it was going to get to 1000 b4 it was solved . Well done . I discovered this by examining the properties of the "Quarter Sequence" .I did not try to invent it , i just kind of stumbled upon it .

It has other properties too .I am trying to decide whether or not to put up another challange or just post it as a part of my work .

[beerbelly777]

thanx "mate",

Your work sounds interesting. you should make a video explaining your sequence and it's properties.

...I like these math challenges, I may come up with my own challenge and put it up.

[PYTHAGORAS101]

.If you subscribe to my channel it would make it easier 4 me to keep you updated on what my work is all about .

I have over 50 subscribers ,

so when i post stuff like this i click "all contacts" .

I am still bewildered ,

i thought at some point i was going to have to just tell everyone .

I am still not quite sure how you did it .

i am about to become your 1st subscriber.

Cheers mate

[beerbelly777]

so, what Math are you working on? How does this challenge relate?

oh, did I mention I like drinking beer?

[beerbelly777]

I only browse youtube videos, I never needed an account for posting. I stumbled on your page yesterday, and decided to solve the challenge. So it was neccesary to create an account for the purpose of posting my solution.

[beerbelly777]

an alternative version of the formula for the sequence if you must have each term in the form of the fraction would be:

a_n= (n+3)/4

This formula gives it in an immediate fraction form. My other formula requires work to convert the answer into fraction form, but you would conveniently get a decimal answer if you computed a term with a calculator.

[PYTHAGORAS101]

check out YouTube Math challenge 2

[Drakeman8]

This reminds me of Pythagoras' work with gnomons. It does not appear that it can be an integer series as a result of the fact that when using integers, the gnomon (i.e. b) will always be greater than a and c. Therefore the b will be unable to become the next a as a result of it being bigger than the next b(i.e. the original c). Correct me if I'm wrong. Maybe something with fractions?

[PYTHAGORAS101]

You are correct in that it is not a integer series . I am amazed that beerbelly777 got it ,but he made a slight error ,check out my response

[PYTHAGORAS101]

check out YouTube Math challenge 2

[talkingdonut]

Im busy on other work, but here is a thought. Construct the sequence begenning 1,2,3. Write a program to map the sequence out "really far" or kill some cosmic time with a pencil and paper. Look for autoregressive properties in the data. Forcast a restricted sequence until it converges using your favorite information criterian as a guideline. Play with different initial guesses. Wait until brown around edges and remove from oven.

[talkingdonut]

What you want to chase is a^n + b^n = c^n, where a not=b not=c, and n greater than 2. Its been proven but its sloppy and needs improvement. I watched lecture 28, multiple integrals, had not seen a jacobian in a long time, though I've had to use a few Hessians. I dont think your series is special, but if your inerested, there is a way to cheat on these kinds of problems using recursion and, basically, OLS estimation.

[PYTHAGORAS101]

What you are talking about is Furmuts (can't spell ) last theorem .

This is the Pythagoras theorem with a twist ,

where b is not squared and has nothing to do with triangles .

I honestly believe i have a proof of the Pythagoras theorem ,

and this is a part of it .

So i think it is very special.

There is no need to cheat or use any form of estimation .

All numbers in my sequence are rational .

If you have time ,

give it your best shot .

[PYTHAGORAS101]

LOL ,

1,2,3.PiE,M,C,2,3,4,PiE,M,....­.Cooked

Hows that?

Seriously now .the sequence exists and you will be amased when you find out what it is .

Thanks 4 posting ,i will send the answer to you when the time is right .

[PassionOfTheChrisT01]

Personally the answer to this isn't really that difficulty. If you were to think about it, since 0^2 + 1 =/= 2^2 and if the later amounts only amount to a higher difference. We have to look forward into the negatives. If B was to be 0, what consecutive numbers when squared would be the same answer. Therefore the answer to this "math challenge" is simply -1, 0, 1 a=-1, b=0, c=1

[PYTHAGORAS101]

Thanks for having a go and commenting .

The problem with your answer is that it is not a continuous (infinite) sequence .If you want to try again i can tell you there are no negative numbers and no number is repeated .

[PYTHAGORAS101]

check out YouTube Math challenge 2

[PYTHAGORAS101]

Please , if anyone is trying to solve this , please comment .