Question But Not A Doubt 1

Given \(a_{1},a_{2},...,a_{n}\) satisfies, \[a_{1}^2+a_{2}^2+...+a_{m}^2=a_{m+1}^2+a_{m+2}^2+...+a_{n}^2\] find one such solution. Here, \(a_{1},a_{2},...,a_{n}\) are positive integers and are distinct

Note by Mohammed Imran
4 months, 2 weeks ago

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Find a suitable xx such that a12+a22=x2a_{1}^2+a_{2}^2=x^2 then the problem reduces to x2+a32+...+am2=am+12+...+an2x^2+a_{3}^2+...+a_{m}^2=a_{m+1}^2+...+a_{n}^2 then find a suitable yy such that am+12+am+22=y2a_{m+1}^2+a_{m+2}^2=y^2. So, doing this process repeatedly on both the L.H.S and the R.H.S until you reach a point where your equation will be of the form: a2+am2=an2a^2+a_{m}^2=a_{n}^2. Now, you know a1,...,am1,am+1,...,ana_{1},...,a_{m-1},a_{m+1},...,a_{n} because we have been substituting suitable values. So, we will have an equation like this: a2+am2=an2a^2+a_{m}^2=a_{n}^2. Now, we can easily solve this using pythagorean triples formula, because we know aa. Hence we have found a suitable solution.

Mohammed Imran - 4 months, 2 weeks ago

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Such that what?

Steven Jim - 4 months, 2 weeks ago

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Such that a12+a22=x2a_{1}^2+a_{2}^2=x^2

Mohammed Imran - 4 months, 2 weeks ago

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Huh, that's actually a good idea. The only thing I can see here is that you haven't proved that it's always suitable to find xx all the time, thus we can't be completely sure about having infinite solutions.

So, here's my question: How can you prove that there are infinitely many solutions?

Steven Jim - 4 months, 2 weeks ago

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Uhh, this is a pretty vague question I believe.

What is mm? What is nn? Must aia_i be an integer? Can it be zero? If it is then I can always make a trillion solutions just by saying aia_i = 0 for all ii ranging from 1 to nn.

Even if the numbers must be integers and positive, by letting nn divisible by 2, I can make infinitely many solutions. (How, I'll leave you to it.)

So, the takeaway? Restrictions!

P/S: Good ideas though, maybe you can capitalize on it by forcing aia_i to be all different; that way you can make a pretty fun problem.

Steven Jim - 4 months, 2 weeks ago

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Yeah. yes ai is an integer. No, the question is not to find the number of solutions, but one such solution. m,n can be any positive integers you want. I just want a general solution.

Mohammed Imran - 4 months, 2 weeks ago

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Even then your question is still too vague.

If 2n2|n then m=n2m=\frac{n}{2} and ai=a(m+i)a_i=a_{(m+i)}.

If n=2k+1n=2k+1 (with k>1k>1) then m=k+2m=k+2, ai=1a_i=1 with ii ranging from 11 to n1n-1 and an=4a_n=4.

If n=3n=3... I don't think I need to introduce to you Pythagorean triples, no?

Even as a bonus, if n=b2+c2n=b^2+c^2 then m=b2m=b^2 and ai=ca_i=c with ii ranging from 11 to mm and aj=ba_j=b with jj ranging from m+1m+1 to nn.

As long as nn is larger than 1... The sky is endless.

Steven Jim - 4 months, 2 weeks ago

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Sorry. All numbers are distinct

Mohammed Imran - 4 months, 2 weeks ago

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There we go. That should be enough constraints for now. Now we're getting to the fun part.

I guess you can agree with me that there exists infinitely many Pythagorean triples.

Now, what about Pythagorean quadruples? Read it carefully; your answer lies in it.

Assume now that you agree with me that there exists infinitely many Pythagorean quadruples. Since we have both infinite triples and quadruples, I guess we can find different numbers to just throw in.

For n<3n<3, obviously no.

For n=3n=3 and n=4n=4, sure, why not?

n=5n=5 is the tricky part. Back to the drawing table for now. (Hint: m=4m=4, and you're using 33 triples.)

Update: After doing extra research, I can see that there might exist infinitely many answers for n=5n=5, but the general formula is unknown to me. However, here is just a solution to n=5n=5 to satisfy my curiosity: 1952+1042+362+482=2292195^2+104^2+36^2+48^2=229^2.

Update 2: Let's call this triple x2+y2=z2x^2+y^2=z^2 for the sake of clarity. Now, you can see there are infinitely many solutions so that xx is a multiple of 5. Needless to say, xx can be partitioned to 22 other squares. (The reason, I'll leave you to finding out.) What about yy? This is a more difficult question, and I haven't had yet an idea to answer. However, you can brute force them to find a solution. And based on my knowledge, there should exist infinitely many solutions (hopefully).

For n>5n>5, you can just add in triples and quadruples until you get the necessary value of nn. Voilà.

Steven Jim - 4 months, 2 weeks ago

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Uhh that is a very tough method

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran Well, as I have mentioned above, it's almost impossible to point out a "general formula" here (considering the lack of definitive nn), but those are the clues to at least show that there are infinitely many answers.

Also, can I know why you find it tough? To me this is completely undestandable, actually.

Steven Jim - 4 months, 2 weeks ago

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But there is a very easy method!!!

Mohammed Imran - 4 months, 2 weeks ago

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Hmm... I do not know, but for me I am heading towards finding all possible solutions, hence the added difficulty.

However, I still don't regard my method as too difficult. I would love to listen on how there's one general formula that actually holds true for all nn.

Steven Jim - 4 months, 2 weeks ago

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There is no general formula, there are infinite solutions. So, maybe I should tell my method now!!!

Mohammed Imran - 4 months, 2 weeks ago

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Yes, I am more than willing to listen.

Steven Jim - 4 months, 2 weeks ago

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ok sure!!!

Mohammed Imran - 4 months, 2 weeks ago

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Just us pythagorean triplet formula

Mohammed Imran - 4 months, 2 weeks ago

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The idea of the formula is still insufficient. While there are infinitely many Pythagorean numbers, there are not infinitely many triples for a certain number. This is the dead end here.

Steven Jim - 4 months, 2 weeks ago

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No I am sorry for partially giving a solution

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran Don't be sorry. This is the learning experience. The more you learn the more you know.

Again, your solution does not even guarantee there would be a solution for mm in any nn, for the reasons already said above.

Steven Jim - 4 months, 2 weeks ago

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@Steven Jim No. I have presented them at a conference

Mohammed Imran - 4 months, 2 weeks ago

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@Steven Jim So, please give me some time and I'll definitely post the proper solution

Mohammed Imran - 4 months, 2 weeks ago

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Also, I am quite busy now

Mohammed Imran - 4 months, 2 weeks ago

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But I'll definitely post the proper solution for you after a while

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran I am very grateful for that. Thanks.

Steven Jim - 4 months, 2 weeks ago

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@Steven Jim Welcome!!!!!!!

Mohammed Imran - 4 months, 2 weeks ago

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Good bye!!!

Mohammed Imran - 4 months, 2 weeks ago

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You posted a problem related to this right?

Mohammed Imran - 4 months, 2 weeks ago

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Yes I did. I figure that your idea is good, so I decided to post it and attribute you to it.

Steven Jim - 4 months, 2 weeks ago

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Thank you very much!!!

Mohammed Imran - 4 months, 2 weeks ago

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That is just awesome!!!

Mohammed Imran - 4 months, 2 weeks ago

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You do have brilliant thinking skills on brilliant!!!

Mohammed Imran - 4 months, 2 weeks ago

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Thank you.

Steven Jim - 4 months, 2 weeks ago

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Welcome!!!

Mohammed Imran - 4 months, 2 weeks ago

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maybye I will eplain you with an eample. Ok?!

Mohammed Imran - 4 months, 2 weeks ago

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Yeah sure

Steven Jim - 4 months, 2 weeks ago

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Example: Find one 5-tuple (a,b,c,d,e)(a,b,c,d,e) such that, a2+b2+c2=d2+e2a^2+b^2+c^2=d^2+e^2

Solution: Find a suitable xx such that a2+b2=x2a^2+b^2=x^2. One such solution is a=3,b=4,x=5a=3, b=4, x=5. So, the problem reduces to 52+c2=d2+e25^2+c^2=d^2+e^2. Let e=13e=13 then we have c2d2=144c^2-d^2=144. So, (c+d)(cd)=144(c+d)(c-d)=144 which implies that c=37,d=3c=37, d=3. So, one solution set is (a,b,c,d,e)=(3,4,37,35,13)(a,b,c,d,e)=(3,4,37,35,13)

Mohammed Imran - 4 months, 2 weeks ago

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Well, try n=10n=10. Or n=100n=100. How can you ensure there would be a solution?

Generalize it and you'll see some problems.

Steven Jim - 4 months, 2 weeks ago

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No. There are infinite pythagorean triples and quadruples. So, there will not be any problems

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran If you do it the way I do, there would be zero problems.

The problem with your method is that you can't ensure that you will be able to find a suitable xx, and if you can prove that you can always find a suitable xx then your proof will be complete.

Steven Jim - 4 months, 2 weeks ago

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@Steven Jim Your way is good, but my way is also similar

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran Actually I hoped it was similar. You were a bit mistaken here.

Your method should be easier if there isn't a small problem, which I will restate again: how do you find a suitable xx? Is it always possible?

Focus on solving that problem, then your proof will be complete.

Steven Jim - 4 months, 2 weeks ago

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@Steven Jim Yes it is always possible

Mohammed Imran - 4 months, 2 weeks ago

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@Steven Jim As I told earlier, I am not able to express my ideas perfectly

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran You can state your ideas, if any. For me, there's little light in the end of the room.

Steven Jim - 4 months, 1 week ago

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@Steven Jim But you don't need to use quadruples. Triples are more than enough

Mohammed Imran - 4 months, 2 weeks ago

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@Steven Jim This way is almost your way but the only difference is there are no quadruples involved

Mohammed Imran - 4 months, 2 weeks ago

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I am not able to explain my ideas through typing. Is there any other way to do?

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran Don't fear. Learn to write. I learnt it the hard way; guess you could start too.

Steven Jim - 4 months, 2 weeks ago

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@Steven Jim Ok sure

Mohammed Imran - 4 months, 2 weeks ago

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Please let me know

Mohammed Imran - 4 months, 2 weeks ago

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I am sorry I made a mistake when writing the general solution. This solution is only properly correct

Mohammed Imran - 4 months, 2 weeks ago

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I was in a hurry to post that solution. So a small error occured.

Mohammed Imran - 4 months, 2 weeks ago

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@Mohammed Imran I will decide to move to this comment for the sake of clarity.

Again, I have still not understood how you would choose the suitable xx. State your ideas here, if you have any.

Steven Jim - 4 months, 1 week ago

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What do you mean new comment?

Mohammed Imran - 4 months, 1 week ago

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State your ideas in a new comment here.

Steven Jim - 4 months, 1 week ago

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@Steven Jim ok !!!!!!!!!!!!!!!!!!!!

Mohammed Imran - 4 months, 1 week ago

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We want one solution to the equation: a12+a22+...+am2=am+12+...+an2a_{1}^2+a_{2}^2+...+a_{m}^2=a_{m+1}^2+...+a_{n}^2.

Now, choose a suitable xx such that (a1,a2,x)(a_{1},a_{2},x) forms a Pythagorean triple. Now, the problem reduces to finding one solution to x2+a32+..+am2=am+12+...+an2x^2+a_{3}^2+..+a_{m}^2=a_{m+1}^2+...+a_{n}^2. Now, choose a suitable x1x_{1} such that (x,a3,x1)(x,a_{3},x_{1}) forms a Pythagorean triple. Keep doing this process on both sides until we land on an equation of the form a2+am2=b2+an2a^2+a_{m}^2=b^2+a_{n}^2. Now, since we have been substituting values for a1,a2,..,am1a_{1},a_{2},..,a_{m-1}, we know the values of a,ba,b. So, we can rewrite the equation as a2b2=an2am2a^2-b^2=a_{n}^2-a_{m}^2 and solve the equation by "Simon's Factorisation". So, we have found a solution set.

Mohammed Imran - 4 months, 1 week ago

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I think I might have just found a way to ensure you can always find a suitable xx.

Start with an arbitrary xx larger than 100100. We need to find a suitable x1x_1, as stated above. Now, do either of these steps:

a/ If x2x^2 is odd, partition it into two numbers, 11 and itself, then use the equation x2=(x1a3)(x1+a3)x^2=(x_1-a_3)(x_1+a_3) to attribute 11 to x1a3x_1-a_3 and x2x^2 to x1+a3x_1+a_3. Obviously, x1x_1 and a3a_3 are both larger than xx.

b/ If x2x^2 is even, divide by 44 until the number is odd. Then do as above. Again, obviously, x1x_1 and a3a_3 are both larger than xx (why, I will let you find out).

As the numbers increase over time, there can't be any repetition in numbers between variables. Thus, we can always find a suitable xx satisfying the equation.

Q.E.D :)

@Mohammed Imran There you go, got it done.

Note: You will still need to complete the details of this prove, but I can guarantee you that this is right.

Steven Jim - 4 months, 1 week ago

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Yeah, that is what I was telling you that you can find suitable x

Mohammed Imran - 4 months, 1 week ago

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@Mohammed Imran You never stated it out carefully however, did you?

This is not trivial after all; you will still need to state the dynamics of your answer.

But yes, now that your answer is complete, you can proudly show it out to everyone.

Steven Jim - 4 months, 1 week ago

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@Steven Jim Yes. But I thought that this is trivial

Mohammed Imran - 4 months, 1 week ago

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@Mohammed Imran This is not trivial at all. Especially this statement: "As the numbers increase over time, there can't be any repetition in numbers between variables." You can't ensure that repetition is impossible. Hence the non-triviality.

Steven Jim - 4 months, 1 week ago

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@Steven Jim oh!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mohammed Imran - 4 months, 1 week ago

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@Steven Jim Sorry for wasting your time on this!!!

Mohammed Imran - 4 months, 1 week ago

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@Steven Jim Thank you once again!!!!!!!!!!

Mohammed Imran - 4 months, 1 week ago

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@Mohammed Imran You're welcome.

Steven Jim - 4 months, 1 week ago

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Thank you very much

Mohammed Imran - 4 months, 1 week ago

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Do you want to check my other notes out???

Mohammed Imran - 4 months, 1 week ago

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Steven Jim, do you have any comments??

Mohammed Imran - 4 months, 1 week ago

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Steven Jim, if you are interested for general divisibility criteria of odd numbers, please have a look at my note: "A Question But Not A Doubt 2"

Mohammed Imran - 4 months, 1 week ago

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