# 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
1 year, 2 months ago

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## Comments

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Find a suitable $x$ such that $a_{1}^2+a_{2}^2=x^2$ then the problem reduces to $x^2+a_{3}^2+...+a_{m}^2=a_{m+1}^2+...+a_{n}^2$ then find a suitable $y$ such that $a_{m+1}^2+a_{m+2}^2=y^2$. 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: $a^2+a_{m}^2=a_{n}^2$. Now, you know $a_{1},...,a_{m-1},a_{m+1},...,a_{n}$ because we have been substituting suitable values. So, we will have an equation like this: $a^2+a_{m}^2=a_{n}^2$. Now, we can easily solve this using pythagorean triples formula, because we know $a$. Hence we have found a suitable solution.

- 1 year, 2 months ago

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Such that what?

- 1 year, 2 months ago

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Such that $a_{1}^2+a_{2}^2=x^2$

- 1 year, 2 months 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 $x$ 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?

- 1 year, 2 months ago

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Uhh, this is a pretty vague question I believe.

What is $m$? What is $n$? Must $a_i$ be an integer? Can it be zero? If it is then I can always make a trillion solutions just by saying $a_i$ = 0 for all $i$ ranging from 1 to $n$.

Even if the numbers must be integers and positive, by letting $n$ 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 $a_i$ to be all different; that way you can make a pretty fun problem.

- 1 year, 2 months 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.

- 1 year, 2 months ago

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Even then your question is still too vague.

If $2|n$ then $m=\frac{n}{2}$ and $a_i=a_{(m+i)}$.

If $n=2k+1$ (with $k>1$) then $m=k+2$, $a_i=1$ with $i$ ranging from $1$ to $n-1$ and $a_n=4$.

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

Even as a bonus, if $n=b^2+c^2$ then $m=b^2$ and $a_i=c$ with $i$ ranging from $1$ to $m$ and $a_j=b$ with $j$ ranging from $m+1$ to $n$.

As long as $n$ is larger than 1... The sky is endless.

- 1 year, 2 months ago

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Sorry. All numbers are distinct

- 1 year, 2 months 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 $n<3$, obviously no.

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

$n=5$ is the tricky part. Back to the drawing table for now. (Hint: $m=4$, and you're using $3$ triples.)

Update: After doing extra research, I can see that there might exist infinitely many answers for $n=5$, but the general formula is unknown to me. However, here is just a solution to $n=5$ to satisfy my curiosity: $195^2+104^2+36^2+48^2=229^2$.

Update 2: Let's call this triple $x^2+y^2=z^2$ for the sake of clarity. Now, you can see there are infinitely many solutions so that $x$ is a multiple of 5. Needless to say, $x$ can be partitioned to $2$ other squares. (The reason, I'll leave you to finding out.) What about $y$? 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 $n>5$, you can just add in triples and quadruples until you get the necessary value of $n$. Voilà.

- 1 year, 2 months ago

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Uhh that is a very tough method

- 1 year, 2 months ago

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Well, as I have mentioned above, it's almost impossible to point out a "general formula" here (considering the lack of definitive $n$), 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.

- 1 year, 2 months ago

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But there is a very easy method!!!

- 1 year, 2 months 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 $n$.

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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Yes, I am more than willing to listen.

- 1 year, 2 months ago

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ok sure!!!

- 1 year, 2 months ago

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Just us pythagorean triplet formula

- 1 year, 2 months 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.

- 1 year, 2 months ago

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No I am sorry for partially giving a solution

- 1 year, 2 months ago

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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 $m$ in any $n$, for the reasons already said above.

- 1 year, 2 months ago

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No. I have presented them at a conference

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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Also, I am quite busy now

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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I am very grateful for that. Thanks.

- 1 year, 2 months ago

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Welcome!!!!!!!

- 1 year, 2 months ago

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Good bye!!!

- 1 year, 2 months ago

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You posted a problem related to this right?

- 1 year, 2 months 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.

- 1 year, 2 months ago

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Thank you very much!!!

- 1 year, 2 months ago

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That is just awesome!!!

- 1 year, 2 months ago

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You do have brilliant thinking skills on brilliant!!!

- 1 year, 2 months ago

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Thank you.

- 1 year, 2 months ago

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Welcome!!!

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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Yeah sure

- 1 year, 2 months ago

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Example: Find one 5-tuple $(a,b,c,d,e)$ such that, $a^2+b^2+c^2=d^2+e^2$

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

- 1 year, 2 months ago

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

Generalize it and you'll see some problems.

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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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 $x$, and if you can prove that you can always find a suitable $x$ then your proof will be complete.

- 1 year, 2 months ago

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Your way is good, but my way is also similar

- 1 year, 2 months ago

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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 $x$? Is it always possible?

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

- 1 year, 2 months ago

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Yes it is always possible

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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Ok sure

- 1 year, 2 months ago

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Please let me know

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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

- 1 year, 2 months 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 $x$. State your ideas here, if you have any.

- 1 year, 2 months ago

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What do you mean new comment?

- 1 year, 2 months ago

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State your ideas in a new comment here.

- 1 year, 2 months ago

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ok !!!!!!!!!!!!!!!!!!!!

- 1 year, 2 months ago

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We want one solution to the equation: $a_{1}^2+a_{2}^2+...+a_{m}^2=a_{m+1}^2+...+a_{n}^2$.

Now, choose a suitable $x$ such that $(a_{1},a_{2},x)$ forms a Pythagorean triple. Now, the problem reduces to finding one solution to $x^2+a_{3}^2+..+a_{m}^2=a_{m+1}^2+...+a_{n}^2$. Now, choose a suitable $x_{1}$ such that $(x,a_{3},x_{1})$ forms a Pythagorean triple. Keep doing this process on both sides until we land on an equation of the form $a^2+a_{m}^2=b^2+a_{n}^2$. Now, since we have been substituting values for $a_{1},a_{2},..,a_{m-1}$, we know the values of $a,b$. So, we can rewrite the equation as $a^2-b^2=a_{n}^2-a_{m}^2$ and solve the equation by "Simon's Factorisation". So, we have found a solution set.

- 1 year, 2 months ago

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

Start with an arbitrary $x$ larger than $100$. We need to find a suitable $x_1$, as stated above. Now, do either of these steps:

a/ If $x^2$ is odd, partition it into two numbers, $1$ and itself, then use the equation $x^2=(x_1-a_3)(x_1+a_3)$ to attribute $1$ to $x_1-a_3$ and $x^2$ to $x_1+a_3$. Obviously, $x_1$ and $a_3$ are both larger than $x$.

b/ If $x^2$ is even, divide by $4$ until the number is odd. Then do as above. Again, obviously, $x_1$ and $a_3$ are both larger than $x$ (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 $x$ 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.

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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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.

- 1 year, 2 months ago

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Yes. But I thought that this is trivial

- 1 year, 2 months ago

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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.

- 1 year, 2 months ago

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oh!!!!!!!!!!!!!!!!!!!!!!!!!!!

- 1 year, 2 months ago

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Sorry for wasting your time on this!!!

- 1 year, 2 months ago

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Thank you once again!!!!!!!!!!

- 1 year, 2 months ago

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You're welcome.

- 1 year, 2 months ago

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Thank you very much

- 1 year, 2 months ago

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

- 1 year, 2 months ago

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Steven Jim, do you have any comments??

- 1 year, 2 months 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"

- 1 year, 2 months ago

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