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

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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.

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?

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.

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.

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.

@Mohammed Imran
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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.

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 $n$.

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.

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 $(a,b,c,d,e)=(3,4,37,35,13)$

@Mohammed Imran
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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.

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.

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.

@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.

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## Comments

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TopNewestFind 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.

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Such that what?

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

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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?

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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.

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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.

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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.

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Sorry. All numbers are distinct

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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

mightexist 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

shouldexist infinitely many solutions (hopefully).For $n>5$, you can just add in triples and quadruples until you get the necessary value of $n$. Voilà.

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Uhh that is a very tough method

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$n$), but those are the clues to at least show that there are infinitely many answers.

Well, as I have mentioned above, it's almost impossible to point out a "general formula" here (considering the lack of definitiveAlso, can I know why you find it tough? To me this is completely undestandable, actually.

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But there is a very easy method!!!

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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 $n$.

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

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Yes, I am more than willing to listen.

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ok sure!!!

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Just us pythagorean triplet formula

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The idea of the formula is still insufficient. While there are infinitely many Pythagorean numbers, there are not

infinitelymany triples for a certain number. This is the dead end here.Log in to reply

No I am sorry for partially giving a solution

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Again, your solution does not even

guaranteethere would be a solution for $m$ in any $n$, for the reasons already said above.Log in to reply

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Also, I am quite busy now

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

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Good bye!!!

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You posted a problem related to this right?

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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.

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Thank you very much!!!

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That is just awesome!!!

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You do have brilliant thinking skills on brilliant!!!

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Thank you.

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Welcome!!!

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

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Yeah sure

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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 $(a,b,c,d,e)=(3,4,37,35,13)$

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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.

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

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The problem with your method is that you can't ensure that you

will be ableto find a suitable $x$, and if you can prove that you canalwaysfind a suitable $x$ then your proof will be complete.Log in to reply

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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

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

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

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Please let me know

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

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

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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.

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What do you mean new comment?

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State your ideas in a new comment here.

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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.

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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.

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

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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.

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nottrivial 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.Log in to reply

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Thank you very much

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

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Steven Jim, do you have any comments??

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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"

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