# 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
7 months, 1 week ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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.

- 7 months, 1 week ago

Such that what?

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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?

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

Sorry. All numbers are distinct

- 7 months, 1 week ago

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.

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.

- 7 months, 1 week ago

Uhh that is a very tough method

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

But there is a very easy method!!!

- 7 months, 1 week ago

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 $n$.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Yes, I am more than willing to listen.

- 7 months, 1 week ago

ok sure!!!

- 7 months, 1 week ago

Just us pythagorean triplet formula

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

No I am sorry for partially giving a solution

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

No. I have presented them at a conference

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Also, I am quite busy now

- 7 months, 1 week ago

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

- 7 months, 1 week ago

I am very grateful for that. Thanks.

- 7 months, 1 week ago

Welcome!!!!!!!

- 7 months, 1 week ago

Good bye!!!

- 7 months, 1 week ago

You posted a problem related to this right?

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Thank you very much!!!

- 7 months, 1 week ago

That is just awesome!!!

- 7 months, 1 week ago

You do have brilliant thinking skills on brilliant!!!

- 7 months, 1 week ago

Thank you.

- 7 months, 1 week ago

Welcome!!!

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Yeah sure

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

Generalize it and you'll see some problems.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

Your way is good, but my way is also similar

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

Yes it is always possible

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Ok sure

- 7 months, 1 week ago

Please let me know

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

@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.

- 7 months, 1 week ago

What do you mean new comment?

- 7 months, 1 week ago

State your ideas in a new comment here.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

Yes. But I thought that this is trivial

- 7 months, 1 week ago

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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Sorry for wasting your time on this!!!

- 7 months, 1 week ago

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

- 7 months, 1 week ago

You're welcome.

- 7 months, 1 week ago

Thank you very much

- 7 months, 1 week ago

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

- 7 months, 1 week ago

Steven Jim, do you have any comments??

- 7 months, 1 week ago

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"

- 7 months, 1 week ago