Anyone able to solve this equation?

[Ethan3141]

Hello, I am looking for a method in order to find integer solutions to this equation. Not sure if this is possible (methods other than just brute-forcing) however I have heard of equations called diophantine equations which are a system of equations in which you have fewer equations that variables and can solve for integer solutions.

[MATH]\frac{8+n}{4n-1} =k[/MATH]
I would also like to clarify I am looking for a full solution not a hint or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved).

Thanks, Ethan.

 

[Dr.Peterson]

I would also like to clarify I am looking for a full solution not a hint or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved).

Did you read our guidelines? See especially #3.

This site is a HELP site, not a DO EVERYTHING FOR ME site. At the least, before demanding a complete solution, you should tell us why you are not willing to do any work yourself.

 

[Steven G]

I would also like to clarify I am looking for a full solution not a hint or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved).

The funny thing is that you are serious. I had to look at my calendar to make sure it was not April 1st.

 

[Ethan3141]

Considering the solution is clear after multiplying by 4 I really wasn't asking for much. I didn't mean to cause offence and frankly didn't even think it was possible to cause offence. Instead of attacking me for my request you could've instead gave your interpretation of help. I mean you point out the fact this site is for helping and then you attack me instead of helping me. Well thanks for nothing, great help. I will definitely be using this forum more in the future!

 

[tkhunny]

Considering the solution is clear after multiplying by 4 I really wasn't asking for much. I didn't mean to cause offence and frankly didn't even think it was possible to cause offence. Instead of attacking me for my request you could've instead gave your interpretation of help. I mean you point out the fact this site is for helping and then you attack me instead of helping me. Well thanks for nothing, great help. I will definitely be using this forum more in the future!

This is a common complaint from those who do not actually want help. Personally, I do not believe "it would be very helpful to [you]" if we do all the work for you. That sort of thing doesn't fall under my definition of "helping".

Welcome, and let's learn some mathematics, shall we? It will be most helpful if you follow the guidelines.

 

[Dr.Peterson]

Considering the solution is clear after multiplying by 4 I really wasn't asking for much. I didn't mean to cause offence and frankly didn't even think it was possible to cause offence. Instead of attacking me for my request you could've instead gave your interpretation of help. I mean you point out the fact this site is for helping and then you attack me instead of helping me. Well thanks for nothing, great help. I will definitely be using this forum more in the future!

You're right that it isn't a hard problem (particularly if you were specifically looking for positive integer solutions). But in demanding (not just requesting) a full solution, you refused what we would like, which is a dialogue.

Can you explain why you added that line to what would otherwise have been a very reasonable question? Frankly, it is offensive, whether intended that way or not.

But let's set that aside ...

The equation can be solved most easily by what you might call "brute force" - just observing what few cases need to be tested, and doing so. There are probably several other ways. Want to discuss it? Show the specifics of the solution(s) you have in mind, and we can talk about alternatives.

 

[Deleted member 4993]

Considering the solution is clear after multiplying by 4 I really wasn't asking for much. I didn't mean to cause offence and frankly didn't even think it was possible to cause offence. Instead of attacking me for my request you could've instead gave your interpretation of help. I mean you point out the fact this site is for helping and then you attack me instead of helping me. Well thanks for nothing, great help. I will definitely be using this forum more in the future!

"....I really wasn't asking for much" - yes - but your "demand" was to act against our policies.

There are paid-websites that will provide you with answers - but we don't get paid. So we don't "have to" bow down to your demand.

Be polite - specially when it is you who is asking a favor......

 

[pka]

Hello, I am looking for a method in order to find integer solutions to this equation. Not sure if this is possible (methods other than just brute-forcing) however I have heard of equations called diophantine equations which are a system of equations in which you have fewer equations that variables and can solve for integer solutions.
[MATH]\frac{8+n}{4n-1} =k[/MATH]I would also like to clarify I am looking for a full solution not a hint or anything like that.

To Ethan, I am not doing any of that.
However, I you are still interested then LOOK @ THIS LINK.
The solution is not on that particular page. But if you make adjustments in the range of \(\displaystyle n\) you can find two solutions.

 

[Dr.Peterson]

I avoided brute force (assuming positive variables) by first solving [MATH]\frac{n+8}{4n-1} \ge 1[/MATH], since k is a positive integer. That cut down the search space very nicely.

I also checked k=0 and found that that is a solution (with negative n).

If k can be negative, examining the behavior of the LHS reveals that there are no such solutions. (Sketch the graph, for instance.)

It's also interesting to solve for n, which reveals a symmetry in the problem.

None of this is brute force; none of it is fancy number theory either. And all of it is hint, rather than full solution or standard method, because it is the former that is most interesting!

 

[Ethan3141]

I understand that the request of a full solution was wrong but that could've been clarified to me by a single sentence by one person. At the same time you could've provided your idea of help which would familiarise me with the system of this forum and how it works. Regardless what is done is done so I would just like to explain the solution for anyone else who might find it useful:

[MATH]\frac{33+4n-1}{4n-1}=4k[/MATH]
[MATH]\frac{33}{4n-1}=4k-1[/MATH]
Thus we know [MATH]4n-1 \vert 33 {\implies}4n-1\in[\pm1,\pm3,\pm11\pm33][/MATH]
Solving for n against each factor will find all integer solutions for n and k.

 

[Ethan3141]

Any help on how I could solve this equation for integer solutions for x and y (positive and negative) would be greatly appreciated. It is relevant to this question since it is using [MATH]n=x-y[/MATH] and [MATH]k=x+y[/MATH] with abit of rearrangement you get:

[MATH](4x-1)^2-(4y)^2=33[/MATH]
My attempt so far:

[MATH]y =\pm\sqrt\frac{2x^2-x-4}{2}[/MATH]
So far I can narrow it down to [MATH]x=2n_1 {\implies} y=2n_2[/MATH] and [MATH]2x^2-x-4{\ge}2{\implies}x{\ge}2{\vee}x{\lt}\frac{-3}{2}[/MATH] Is there anyway I can further restrict it (maybe an upperbound of for the two x cases? or something more to do with the squareroot having to be an integer?)

I hope this is more inline with the guidelines than the initial question. I really do apologise for the way I entered this forum with no respect for its guidelines or the forum users and hope to learn from my mistake

 

[Cubist]

I will definitely be using this forum more in the future!

I genuinely hope that you will use this forum again. All the people above want to help you. I suspect that you might have become too frustrated with this problem before you first came here for help, and you no longer wanted to look at the question yourself? If so, then I recommend that you ask for help a lot sooner next time (or move to another question). It isn't always productive to be "stuck" (making no progress) on a problem for too long.


Dr Peterson spotted the limited domain of n values...

I avoided brute force (assuming positive variables) by first solving [MATH]\frac{n+8}{4n-1} \ge 1[/MATH], since k is a positive integer. That cut down the search space very nicely.

...but if k is not limited to +ve integers, then you can simply extend this idea to...

[MATH]\frac{n+8}{4n-1} \ge 1[/MATH] or [MATH]\frac{n+8}{4n-1} = 0[/MATH] or [MATH]\frac{n+8}{4n-1} \le -1[/MATH]
...obviously subject to n being integer too. This is still a very limited search space. Another way of spotting this is to think about what happens if n becomes large, because the denominator would grow more quickly than the numerator, and would this scenario ever be able to produce an integer k value as n increases further?

 

[HallsofIvy]

You wrote "I would also like to clarify I am looking for a full solution not a hint or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved)."

Is it possible that you simply got those two clauses backwards and meant to write
"I would also like to clarify I am looking for a hint not a full solution or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved)"?

 

[Ethan3141]

I genuinely hope that you will use this forum again. All the people above want to help you. I suspect that you might have become too frustrated with this problem before you first came here for help, and you no longer wanted to look at the question yourself? If so, then I recommend that you ask for help a lot sooner next time (or move to another question). It isn't always productive to be "stuck" (making no progress) on a problem for too long.


Dr Peterson spotted the limited domain of n values...

...but if k is not limited to +ve integers, then you can simply extend this idea to...

[MATH]\frac{n+8}{4n-1} \ge 1[/MATH] or [MATH]\frac{n+8}{4n-1} = 0[/MATH] or [MATH]\frac{n+8}{4n-1} \le -1[/MATH]
...obviously subject to n being integer too. This is still a very limited search space. Another way of spotting this is to think about what happens if n becomes large, because the denominator would grow more quickly than the numerator, and would this scenario ever be able to produce an integer k value as n increases further?

I already solved the initial problem in the initial form but it is not nearly as useful to me as I originally thought so now I am trying to develop a solution in the form of a polynomial (The messages stating this appears above yours for me but it is awaiting approval, in these messages I apologise for how I disrespected the forum and forum users). I would also like to clarify that I did look at the problem prior to asking this: I coded a python program to brute force solutions as I simply didn't know where to start with developing a method to solve it, however I asked here for a method to solve it as it would be much more efficient to implement in my program than just using brute-force.

This whole awaiting approval is making this conversation very difficult, is there perhaps a way I can get rid of this restriction?

Solution mentioned above:

I understand that the request of a full solution was wrong but that could've been clarified to me by a single sentence by one person. At the same time you could've provided your idea of help which would familiarise me with the system of this forum and how it works. Regardless what is done is done so I would just like to explain the solution for anyone else who might find it useful:

[MATH]\frac{33+4n-1}{4n-1}=4k[/MATH]
[MATH]\frac{33}{4n-1}=4k-1[/MATH]
Thus we know [MATH]4n-1 \vert 33 {\implies}4n-1\in[\pm1,\pm3,\pm11\pm33][/MATH]
Solving for n against each factor will find all integer solutions for n and k.

Alternative solution (work in progress - need help):

Any help on how I could solve this equation for integer solutions for x and y (positive and negative) would be greatly appreciated. It is relevant to this question since it is using [MATH]n=x-y[/MATH] and [MATH]k=x+y[/MATH] with abit of rearrangement you get:

[MATH](4x-1)^2-(4y)^2=33[/MATH]
My attempt so far:

[MATH]y =\pm\sqrt\frac{2x^2-x-4}{2}[/MATH]
So far I can narrow it down to [MATH]x=2n_1 {\implies} y=2n_2[/MATH] and [MATH]2x^2-x-4{\ge}2{\implies}x{\ge}2{\vee}x{\lt}\frac{-3}{2}[/MATH] Is there anyway I can further restrict it (maybe an upperbound of for the two x cases? or something more to do with the squareroot having to be an integer?)

I hope this is more inline with the guidelines than the initial question. I really do apologise for the way I entered this forum with no respect for its guidelines or the forum users and hope to learn from my mistake

 

[Ethan3141]

You wrote "I would also like to clarify I am looking for a full solution not a hint or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved)."

Is it possible that you simply got those two clauses backwards and meant to write
"I would also like to clarify I am looking for a hint not a full solution or anything like that, it would be very helpful to me if you could solve this (assuming it can be solved)"?

I wish that was what I meant but I did mean what I initially wrote at that moment, however I now realise how naive, disrespectful and wrong it was of me to request this.

 

[Cubist]

This whole awaiting approval is making this conversation very difficult, is there perhaps a way I can get rid of this restriction?

The restriction only applies to your first 5 posts, so it should not be a problem any more.

 

[Deleted member 4993]

Any help on how I could solve this equation for integer solutions for x and y (positive and negative) would be greatly appreciated. It is relevant to this question since it is using [MATH]n=x-y[/MATH] and [MATH]k=x+y[/MATH] with abit of rearrangement you get:

[MATH](4x-1)^2-(4y)^2=33[/MATH]
My attempt so far:

[MATH]y =\pm\sqrt\frac{2x^2-x-4}{2}[/MATH]
So far I can narrow it down to [MATH]x=2n_1 {\implies} y=2n_2[/MATH] and [MATH]2x^2-x-4{\ge}2{\implies}x{\ge}2{\vee}x{\lt}\frac{-3}{2}[/MATH] Is there anyway I can further restrict it (maybe an upperbound of for the two x cases? or something more to do with the squareroot having to be an integer?)

I hope this is more inline with the guidelines than the initial question. I really do apologise for the way I entered this forum with no respect for its guidelines or the forum users and hope to learn from my mistake

[MATH](4x-1)^2-(4y)^2=33[/MATH]
[MATH](x - 1/4)^2 - (y)^2 = \ \left(\frac{\sqrt{33}}{2}\right)^2[/MATH]
This is an equation of a hyperbola.

Plot it using pka's favorite website (Wolframalfa).

You would see that the range of the solution space is unlimited.

You asked:

".... further restrict it (maybe an upperbound of for the two x cases"

I do not see any upperbound restriction on |x|.

 

[Dr.Peterson]

I understand that the request of a full solution was wrong but that could've been clarified to me by a single sentence by one person. At the same time you could've provided your idea of help which would familiarise me with the system of this forum and how it works. Regardless what is done is done so I would just like to explain the solution for anyone else who might find it useful:

[MATH]\frac{33+4n-1}{4n-1}=4k[/MATH]
[MATH]\frac{33}{4n-1}=4k-1[/MATH]
Thus we know [MATH]4n-1 \vert 33 {\implies}4n-1\in[\pm1,\pm3,\pm11\pm33][/MATH]
Solving for n against each factor will find all integer solutions for n and k.

Thanks. Actually, I usually try to give a hint along with a request for information, and in this case I started to edit my initial response to do so, but decided that my suggestion might not be in line with what you wanted. (I am not very good at Diophantine equations in general, and my domain-limiting idea is not a general technique you might be looking for.)

Interestingly, your more sophisticated approach requires you to check more cases (8) than mine (4 or 5)! But that doesn't surprise me. Solutions in this area often work that way, and also can depend on creativity.

 

[Dr.Peterson]

Any help on how I could solve this equation for integer solutions for x and y (positive and negative) would be greatly appreciated. It is relevant to this question since it is using [MATH]n=x-y[/MATH] and [MATH]k=x+y[/MATH] with abit of rearrangement you get:

[MATH](4x-1)^2-(4y)^2=33[/MATH]

This new equation can be solved easily by one of the few methods I know ...

Factor the LHS as [MATH]((4x-1)+4y)((4x-1)-4y)=33[/MATH], and consider each possible factorization of the RHS. This gives you eight systems of equations, four of which have integer solutions.

 

[Deleted member 4993]

[MATH](4x-1)^2-(4y)^2=33[/MATH]
[MATH](x - 1/4)^2 - (y)^2 = \ \left(\frac{\sqrt{33}}{2}\right)^2[/MATH]
This is an equation of a hyperbola.

Plot it using pka's favorite website (Wolframalfa).

You would see that the range of the solution space is unlimited.

You asked:

".... further restrict it (maybe an upperbound of for the two x cases"

I do not see any upperbound restriction on |x|.

Nope - there is an upper-limit (see response #19 above).

I had forgotten the biggest restriction of all - it is a Diophantine problem!!!