Solve equation of arithmetic progression

[Mondo]

I am trying to find a solution/s for the following equation:

[imath]x^2 - y^2 -z^2 = 20[/imath]
knowing that x,y,z are consecutive, positive integers of some arithmetic progression.

So we can see it must be decreasing sequence, otherwise the result would be negative.
I wrote following equations:
[math]20 = x^2 -y ^2-z^2 \\ x = y + d\\ y = z + d\\ z = x - 2d\\[/math]then substituting (2) and (3) into (1)
[imath]20 = (z+2d)^2 - (z -d)^2 -z^2[/imath]

But even solving this quadratic equation doesn't get me any closer to the solution. Am I missing some equation or it cannot be solve in this way?

Thank you.

 

[Harry_the_cat]

I don't think it will help much but your z-d should be z+d.

 

[Dr.Peterson]

I am trying to find a solution/s for the following equation:

[imath]x^2 - y^2 -z^2 = 20[/imath]
knowing that x,y,z are consecutive, positive integers of some arithmetic progression.

So we can see it must be decreasing sequence, otherwise the result would be negative.
I wrote following equations:
[math]20 = x^2 -y ^2-z^2 \\ x = y + d\\ y = z + d\\ z = x - 2d\\[/math]then substituting (2) and (3) into (1)
[imath]20 = (z+2d)^2 - (z -d)^2 -z^2[/imath]

But even solving this quadratic equation doesn't get me any closer to the solution. Am I missing some equation or it cannot be solve in this way?

Thank you.

You could probably do it this way, but I generally start such a question by supposing that the three numbers in arithmetic progression are x-d, x, and x+d. The symmetry makes the resulting equations easier.

But I seem to have proved that if the numbers are positive, then my d must be negative, so they are in reverse order (x>y>z). And your equations seem to assume this, if your d is positive. Is that the intent of the problem?

Where did the problem come from? Have you quoted it exactly?

 

[BigBeachBanana]

Where did the problem come from?

I've seen a similar question before on Project Euler which is meant as a programming challenge.

 

[Mondo]

@Dr.Peterson the source that @BigBeachBanana gave is a good source of the problem. However here I only focus on the first part with the objective of finding [math]x, y ,z[/math]@Dr.Peterson, the way you wrote the arithmetic progression doesn't make the problem any simples. After all we miss one equation - having more unknowns than equations. So how was the solution obtained?

 

[Dr.Peterson]

@Dr.Peterson, the way you wrote the arithmetic progression doesn't make the problem any simpler. After all we miss one equation - having more unknowns than equations. So how was the solution obtained?

It does make the problem simpler, because there are fewer variables, and some terms cancel out.

This is a Diophantine equation. What have you learned about those? Or if this is a new idea to you, can you see any ideas to use, perhaps involving divisibility? (I didn't say it was easy, just easier!)

Of course, the problem quoted gives the answer to the problem you asked us! But in particular, it does show us that they don't mind having a negative common difference.

 

[Mondo]

Thank you for the hints but I don't see how divisions may help here.
Before I tackled it again myself I wanted to see if wolframalpha can get an amswer and surprisingly no - https://www.wolframalpha.com/input?...ationsCalculator",+"equation4"}+->"z+=+x+-2d"

 

[Dr.Peterson]

Thank you for the hints but I don't see how divisions may help here.
Before I tackled it again myself I wanted to see if wolframalpha can get an amswer and surprisingly no - https://www.wolframalpha.com/input?i=system+equation+calculator&assumption={"F",+"SolveSystemOf4EquationsCalculator",+"equation1"}+->"20+=+x^2+-+y^2+-z^2"&assumption={"F",+"SolveSystemOf4EquationsCalculator",+"equation3"}+->"y+=+z+++d"&assumption="FSelect"+->+{{"SolveSystemOf4EquationsCalculator"},+"dflt"}&assumption={"F",+"SolveSystemOf4EquationsCalculator",+"equation2"}+->"x+=+y+d"&assumption={"F",+"SolveSystemOf4EquationsCalculator",+"equation4"}+->"z+=+x+-2d"

You need to use the fact that the variables are positive integers; this is not just a system of equations.

Have you looked up "Diophantine equations"? That is where you need to focus your attention (and where divisibility comes in).

 

[Mondo]

@Dr.Peterson, yes with an additional condition [imath]x > 0[/imath] and an integer I was able to find a solution with the help of wolfram.

Have you looked up "Diophantine equations"? That is where you need to focus your attention (and where divisibility comes in).

Yes, here is where I am stuck now:
[math]20 = (x+d)^2 - x^2 - (x-d)^2 \\ 20 = -x^2 + 4xd[/math]But how can I get the greatest common dividor of [imath]x^2[/imath] and [imath]4xd[/imath] ?

 

[Dr.Peterson]

Yes, here is where I am stuck now:
[math]20 = (x+d)^2 - x^2 - (x-d)^2 \\ 20 = -x^2 + 4xd[/math]But how can I get the greatest common dividor of [imath]x^2[/imath] and [imath]4xd[/imath] ?

This is not about a GCD, just about divisors.

If x(4d-x) = 20, then x must be a (positive) divisor of 20. What can it be? Then, what can d be?

 

[Mondo]

This is not about a GCD, just about divisors.

If x(4d-x) = 20, then x must be a (positive) divisor of 20. What can it be? Then, what can d be?

I agree, but is it now about "guessing" the answer or there is a systematic method that will return [imath]x = 10[/imath]?
Also, I need to prove that a given solution is the only one that satisfies given equation.

 

[Dr.Peterson]

I agree, but is it now about "guessing" the answer or there is a systematic method that will return [imath]x = 10[/imath]?
Also, I need to prove that a given solution is the only one that satisfies given equation.

No, this is not mere guessing. It will prove the uniqueness of the answer. What I suggested is very systematic. Please try, and show us your work.

 

[Mondo]

@Dr.Peterson , I am sorry for a late response, I've been sick for a few days.
So I solved two potential x values that solve the initial equation [imath]-x^2+4xd - 20 = 0[/imath]
One x solution is [math]x_1 = 2(d + \sqrt(d^2-5))[/math] the other is [math]x_1 = 2(d - \sqrt(d^2-5))[/math].
Then I took [math]x_1 = 2(d + \sqrt(d^2-5))[/math]] and solved it for
[math]d = \frac{x^2}{4x} - \frac{20}{4x}[/math]And here I am stuck because if I now plug the formula for d into initial equation involving both x and d I will get [imath]20 = 20[/imath]
So how should I proceed now? I have an equation for d that involves only x and I have two equation for x.

Additional questions:
1.
at some point of the calculations I found my self in this situation: [math]-\sqrt{d^2-5} = \frac{x}{2} -d[/math]Now is it legal to raise both sides to a second power to get rid of the square root funtion? My main concern is the leading negative sign which will be lost after a power function was applied.
2.

If x(4d-x) = 20, then x must be a (positive) divisor of 20. What can it be? Then, what can d be?

How can you tell x must be a divisor of 20 at this stage? The fact that it must be positive also doesn't come from this equation, but from the fact we have an arithmetic progression here. So if X is not positive we would ended up with a negative number, and not 20 as we have.

 

[Dr.Peterson]

And here I am stuck because if I now plug the formula for d into initial equation involving both x and d I will get 20=2020 = 2020=20
So how should I proceed now? I have an equation for d that involves only x and I have two equation for x.

Solving for one variable and then plugging that back into the equation is just a way to check the solution; what you observe will always happen.

The trouble is that in this work, you are not using the most important fact, which is that x and d must be integers. Equations alone will not solve the problem.

How can you tell x must be a divisor of 20 at this stage?

Since the three numbers are consecutive positive integers, x and d both have to be integers. So if x(4d-x) = 20, then x must be a (positive) divisor of 20, because x and 4d-x are integer factors of 20.

So you can list all possible values of x, and see what d is; if it's an integer, you have an answer. There are only four possible values of x to consider.

The fact that it must be positive also doesn't come from this equation, but from the fact we have an arithmetic progression here. So if X is not positive we would ended up with a negative number, and not 20 as we have.

Actually, the fact that x must be positive comes from the statement of the problem, not from being an arithmetic progression (which doesn't even require integers): "knowing that x,y,z are consecutive, positive integers of some arithmetic progression".

 

[Mondo]

So you can list all possible values of x, and see what d is; if it's an integer, you have an answer. There are only four possible values of x to consider.

So I was right with my "guessing" comment above - how do you see only fours possible values of x? I would need to literally go from x = 1, d =1 in a loop and see what potential values can return 20.
Also, d=0.5 multiplied by a reasonable x will also return an integer

 

[Dr.Peterson]

So I was right with my "guessing" comment above - how do you see only fours possible values of x? I would need to literally go from x = 1, d =1 in a loop and see what potential values can return 20.
Also, d=0.5 multiplied by a reasonable x will also return an integer

This is a sequence of integers, right? Can the common difference be 0.5?? Read the problem again.

What positive integers are there that are factors of 20? Just list them. There are six. And if you think just a little more deeply, you could reduce the number to try still further; for example, if x(4d-x) = 20, you can conclude that x can't be odd, since then the product would have to be even. (I think that's part of what I was remembering when I said 4.)

So I was right with my "guessing" comment above

I agree, but is it now about "guessing" the answer or there is a systematic method

As I said before, "guessing" can be systematic and not merely "brute force". This is intelligent guessing, and is standard in this sort of problem. So your comment is wrong, in assuming it can't be both.

You must not have done a lot of math if you think anything that doesn't take you directly to a unique answer is not math! We do students a disservice when we give the impression that all of math works like algebra (though even in algebra, we often "guess", e.g. in factoring a polynomial).

 

[Mondo]

This is a sequence of integers, right? Can the common difference be 0.5?? Read the problem again.

Right, it must be an integer, my mistake.

As I said before, "guessing" can be systematic and not merely "brute force". This is intelligent guessing, and is standard in this sort of problem. So your comment is wrong, in assuming it can't be both.

Agree again. I think I got blinded by looking for a very algebraic solution that after reorganization of terms and maybe plugging one equation into the other would yield x = 10; d = 3. I think you suggestion of Diophantine equations confused me in a search for a method that would clearly produce x and d. While here we just narrowed down the set of potential solutions to just a few terms, easy to test.

Also, can you also comment on this:

at some point of the calculations I found my self in this situation: −d2−5=x2−d-\sqrt{d^2-5} = \frac{x}{2} -d−d2−5=2x−dNow is it legal to raise both sides to a second power to get rid of the square root funtion? My main concern is the leading negative sign which will be lost after a power function was applied.

 

[Dr.Peterson]

1.
at some point of the calculations I found my self in this situation: [math]-\sqrt{d^2-5} = \frac{x}{2} -d[/math]Now is it legal to raise both sides to a second power to get rid of the square root funtion? My main concern is the leading negative sign which will be lost after a power function was applied.

Squaring can produce extraneous solutions, which is your concern. (An extraneous solution in this situation is a solution to the equation with the opposite sign.)

It's legal to do this in finding a solution, but not in proving something, because the result is not an equivalent equation. Whether it is useful depends on the details of what you do with it, and since this is not a useful path, as far as I can see, I didn't comment on it.

 

[Mondo]

Makes sense. Thank you very much @Dr.Peterson for a help in this thread. I think all is clear now