[Guest]

The infinite continued fraction: x +1/ (y+ 1/(x + 1/(y+ ....))) is denoted by F(x, y).

Determine all possible quadruplet(s)(p, q, r, s) of positive integers that satisfy this equation:

(F(p, q))^-1 - 3*(F(r, s))^-1 = 1/2, where p, q, r and s are distinct.

[Guest]

I'm looking forward to tackling this. Thanks, KS.

[Guest]

Using this representation of the function, F = (x + 1) / ((y + 1) / F)), I have exhausted all possible distinct quadruplets of positive integers. Answer: none exists. If this formula is wrong please post a spoiler for me. Thanks.

[bushindo]

Using this representation of the function, F = (x + 1) / ((y + 1) / F)), I have exhausted all possible distinct quadruplets of positive integers. Answer: none exists. If this formula is wrong please post a spoiler for me. Thanks.

This may be the reason for your results

F = x +1/ (y+ 1/(x + 1/(y+ ....))) = x + 1/(y + 1/F), which is an entirely different expression than the one in your post, F = (x + 1) / ((y + 1) / F)).

Edited August 19, 2009 by bushindo

[Guest]

Based on the recursive form:

f(x, y) = x + (1 / y + (1 / f(x, y)))

I was able to reduce it to:

f(x, y) = (x(y + 1)) / y

then substituting in the equation

(p(q+1))/q - (3r(s+1))/s = 1/2

finally I reduce it to the following equation

2ps(q+1) - 6rq(s+1) = qs

So any combination of p, q, r and s that satisfies the last equation is a possible answer.

The problem is that I don't know how to put it into a generic form of the answer.

[Guest]

Interesting

Edited August 20, 2009 by jerbil

[Guest]

I get F = x + 1/(y + 1/F), which is slightly different.

[Guest]

Using the correct equation, F = x + 1/(y + 1/F) {thanks bushindo and jerbil}, I get this reduction.

F = x + 1/(y + 1/F)

F = x + 1/y + F

F - F = x + 1/y

0 = x + 1/y

Are there any sets of positive integers that fit this equation?

[Guest]

Using the correct equation, F = x + 1/(y + 1/F) {thanks bushindo and jerbil}, I get this reduction.

F = x + 1/(y + 1/F)

F = x + 1/y + F

F - F = x + 1/y

0 = x + 1/y

Are there any sets of positive integers that fit this equation?

Actually your doing some reductions wrong... I think

How do you go from F = x + 1/(y + 1/F) to F = x + 1/y + F

Don't take it the wrong way but I think you're doing some magic there I don't know about.

[Guest]

By the way, I found my solution wrong... sorry 4 that

Based on the recursive form:

f(x, y) = x + (1 / y + (1 / f(x, y)))

I was able to reduce it to:

f(x, y) = (x(y + 1)) / y

then substituting in the equation

(p(q+1))/q - (3r(s+1))/s = 1/2

finally I reduce it to the following equation

2ps(q+1) - 6rq(s+1) = qs

So any combination of p, q, r and s that satisfies the last equation is a possible answer.

The problem is that I don't know how to put it into a generic form of the answer.

I rechecked the algebra and I got stuck "somewhere"

Having that

f(x, y) = x + 1/(y + 1/f(x, y)) ......... 1

I reduce as follows, .... I replaced f(x, y) with only f

f = x + 1/(y + 1/f) .................. 1

f = x + 1/((yf + 1)/f)

f = ((xyf + x)/f + 1)/((yf + 1)/f)

f = ((xyf + x + f)/f)/((yf + 1)/f)

f = (xyf + x + f)/(yf + 1)

yf² + f = xyf + x + f

yf² = xyf + x

yf² - xyf - x = 0 .......... 2

I believe equation 2 (the last reduction) can be factorized to have an expression f = g(x, y) where g(x, y) doesn't have f as part of it. Unfortunately, I haven't been able to factorize the expression, so I'm not even sure if that's possible.

If possible, the f = g(x, y) could then be sustituted in the f(p, q) - 3f(r, s) = 1/2 expression, getting an equation involving the p, q, r, s parameters and then any combination of values for p, q, r, s that satisfies the equation is a solution.

If anyone is able to factorize equation 2 please let me know.

*Note: Is easier to follow the reductions I did if writing them into paper to see the algebraic form. All the parenthesis in here make it harder to follow.

Edited August 20, 2009 by VidalPhoenix

[superprismatic]

By the way, I found my solution wrong... sorry 4 that

I rechecked the algebra and I got stuck "somewhere"

Having that

f(x, y) = x + 1/(y + 1/f(x, y)) ......... 1

I reduce as follows, .... I replaced f(x, y) with only f

f = x + 1/(y + 1/f) .................. 1

f = x + 1/((yf + 1)/f)

f = ((xyf + x)/f + 1)/((yf + 1)/f)

f = ((xyf + x + f)/f)/((yf + 1)/f)

f = (xyf + x + f)/(yf + 1)

yf² + f = xyf + x + f

yf² = xyf + x

yf² - xyf - x = 0 .......... 2

I believe equation 2 (the last reduction) can be factorized to have an expression f = g(x, y) where g(x, y) doesn't have f as part of it. Unfortunately, I haven't been able to factorize the expression, so I'm not even sure if that's possible.

If possible, the f = g(x, y) could then be sustituted in the f(p, q) - 3f(r, s) = 1/2 expression, getting an equation involving the p, q, r, s parameters and then any combination of values for p, q, r, s that satisfies the equation is a solution.

If anyone is able to factorize equation 2 please let me know.

*Note: Is easier to follow the reductions I did if writing them into paper to see the algebraic form. All the parenthesis in here make it harder to follow.

Use the quadratic formula to get f=(xy + sqrt(x²y² + 4xy))/(2y).

[Guest]

Use the quadratic formula to get f=(xy + sqrt(x²y² + 4xy))/(2y).

Oh that's true, I had forgotten about the quadratic formula, thx.

But using the formula will create 2 different set of answers

for:

* f=(xy + sqrt(x²y² + 4xy))/(2y)

* f=(xy - sqrt(x²y² + 4xy))/(2y)

So it's starting to get more complex, but I'll give it a try

[Guest]

Well I guess I have the solution, though it gets kinda messy considering divitions can't be seen like on paper, but here it is:

Having originally reached to yf² - xyf - x = 0

I used the quadratic formula to isolate f getting f = (xy +/- <sqrt>(x²y² + 4xy)) / 2y

Then since the equation we have to evaluate uses f ^-1(x, y) rather than f(x, y) I calculated the value of f ^-1 being f ^-1 = 2y / (xy +/- <sqrt>(x²y² + 4xy))

Then considering the original equation to solve f ^-1(p, q) - 3f ^-1(r, s) = 1/2

I substituted the values, resulting in (2q / (pq +/- <sqrt>(p²q² + 4pq))) - (6s / (rs +/- <sqrt>(r²s² + 4rs))) = 1/2

Now this equation can provide us every posible solution for the problem as long as we keep in mind that p,q,r,s > 0 and they're all different from each other. If the values chosen for p,q,r,s stick to those rules and the equation, then those values are one of the solutions to the problem.

* The +/- means that you have to consider both addition and substraction, this is derived from the quadratic formula x = (-b +/- <sqrt>(b² - 4ac))/2a

*** Thanks to Superprismatic for the quadratic formula idea

[superprismatic]

Oh that's true, I had forgotten about the quadratic formula, thx.

But using the formula will create 2 different set of answers

for:

* f=(xy + sqrt(x²y² + 4xy))/(2y)

* f=(xy - sqrt(x²y² + 4xy))/(2y)

So it's starting to get more complex, but I'll give it a try

You can forget the second one. After all, it is negative and it is clear that the solution must be positive.