- #1

- 84

- 0

Is there a simple way to find the integer values of f(x)=(a+5x)/(6x+1) with a, x integers.

- Thread starter epsi00
- Start date

- #1

- 84

- 0

Is there a simple way to find the integer values of f(x)=(a+5x)/(6x+1) with a, x integers.

- #2

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

[tex]a = (6x + 1) n - 5x[/tex]

to get any integer

Is that what you meant by "finding the integer values"?

- #3

- 84

- 0

In fact, it's exactly the opposite I want. a is a given integer and I am looking for the value(s) of x that make f(x) take integer values. It's simple enough for small values of a but for large values, I cannot afford to check the values of x one by one.x, you can set

[tex]a = (6x + 1) n - 5x[/tex]

to get any integernyou like.

Is that what you meant by "finding the integer values"?

- #4

- 695

- 2

At least this cuts the search up to something on the order of the sqrt(a): now you need to iterate, from k=1 to a maximum of k=[itex]\lfloor \frac {2 + \sqrt{4+6a}} 6 \rfloor[/itex], checking if [itex]a \equiv k+1 \pmod {6k+1}[/itex], in which case both x=k and x=[itex]\left( \frac {5k+a} {6k+1} - 1 \right)[/itex] are two new solutions (or one, if these two "dual" solutions happen to be the same).

- #5

CompuChip

Science Advisor

Homework Helper

- 4,302

- 47

If n = (a + 5x) / (6x + 1), then you can solve x from this equation. That should give you something like x = (n - a) / (5 - 6n).

This will give integer values for x when (n - a) is a multiple of (5 - 6n), from which you will get the allowed values of a that will give you f(x) = n for your favorite integer n.

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 2K

- Replies
- 9

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 22

- Views
- 4K

- Replies
- 3

- Views
- 2K

- Replies
- 9

- Views
- 602

- Last Post

- Replies
- 5

- Views
- 2K

- Replies
- 1

- Views
- 1K