Find the total number of positive integer solutions of:

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- Jun 30th 2009, 02:49 PMperashFind the number
Find the total number of positive integer solutions of:

- Jun 30th 2009, 09:35 PMProve It
- Jun 30th 2009, 10:54 PMdhiab
**HEllo : remark**

http://www.mathramz.com/xyz/latexren...320f38cb95.png

**is not :**

http://www.mathramz.com/xyz/latexren...45b369c0d5.png

**conclusion :**

http://www.mathramz.com/xyz/latexren...bec49c6522.png

**THANKS (Evilgrin)** - Jul 1st 2009, 01:03 AMMoo

So here, we have

And since m and n are positive integers, we can choose

We get

For m to be an integer, we need 4012-n to divide the numerator.

I conjecture that the number of possibilities is the number of divisors of (more or less, because we have to remember that n<4012)

But I really don't know yet how to prove it :(

Quote:

- Jul 1st 2009, 05:22 AMrunning-gag
- Jul 1st 2009, 08:37 AMBobP

To begin with, since and are positive integers, neither fraction on the LHS can be greater than , (because then, the other would have to be negative).

So, if then and similarly .

Multiplying throughout by , , so must be an integer multiple of 2006, (since the LHS is an integer).

Let where is an integer, then .

Substituting, , so .

Solving this for . What we need now is that should be an integer and in such a way that the resulting values of and are both integers.

One obvious possibility is , leading to and , and this produces the solution .

Sadly, at this point I grind to a halt (Worried), I don't know of a systematic way of finding further values of such that is a perfect square. (Help anyone ?)

However, notice that the calculated solution cancels down to , which is kind of obvious and does suggest a second solution.

. - Jul 1st 2009, 08:48 AMBobP
Ooops !! (Giggle).

I should have said that one obvious value of was leading to .

(Itwasntme) - Jul 2nd 2009, 01:34 AMmalaygoel
Let be .

Then

now, if

or

then,

hence, it comes down down to factorise 12036 into two factors(u,v), and for each pair we will get two values of k(one positive and one negative).

Since we are interested only in positive values,

Prime factorisation of 12036 will be helpful for finding factors:

Example:

u=102

v=118

gives k=48400,r=3520 - Jul 4th 2009, 01:58 AMBobP
Thanks for that, a useful idea that I don't recall seeing before.

It does produce further solutions, m=12980 n=7480, m=2018036 n=4024, m=8054090 n =4015. (I don't know whether I got all of them !)

I've still not got my head round it totally though, since this method doesn't seem to pick up my original solution of m=12036 n=8024, and I'm not sure whether there might not be others. - Jul 4th 2009, 02:16 AMmalaygoel
- Jul 4th 2009, 09:53 AMSoroban
Hello, perash!

I had a good start on it, but I haven't finished it yet . . .

Quote:

Find the number of positive integer solutions of: .

Multiply by

Solve for

Since is a positive integer, then there are two cases:

. . [1] is a factor of 4012

. . [2] is a factor of

Case [1]: . . . which has 12 divisors.

. . So we have 12 solutions:

. . . . .

Case [2]: I found three more solutions . . . so far.

. .

At this point, I desperately needed a nap . . .