Re: diophantine equation 1/x + 1/y = 1/n

On Aug 22, 2:40 pm, "Philippe 92" <nos...@xxxxxxxxxxxx> wrote:
mukesh tiwari wrote :





On Aug 22, 1:03 pm, Proginoskes <CCHeck...@xxxxxxxxx> wrote:
On Aug 22, 12:32 am, mukesh tiwari <mukeshtiwari.ii...@xxxxxxxxx>
wrote:

hello everybody i want to know how many distinct solution of
equation(1/x+1/y=1/n) for given value of n .

If you are allowing negative integers, the answer is the number of
factors of (n^2). If you want only positive solutions, take the number
of factors of n^2, add 1, and then divide the whole thing by 2.

for example if n=4 then
three distinct solution (5,20)(6,12) and (8,8).
plz tell me the algorithm to solve this problem .

Basically, you combine the terms of 1/n - 1/y and use the fact that
this fraction reduces to 1/x.

negative numbers are not allowed . actually initially i was also using
the same algorithm but i think it seems to be failed for n=1260
answer is 113 . chk out this link .
<http://projecteuler.net/index.php?section=problems&id=110>

Hi,

1260^2 = 2^4 * 3^4 * 5^2 * 7^2 has
5*5*3*3 = 225 divisors (including 1 and 1260^2)
(225+1)/2 = 113

This is not a too large number, and the brute force agrees and gives
the full list of the 113 solutions.

Where is the failure ?

(your direct link to your pb110 fails, so does the internal link
from pb110 to pb108. This allways goes back to your index page)

And this reformulates your problems as :
"find the smallest square whose number of divisors is ..."

See also my own version of this problemhttp://chephip.free.fr/pba_en/pb036.html

and the topic of last month
"A lost treasure (Series within Parallel resistor combinations.)"
by Quentin Grady on sci.math
Message-ID: <ghhr93lgbf8r1vboga40422v2aecl4q0ig@xxxxxxx>

Regards.

--
Philippe C., mail : chephip+n...@xxxxxxx
site :http://chephip.free.fr/ (recreational mathematics)

thnkx a lot may be i confused and i have calculated only primefactor
of 1260 . thnkx again ....

.

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