- 29 Jun '12 11:58Rearranging gives

y = 11*x/(x-11)

we need whole number values for y, most obvious is x = 12 which gives y = 132, there's also x = 22, y = 22.

For the general case, I guess for prime numbers the only values would be x = a + 1, y = (a+1)*a and x = 2*a, y = 2*a. Still working on the details of testing for things being coprime! - 29 Jun '12 20:42That would just about cover it, yes. ðŸ™‚

1/x + 1/y = 1/11

0 = xy - 11x - 11y

121 = xy - 11x - 11y + 121

121 = (x - 11) (y - 11)

121 can be a factor of two integers as

11 x 11 ; which gives 1 / 22 + 1 / 22 = 1 / 11

121 x 1 ; which gives 1 / 132 + 1 / 12 = 1 / 11

(-121) x (-1) ; which gives 1 / (-110) + 1 / 10 = 1 / 11

(-11) x (-11) ; which gives no solution as x and y can't be zero.

The same as you pointed out works for all primes. Composite numbers can be a bit trickier as there are so many ways to derive them by multiplying two integers together. - 30 Jun '12 14:32 / 3 edits

rearrange to y = 11x/(x-11)*Originally posted by talzamir***That would just about cover it, yes. ðŸ™‚**

1/x + 1/y = 1/11

0 = xy - 11x - 11y

121 = xy - 11x - 11y + 121

121 = (x - 11) (y - 11)

121 can be a factor of two integers as

11 x 11 ; which gives 1 / 22 + 1 / 22 = 1 / 11

121 x 1 ; which gives 1 / 132 + 1 / 12 = 1 / 11

(-121) x (-1) ; which gives 1 / (-110) + 1 / 10 = 1 / 11

(-11) x (-11) ; which gives e a bit trickier as there are so many ways to derive them by multiplying two integers together.

if x is n*11 we have:

Y = 11*11*n/(n-1)*11

= 11n/(n-1)

n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:

x = 0

x = 22

x = -110

x = 132

let us try for solutions where x = n*11 + b, where b is between 1 and 10

then y = 11*(n*11 + b)/(n*11+b - 11)

Here the top is clearly a multiple of11, and the bottom is clearly NOT a multiple of 11, so this only has integer solutions when the bottom is plus or minus 1

so this gives us the solutions

n=0, b = 10, x = 10

n=1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132 - 30 Jun '12 17:16 / 2 edits

Interesting that in the original equation, x=0 can't be a solution, but it can once you solve the equation for y. Once solved for y, x can't be 11. But x=11 was not going to produce a=11 in the original equation, so I guess we just manipulate things as needed to give the highest number of solutions. ðŸ˜›*Originally posted by iamatiger***rearrange to y = 11x/(x-11)**

if x is n*11 we have:

Y = 11*11*n/(n-1)*11

= 11n/(n-1)

n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:

x = 0

x = 22

x = -110

x = 132

let us try for solutions where x ...[text shortened]... =1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132 - 01 Jul '12 12:37

Yes 0 doesn't work as a solution because in the following:*Originally posted by iamatiger***rearrange to y = 11x/(x-11)**

if x is n*11 we have:

Y = 11*11*n/(n-1)*11

= 11n/(n-1)

n is divisible by n-1, only for n=0 or n= 2 and there are no integers n-1 for which 11/(n-1) is an integer, except for n= 0 , n = 2, n = -10 and n=12

so where x is 0 mod 11 we have the solutions:

x = 0

x = 22

x = -110

x = 132

let us try for solutions where x ...[text shortened]... =1, b = 1, x = 12

i.e. the only solutions are x = -110, x=0, x = 10, x = 12, x=22 and x = 132

1/x + 1/y = 1/11

to simplify this, we take 1/x from both sides, this doesn't work if x is 0 because then we are taking infinity from both sides, so we have assumed x /= 0, so we get to

1/y = 1/11 - 1/x {x =/ 0}

that assumption then carries through, so I should have got the pairs:

-1/110 + 1/10 = 1/11

1/12+ 1/132 = 1/11

1/22 + 1/22 = 1/11

which are the only possibilities - 01 Jul '12 19:01 / 4 edits

No one has posted the generic solution yet, so here it goes.*Originally posted by talzamir***Could a generic solution be found for a 1 / x + 1 / y = 1 / a where a is some integer > 0 ?**

Solving the initial equation for y, we get:y=ax/(x-a) [x!=0, y!=0]

to get a whole number for y, x = a+1 or a-1, or

x is some multiple of a, say ca. theny=ca^2/(ca-a)

...such that the fraction makes a whole number

=ca^2/a(c-1)

=ca/(c-1)

obviously c=2 works, but not c=0 because that would make y=0

the only other way to get a whole number is to make the denominator a or -a. Making the denominator magnitude any bigger means that c/(c-1) is forever (in magnitude) approaching 1 without actually hitting it; thus not evenly divisible.

So,c=2 (again) and c = 1 - a

...and that is all the solutions, for any integer*a*. [a!=0]

x=a+1 y=a^2+a

x=a-1 y=a-a^2

x=2a y=2a

x=a-a^2 y=a-1 - 01 Jul '12 22:57 / 1 edit

Generalising talzamir's approach*Originally posted by SwissGambit***No one has posted the generic solution yet, so here it goes.**

Solving the initial equation for y, we get:y=ax/(x-a) [x!=0, y!=0]

to get a whole number for y, x = a+1 or a-1, or

x is some multiple of a, say ca. theny=ca^2/(ca-a)

...such that the fraction makes a whole number

=ca^2/a(c-1)

=ca/(c-1)

obviousl ...[text shortened]...

x=a+1 y=a^2+a

x=a-1 y=a-a^2

x=2a y=2a

x=a-a^2 y=a-1

1/x + 1/y = 1/a

0 = xy - ax - ay

a^2 = xy - ax - ay + a^2

a^2 = (x - a) (y - a)

So take all the ways of factoring a^2 into two integers, and in each case add A to each factor to get x and y.

Lets try 1/x + 1/y = 1/6

6^2 is 36 = 2*2*3*3

So factorisations are:

1*36,2*18,3*12,4*9,6*6, plus all of those combos again with both numbers negated

Giving the combinations x&y of:

1/2-1/3

1/3-1/6

1/4-1/12

1/5-1/30

1/7+1/42

1/8+1/24

1/9+1/18

1/10+1/15

1/12+1/12 - 02 Jul '12 00:28 / 2 edits

Yeah, I guess mine only works when a is a prime number.*Originally posted by iamatiger***Generalising talzamir's approach**

1/x + 1/y = 1/a

0 = xy - ax - ay

a^2 = xy - ax - ay + a^2

a^2 = (x - a) (y - a)

So take all the ways of factoring a^2 into two integers, and in each case add A to each factor to get x and y.

Lets try 1/x + 1/y = 1/6

6^2 is 36 = 2*2*3*3

So factorisations are:

1*36,2*18,3*12,4*9,6*6, plus all of those c ...[text shortened]... y of:

1/2-1/3

1/3-1/6

1/4-1/12

1/5-1/30

1/7+1/42

1/8+1/24

1/9+1/18

1/10+1/15

1/12+1/12

Jeez, and I missed -110, 10. Really cocked this up. ðŸ˜ž - 02 Jul '12 18:20

Oh dear. I have to have missed something because I should have five solutions.*Originally posted by iamatiger***No, you got -110,10**

that is your:*x=a-a^2 y= a-1*

don't despair! ðŸ™‚

So which one did I not swap x and y for?

x=a^2+a y=a+1

x=132, y=12

Yeah, that's the one I missed. I guess I should have read your earlier post. You had five solutions. I should have known my generalization was wrong just from that.