Rearranging 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!
That would just about cover it, yes. 🙂
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.
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3 edits
Originally posted by talzamirrearrange to y = 11x/(x-11)
That would just about cover it, yes. 🙂
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
2 edits
Originally posted by iamatigerInteresting 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. 😛
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
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Originally posted by iamatigerYes 0 doesn't work as a solution because in the following:
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
4 edits
Originally posted by talzamirNo one has posted the generic solution yet, so here it goes.
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. then
y=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+1y=a^2+a
x=a-1y=a-a^2
x=2ay=2a
x=a-a^2y=a-1
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Originally posted by SwissGambitGeneralising talzamir's approach
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+1y=a^2+a
x=a-1y=a-a^2
x=2ay=2a
x=a-a^2y=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
2 edits
Originally posted by iamatigerYeah, I guess mine only works when a is a prime number.
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. 😞
Originally posted by iamatigerOh dear. I have to have missed something because I should have five solutions.
No, you got -110,10
that is your:
x=a-a^2 y= a-1
don't despair! 🙂
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.