Find positive integers x,y,z in 1/x+1/y=7/8(x,y∈N)

[Albert1]

$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$

 

[kaliprasad]

$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$

without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

Spoiler

$\frac{7}{8} > \frac{1}{x} >= \frac{7}{16}$ or so x = 2 and it does not give y integer so no solution

for (2)

Spoiler

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}$

or $\dfrac{1}{2} > \dfrac{1}{x} >= \dfrac{1}{4}$ x = 4 or 3 , x=4 gives y =4 and x= 3 gives y = 6 so solution set(4,4), (3,6), (6,3)

 

[Albert1]

without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

Spoiler

$\frac{7}{8} > \frac{1}{x} >= \frac{7}{16}$ or so x = 2 and it does not give y integer so no solution

for (2)

Spoiler

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}$

or $\dfrac{1}{2} > \dfrac{1}{x} >= \dfrac{1}{4}$ x = 2 or 3 , x=2 gives y =2 and x= 3 gives y = 6 so solution set(2,2), (3,6), (6,3)

Spoiler

set (x,y)=(2,2) is not a solution ,it should be (x,y)=(4,4)

 

[kaliprasad]

Spoiler

set (x,y)=(2,2) is not a solution ,it should be (x,y)=(4,4)

Spoiler

you are right I have done correction inline . Thanks

 

[Albert1]

$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$

hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

Spoiler

let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy

 

[kaliprasad]

hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

Spoiler

let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy

The above hint is not correct because

Spoiler

x= 6, y = 16, z= 48 is solution not provided by above

 

[Opalg]

hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$

Spoiler

let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy

[sp]But that method does not find all the solutions, for example $\dfrac6{24} = \dfrac15 + \dfrac1{21} + \dfrac1{420}$.[/sp]

 

[Albert1]

Spoiler

yes you are right,that is why I express (3) as:
$\dfrac{6}{24} = \dfrac1{x} + \dfrac1{y} + \dfrac1{z}$
if I express (3) as :
$\dfrac{12}{48} = \dfrac1{x} + \dfrac1{y} + \dfrac1{z}$
and using the same method we will get more solutions,up to now I did not see
any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me
(if both numerator and denominator are fixed,then the solutions will be fixed,if both
are multiplied by some factor k,then may be we will get more answers)

 

[Opalg]

any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me

These are known as Egyptian fractions.

 

[Albert1]

These are known as Egyptian fractions.

Spoiler

Can we get all the amounts of solutions of this kind ?
Is there any formula to express it ?