The problem of 7/5

Find all integral solutions to the above equation.

Solution

The above equation require on elementary mathematics to solve.
It is an equation in 3 variables x,y,z and you require at-least 3 equations to solve it.
But the important thing to notice is that the solution must be all integers and thanks to this constraint, we can find all possible [x,y,z] even though there is only 1 equation to work with here.

Let's begin.
As part one, I will consider only +ve integers.
Solving for -ve integers is left as an exercise if you are interested.

1/x + 1/y + 1/z = 7/5 = 1.4

Lets look at possible values that 1/(Integer)  can take
1/1 = 1
1/2 = 0.5
1/3 = 0.33
1/4 = 0.25
1/5 = 0.20
1/6 = 0.1667

 ..... as so on...

If we choose all x,y,z >=3, then the max value of LHS is 0.33 + 0.33 + 0.33 = 0.99
This implies that possible solution must include 1 or 2 in the solution set.

Now lets break it down into 2 case -
Case 1 : The solution contains '1' 
Lets fix x=1 and re look at the equation -
1/1+1/y+1/z=1.4
=> 1/y+1/z=0.4
extending similar logic if both x,y >=6 , then max val of 1/y + 1/z is 0.1667+0.1667 = 0.333 < 0.4
Hence at-least one of y or z  has to take on values from [2,3,4,5]
let's assign these values to y and look for a solution.
for y = 2
1/2 + 1/z = 0.4
=> 1/z = -0.1
=> z = -10 [a pointer to solution for -ve integers]
since we are looking for +ve integers, this is inadmissible

for y = 3
1/3 + 1/z = 0.4=4/10
=> 1/z=(12-10)/30=2/30
=> z = 15

possible solution [1,3,15] and its permutations

for y = 4
1/4 + 1/z = 0.4
=> 1/z=0.15
=> z = 2/3 which is not an integer

for y = 5
1/5 + 1/z = 0.4
=> 1/z=0.2
=> z = 5

possible solution [1,5,5] and its permutations

Case 2 : The solution contains '2' 

Lets fix x=2 and re look at the equation -
1/2+1/y+1/z=1.4
=> 1/y+1/z=0.9
if both y & z >=3, then max value of LHS is 0.33 +0.33 = 0.66 < 0.9
Hence at-least one of y or z  has to take on values from [1,2]

let's assign these values to y and look for a solution.
for y = 1
1/1 + 1/z = 0.9
=> 1/z = -0.1
=> z = -10 [a pointer to solution for -ve integers]
since we are looking for +ve integers, this is inadmissible

for y = 2
1/2 + 1/z = 0.9
=> 1/z = 0.4
=> z = 2/5  which is not an integer


Summing it up, solution in +ve integer space consists of following sets-
[1,3,15], [1,15,3], [3,1,15], [3,15,1], [15,1,3], [15,3,1]
& [1,5,5], [5,1,5], [5,5,1]