Question

\(\large\tt \begin{align} \color{black}{
\normalsize \text{prove that the equation}\hspace{.33em}\\~\\
\dfrac{1}{x}+\dfrac{1}{y}=2\hspace{.33em}\\~\\
\normalsize \text{has no integer solution ,except }\hspace{.33em}\\~\\
x=y=1\hspace{.33em}\\~\\
}\end{align}\)

Answer 1

idk

Answer 2

this equation can be written in the form : x= y/(2y-1)

now you can see that 2y-1 is bigger than y for all integers values but 1.
hence x wont be an integer for y integer other than 1.

Answer 3

note that the same thing can be said for negative y.
and remember x cant be 0 and also y cant be 0.

Answer 4

We can be confident that no other solutions can be found when both have the same sign, because a fractional value is always < 1 for positive integers and in the case of negative, it's pretty obvious how the left-hand-side is negative unlike the right side. Let's say that x and y are of opposite signs where \(x>0, y<0\) WLOG.\[\dfrac{1}{x} = 2 - \frac{1}{y}\]LHS < 1 and RHS > 2.

Hence proved.

Answer 5

Case 1: x > 0, y > 0.
LHS \(\le 2\).

Case 2: x < 0, y < 0.
LHS \(< 0\).

Case 3: x < 0, y > 0.
As shown above.

Answer 6

1/x+1/y=2
multiply both sides by xy
y+x=2xy
add negative y for both sides
x=2xy-y
x=y(2x-1)
y=x/(2x-1)

Answer 7

we can perform a demonstration by absurd, as below:
let's suppose that x is an integer m which is greater than 1, namely x=m>1
and y is another integer equal to 1, namely y=1.
Now we have:
\[\frac{ 1 }{ m }<1\]
on the other hand if I substitute those above quantities into the starting expression, I get:
\[\frac{ 1 }{ m }+1=2\]
or:
\[\frac{ 1 }{ m }=1\]
which is a contraddiction.

Answer 8

thanks all