Question 805934
<pre>
(x+1)(y+1) = 2xy
Multiply the left side out:
xy + x + y + 1 = 2xy
Subtract xy from both sides:
x + y + 1 = xy
Isolate the terms in y
x + 1 = xy - y
Factor y out on the right
x + 1 = y(x - 1)
Divide both sides by (x-1)
{{{(x+1)/(x-1)}}} = y
I like y on the left
y = {{{(x+1)/(x-1)}}}
Divide that fraction out by long division

   <u>   1</u>
x-1)x+1
    <u>x-1</u>
      2

y = QUOTIENT + {{{REMAINDER/(DIVISOR)}}}
y = 1 + {{{2/(x-1)}}}
In order for y to be an integer, the fractional expression
{{{2/(x-1)}}} must equal to an integer.  That means the 
denominator x-1 must be a factor of the numerator, 2.

The only factors of 2 are 1 and 2. So we set the
denominator x-1 equal to each of those:

If x-1 = 1
     x = 2

y = {{{(x+1)/(x-1)}}}
y = {{{(2+1)/(2-1)}}} 
y = {{{3/1}}}
y = 3

So one possible pair of integers is x=2 and y=3.

If x-1 = 2
     x = 3

y = {{{(x+1)/(x-1)}}}
y = {{{(3+1)/(3-1)}}} 
y = {{{4/2}}}
y = 2

So the other possible pair of integers is x=3 and y=2.

Edwin</pre>