Question 1138622
<pre><font size=5><b>     Here's the strategy 
greenestamps was looking for:</b></font>
</pre>
Find all positive integers a and b so that (a+1)/b and (b+2)/a are simultaneously
positive integers.
<pre>
{{{matrix(1,4,(a+1)/b,"","", (b+2)/a))}}}
Those are positive integers, so
{{{system(1<=(a+1)/b, 1<=(b+2)/a)}}}
{{{system(b<=a+1, a<=b+2)}}}
Add 1 to both sides of the second inequality
{{{system(b<=a+1, a+1<=b+3)}}}
Those inequalities can be combined as
{{{b<=a+1<=b+3}}}
So we have four cases: a+1=b, a+1=b+1, a+1=b+2, a+1=b+3
which simplify to a=b-1, a=b, a=b+1, a=b+2

Case 1:  a=b-1 substitute in 
{{{matrix(4,1,
matrix(1,3,(a+1)/b, and, (b+2)/a),
matrix(1,3,(b-1+1)/b, and, (b+2)/(b-1)),
matrix(1,3,b/b, and, ((b-1)+3)/(b-1)),
matrix(1,3,1, and, 1+3/(b-1)) )}}}

For the expression to be a positive integer,
b-1 is either 1 or 3.
For b-1=1 or b=2 and a=b-1=2-1=1, (a,b)=(1,2)
For b-1=3 or b=4 and a=b-1=4-1=3, (a,b)=(3,4)

Case 2:  a=b substitute in 
{{{matrix(4,1,
matrix(1,3,(a+1)/b, and, (b+2)/a),
matrix(1,3,(b+1)/b, and, (b+2)/b),
matrix(1,3,b/b+1/b, and, b/b+2/b),
matrix(1,3,1+1/b, and, 1+2/b ))}}}
Those will both be integers only if b=1. 
a=b=1, (a,b)=(1,1) 

Case 3:  a=b+1 substitute in 
{{{matrix(5,1,
matrix(1,3,(a+1)/b, and, (b+2)/a),
matrix(1,3,(b+1+1)/b, and, (b+2)/(b+1)),
matrix(1,3,(b+2)/b, and, ((b+1)+1)/(b+1)),
matrix(1,3,b/b+2/b, and, (b+1)/(b+1)+1/(b+1)),
matrix(1,3,1+2/b, and, 1+1/(b+1))   )}}}

For the second expression to be a positive integer,
b+1 must be 1, which makes b=0. So we must discard 
Case 3 as impossible. 

Case 4:  a=b+2 substitute in 
{{{matrix(4,1,
matrix(1,3,(a+1)/b, and, (b+2)/a),
matrix(1,3,(b+2+1)/b, and, (b+2)/(b+2)),
matrix(1,3,(b+3)/b, and, 1),
matrix(1,3,1+3/b, and, 1) )}}}
For the first expression to be a positive integer,
b is either 1 or 3.
For b=1, a=b+2=1+2=3, so (a,b)=(3,1) is a solution.
For b=3, a=b+2=3+2=5, so (a,b)=(5,3) is a solution.

So there are 5 solutions:
(a,b)=(1,2)
(a,b)=(3,4) 
(a,b)=(1,1)
(a,b)=(3,1)
(a,b)=(5,3)

Edwin</pre>