158775

Posing the first number as a and the second number as b, the problem becomes:
For which a,b belonging to the real numbers do
a-b=a/b
We get with simple algebra
ab-b^2=a
ab-a=b^2
a(b-1)=b^2
a=b^2/(b-1)
This gives us a formula to generate a's with b's as an input.
Therefore the answer couplets are:
(a,b)=(b^2/(b-1),b)
=(a,b)=(((b+1)(b-1)+1)/(b-1),b)
=(a,b)=((b+1)+1/(b-1),b)
which explains her observation
If we go through the same algebra but instead replace the subtraction sign with an addition sign we get the answer couplets:
(a,b)=(-b^2/(b-1),b)