.
1. = (x+y)*(x-y).
2. 495 = 5*99 = .
Therefore, each presentation = 495 corresponds to one of the system
a) x + y = 495.
x - y = 1;
b) x + y = 165; (= 495/3)
x - y 3;
c) x + y = 55, ( = 495/9)
x - y = 9;
d) x + y = 99, (= 495/5)
x - y = 5;
e) x + y = 45, (=495/11)
x - y = 11;
and so on . . .
Did you get the idea ?
The number of different presentations = 495 is equal to the number of natural divisors of the number 495.
The number of different divisors of the number 495 we can determine based on its prime decomposition
495 = . (1)
It is equal to (2+1)*(1+1)*(1+1) = 3*2*2 = 12.
The numbers in parentheses is 1 (one) PLUS the index of the prime divisors in decomposition (1).
This problem is of the level "Math+", which means the level slightly above the standard school math.
So I assume that your level corresponds to the level of the problem and do not go in further explanations.