Question 1084247
.
<pre>
1.  {{{x^2 - y^2}}} = (x+y)*(x-y).


2.  495 = 5*99 = {{{3^2*5*11}}}.


Therefore, each presentation {{{x^2 - y^2}}} = 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 . . . 
</pre>

Did you get the idea ?


<pre>
     The number of different presentations {{{x^2-y^2}}} = 495 is equal to the number of natural divisors of the number 495.
</pre>

The number of different divisors of the number 495 we can determine based on its prime decomposition 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;495 = {{{3^2*5*11}}}.      &nbsp;&nbsp;&nbsp;&nbsp;(1)


<U>It is equal to (2+1)*(1+1)*(1+1) = 3*2*2 = 12</U>.



<U>The numbers in parentheses is 1 (one) PLUS the index of the prime divisors in decomposition (1)</U>.


<pre>
      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.
</pre>