How many pairs of natural numbers are there so that difference of their squares is 60 ?

• two pairs
  
  If x and y are numbers, then
  (x+y)*(x-y)=60
  possible combinations of natural numbers are
  16,14 and 8,2
• 12 years agoHelpfull: Yes(4) No(3)
• 12 pairs
  
  If x and y are numbers, then
  (x+y)*(x-y)=60
  
  possible combinations of (x+y) and (x-y) are
  60,1
  30,2
  20,3
  15,4
  12,5
  10,6
  
  so total possible combinations are 6*2=12
• 12 years agoHelpfull: Yes(3) No(3)
• 2 pairs.
  
  Let numbers be x and y. Then
  x^2 – y^2 = 60
  (x – y)(x + y) = 60
  
  6 * 10 = 60
  5 * 12 = 60
  4 * 15 = 60
  3 * 20 = 60
  2 * 30 = 60
  1 * 60 = 60
  
  Here, only 2 pairs (2,30) and (6,10)satisfy the given condition
• 12 years agoHelpfull: Yes(3) No(1)
• In one case , values will be positive,
  in second case, values will be negative.
  
  For 60,1 and -60 ,-1
  
  values of x,y will be
  
  (30.5,29.5 ) and (-30.5, -29.5)
• 12 years agoHelpfull: Yes(1) No(4)
• (x^2)-(y^2)=60
  (x+y)*(x-y)=60
  
  possible combinations of (x+y) and (x-y) are
  60,1
  30,2
  20,3
  15,4
  12,5
  10,6
  
  so possible combinations are (16,14)&(6,2).
  
• 12 years agoHelpfull: Yes(1) No(4)