find the sum of all the positive integral value of n, such that 1001+n^2 is a perfect square??plz explain also.......

1) 640
2) 620
3) 684
4) 580

• 640
  
  lt 1001 + n^2 = x^2
  1*7*11*13=x^2 - n^2
  1*7*11*13=(x+n)(x-n)
  there for pairs of (x,n)=(45,32) (51,40) (75,68) (501,500)
  sum of all n = 32+40+68+500 = 640
• 11 years agoHelpfull: Yes(0) No(3)