It is given that n+x and n+y both are perfect square. and x=18 & y=90. How many such number exist for n ?
a) 1
b)3
c) don't know
d)none

• n+18 = a^2
  n+90 = b^2
  subtract the two
  b^2 - a^2 = 72
  (b-a)(b+a) = 72
  Look at factors of 72
  1*72
  2*36
  3*24
  4*18
  6*12
  8*9
  
  How many of these have an a and b such that (b-a) is the small number and (b+a) is the large?
  b-a = 1
  b+a = 72
  add the two
  2b = 73 (no... needs to be whole numbers)
  b-a = 2
  b+a = 36
  2b = 38
  b = 19, a = 17, b^2 = 361, n + 90 = 361 so n = 271
  n should also equal 17^2 - 18 = 271 (so that's 1)
  
  we notice the sum of the factors must be even and the average of the factors is b
  
  3*24 sum is 27
  4*18 sum is 22, the average is 11
  b = 11
  n = 11^2 - 90 = 31
  a = b - small factor = 11-4 = 7
  check : 31+18 = 49 = 7^2 (so that's 2)
  
  6*12 sum is 18, average is 9
  b = 9, a = 9-6 = 3
  n= 9^2 - 90 = -9
  -9 + 18 = 9 = 3^2 (that's 3)
  
  8*9 sum is odd
• 10 years agoHelpfull: Yes(23) No(1)
• b) 3
  31, -9, 271
• 10 years agoHelpfull: Yes(13) No(6)
• how should we know that there are 3 such no. -like we can go on couinting after 271?Where should we stop.Option could be D.none
• 10 years agoHelpfull: Yes(9) No(3)
• @PRANAMI BURAGOHAIN.. your solution was very helpful... Thanks!!!
• 10 years agoHelpfull: Yes(5) No(0)
• Ans) (a) 1
• 10 years agoHelpfull: Yes(1) No(5)
• a) 1
  
  31 + 18 = 49
  31 + 90 = 121
• 10 years agoHelpfull: Yes(1) No(3)
• b) 31,126,271
• 10 years agoHelpfull: Yes(1) No(12)
• a) 49 and 121
  
• 10 years agoHelpfull: Yes(0) No(3)
• 49 and 121,
  9 and 81 where n= -9
• 10 years agoHelpfull: Yes(0) No(2)
• c,as we didn't find the actual no.
  
• 10 years agoHelpfull: Yes(0) No(0)
• the difference between two square is 72.now u hv to find the suc no's of square diffrnce 72.answer wil be 2
• 10 years agoHelpfull: Yes(0) No(0)
• 3--31,-9,271
• 10 years agoHelpfull: Yes(0) No(0)