If n+x and n+y both are perfect square and x=18, y=90, then how many values of n will satisfied above condition

• let n+18= a^2 ........(1)
  n+90= b^2 ........(2)
  
  b^2 = n+90
  = n+18+72
  = a^2+72 (from 1)
  b^2-a^2 = 72
  (b+a)(b-a)=72.......(3)
  factorize 72
  1*72 (a,b will be fractions)
  2*36 [(19-17)(19+17) then substitute a = 17 and b = 19 then n=271]
  3*24 (a,b will be fractions)
  4*18 [(11+7)(11-7) then substitute a = 7 and b = 11 then n=31]
  6*12 [(9+3)(9-3) then substitute a = 3 and b = 9 then n=-9]
  8*9 (a,b will be fractions)
• 11 years agoHelpfull: Yes(77) No(2)
• fOR 3 VALUES OF N.
  
  ie. n=31, n=271, n=-9.
• 11 years agoHelpfull: Yes(24) No(3)
• for 3 values of n.
  i.e. n=31, n=271, n=-9
  for n=31
  31+18=49=7
  31+90=121=11
  
  for n=271
  271+18=289=17
  271+90=361=19
  
  for n=-9
  -9+18=9 =3
  -9+90=81 =9
• 11 years agoHelpfull: Yes(11) No(1)
• for n=31 n+x=49 and n+y=121
  for n=271 n+x=289 and n+y=361
  
• 11 years agoHelpfull: Yes(3) No(6)
• only 1 value i.e. n=31
• 11 years agoHelpfull: Yes(2) No(16)
• n=31 and n=-9
  
• 11 years agoHelpfull: Yes(2) No(7)
• 31+18=49
  31+90=121
  Are the Perfect squares.
  Ans: n=31
• 11 years agoHelpfull: Yes(1) No(7)
• yes only one solution n=31
  for n=271 n+y=351.I wrote 2 soltions earlier by mistake.
  only one solution is possible
• 11 years agoHelpfull: Yes(1) No(9)
• only one ......n =31
  31+18=49 (perfect square of 7)
  31+90=121 (perfect sqaure of 11)
• 11 years agoHelpfull: Yes(1) No(4)