For which of the following “n” is the number 274 + 22058 + 22n a perfect square?

• (2^37)^2+2^2058+(2^n)^2 is equivalent to a^2+2ab+b^2
  So,2*2^37*2^n=2^2058
  2^(38+n)=2^2058
  So,n=2020
• 11 years agoHelpfull: Yes(12) No(4)
• 2^(2*37) + 2^(2058) + 2^2n
  a^2+ 2ab + b^2
  (2^37)^2 + (2^n)^2 + 2*2^37*2^n
  Here a = 2^37, b = 2^n ,2ab= 2*(2^37)*(2^n)
  2^(38+n) = 2^2058
  38+n =2058
  n = 2020
• 11 years agoHelpfull: Yes(3) No(1)
• n=8 then perfect square=150
• 11 years agoHelpfull: Yes(0) No(13)