For which of the following “n” is the number 2^74 +2^2058+2^2n is a perfect square?

• (a+b)^2 = a^2+2ab+b^2
  On comparing
  (2^37 + 2^n)^2 = 2^74 + 2*(2^37)*(2^n) + 2^2n
  
  2*(2^37)*(2^n)= 2^(38+n)
  Comparing the powers
  (38+n)= 2058
  we get n=2020
• 9 years agoHelpfull: Yes(17) No(0)
• i'm sorry,the ans is 2020
  coz on comparing its 38+n=2058
• 9 years agoHelpfull: Yes(2) No(0)
• ANCHAL GUPTA
  
  By looking into equation we can identify
  that by practice it will come don't worry man
  
  (a+b)^2 = a^2+2ab+b^2
  On comparing
  (2^37 + 2^n)^2 = 2^74 + 2*(2^37)*(2^n) + 2^2n
  
  2*(2^37)*(2^n)= 2^(38+n)
  Comparing the powers
  (38+n)= 2058
  we get n=2020
• 9 years agoHelpfull: Yes(2) No(0)
• Among these solutions which one is correct
• 9 years agoHelpfull: Yes(1) No(0)
• convert it to the form a^2+2ab+b^2
  (2^37)^2+2*2^37*2^n+(2^n)^2
  so on comparing powers 37+n=2058
  so n=2021
  
• 9 years agoHelpfull: Yes(0) No(3)
• explain it
  how compare
• 9 years agoHelpfull: Yes(0) No(2)
• ANS WILL BE 3.
• 9 years agoHelpfull: Yes(0) No(0)
• (2^37)^2+2^2058+2^2020+2^2n-2^2020
  then n=1010..absolute..
  
• 9 years agoHelpfull: Yes(0) No(1)