2^74 + 2^2058 + 2^2n. what is the value of n make it perfect square?

• (a+b)^2 = a^2 + 2ab + b^2 :: 2^74 + 2^2058 + 2^2n
  compare it
  a^2 = 2^74 => a = 2^37
  & b^2 = b^2n => b =2^n
  2ab = 2^2058 => 2 * 2^37 * 2^n
  => 2^2058 = 2^(38+n)
  => 38+n = 2058
  => n = 2020
  
• 9 years agoHelpfull: Yes(32) No(0)
• 2^74 + 2^2058 + 2^2n = ((2^37)^2) +(2* (2^37) * (2^2020))+((2^n)^2)
  this is of the form (a+b)^2= a^2+2ab+b^2
  where a=(2^37) and b= (2^2020)=(2^n)
  therefore it can be written as ((2^37)+(2^2020))^2 which is a perfect square
  n=2020
• 9 years agoHelpfull: Yes(1) No(0)
• so,looking at the first glance we see that it is something relative to (a+b)^2
  
  hmm.. now (a+b)^2 = a^2 + b^2 + 2ab
  now match it with the question..
  2^74 = a^2 so what should be a ?? 2^37
  similarly b = 2^n
  now 2ab = 2* 2^37 * 2^n(putting the respective values)
  and this 2ab = 2^2058
  so adding the powers you get 1+37+n
  and that will be equal to 2058.
  ans nikal lo yaar..
• 9 years agoHelpfull: Yes(0) No(0)
• please give the options because the other solutions are right abut if you take a^2=2^74 then a = 2^37 and if b^2=2^2058 then b = 2^1029, then 2ab = 2^1067. therefore 2n =1067 and n can be 533.5. so the answer can be either 2020 or 533.5
• 9 years agoHelpfull: Yes(0) No(0)