2^74+2^2058+2^2n,what is the value of n for which the expression is a perfect square?

• (a^2)+2ab+(b^2)
  (2^37)^2+(2*2^37*2^n)+(2^2n)=(2^74)+(2*2^37*2^2020)+(2^4040)
  hence by comparision
  2n=4040 ; n=2020
• 8 years agoHelpfull: Yes(4) No(1)
• 23 is the correct ans
  4(1066+n)
  n=23
  4(1066+23)
  4(1089)
  which is a perfect square of 33*2=66
  
• 8 years agoHelpfull: Yes(0) No(2)
• a^2 +2ab+b^2
• 8 years agoHelpfull: Yes(0) No(2)
• 2^74 + 2^2058 + 2^2n =(2^37)^2 + (2^1029)^2 + (2^n)^2
  now, if we put 2^37 = a ; 2^1029 = b; Then for the above expression to be perfect square 2^2n must be equal to (2*a*b)= 2*(2^37)*(2^1029);
  ==> 2^2n = 2^(1067)
  ==> 2n = 1067 ,
  but this case is not possible since R.H.S is an odd integer whereas L.H.S is an even integer.
  So , the above mentioned case can't hold.
  
  Now,if we put 2^37 = a; 2^n = b ; So, for the given expression to be perfect square 2^2058 = (2*a*b)= 2*(2^37)*(2^n) = 2^(n+38);
  So, 2058 = (n+38)
  => n = 2020
  So,The answer is n = 2020
• 4 years agoHelpfull: Yes(0) No(0)