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

• We know that a^2 + 2ab + b^2 = (a+b)^2
  Here, 2^74 + 2^2058 + 2^2n = (2^37)^2 + 2*2^37*2^2020 + (2^n)^2
  If the number is a perfect square, then 2^n should equal 2^2020.
  Hence, n=2020
• 9 years agoHelpfull: Yes(13) No(3)
• 992
  here 2058 divided by 2 we got 1029
  1029 is common for a,b
  we know it a=37
  so separate a value from 1029
  we got 992
  
• 9 years agoHelpfull: Yes(0) No(2)