2^74+2^2048+2^2n what is the value of n so that sum will be perfect square

• n=2010
  (a+b)^2=a^2 +b^2 +2ab
  2^74=a^2
  2ab=2^2048
  b^2=2^2n
• 12 years agoHelpfull: Yes(24) No(5)
• since, (a+b)^2=a^2+b^2+2ab
  2^74+2^2n+2^2048=(2^37)^2+(2^n)^2+2^2048
  2*(2^37)*(2^n)=2^2048
  2^(1+37+n)=2^2048
  since base are same ,equating powers,we get
  1+37+n=2048
  38+n=2048
  n=2010
• 12 years agoHelpfull: Yes(16) No(3)
• iam sorry 2010
• 12 years agoHelpfull: Yes(5) No(2)
• (2^34)^2 + 2*2^34*2^n+(2^n)^2 compare with a^2+2ab+b^2
  then 35+n=2048 = n=2013
• 12 years agoHelpfull: Yes(4) No(14)
• 2011
  n can be any odd no.
  
• 12 years agoHelpfull: Yes(3) No(7)
• 2^74=(2^34)^2
  2^2n=(2^n)^2
  2^74+2^2048+2^2n ----------1
  IF(2^34)^2+ 2.2^34.2^n+ (2^n)^2 -------------2
  =(2^34+2^n)^2 so
  from equn 1 &2 2.2^34.2^n must be equal to 2^2048
  so (34+1+n)=2048 so n=2048-35=2013(ANS)
• 12 years agoHelpfull: Yes(2) No(9)
• ans is 2010
  
  (a+b)^2 = a^2 + 2*a*b + b^2
  
  a^2= (2^37)^2 = 2^74
  b^2= (2^2010)^2 = 2^4020
  2*a*b = 2* 2^37 * 2^2010= 2^2048
• 12 years agoHelpfull: Yes(2) No(1)
• sai ji aap hamesa confusion m kyun rehti h?
  
• 12 years agoHelpfull: Yes(2) No(0)
• (A+B)^2=A^2+B^2+2*A*B
  (2^37)^2+(2^n)^2+2*(2^37)*(2^2010)AS SAME AS GIVEN EQ.
  SO n=2010 ( 2048=37+1+2010)
• 12 years agoHelpfull: Yes(1) No(1)
• correct ans is 2013..
  
• 12 years agoHelpfull: Yes(0) No(9)
• correct ans 1699
  
• 12 years agoHelpfull: Yes(0) No(1)