Question

N Is The Number 2^74+2^2058+2^2N Is A Perfect Square

Answer 1

Answer:

I don't know what is the answer

Answer 2

Step-by-step explanation:

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 = 20208