Question

What is the value of n in which $2^{74}+2^{2058}+2^{2n}$ is a perfect square ..........

Answer 1

$2^{7}4+2^{2}058+2^{2n}=(2^{37}{)}^2+(2^{1029}{)}^2+(2^n{)}^n$
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 $(2\times a\times b)=$ $2\times \left(2^{37}\right)\times \left(2^{1029}\right)$
$=2^{2n}=2^{1067}$
$=2n=1067$
But this case is not possible since RHS is an odd integer where as LHS 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 $2^{2058}=(2\times a\times b)=2\times \left(2^{3}7\right)\times \left(2^n\right)=2^{(n+38)}$;
So, $2058=(n+38)$
$n=2020$
So, the answer is $n=2020$