A problem by mridul jain

Let's rewrite the above expression per the following:

$2^{200} - 31 \cdot 2^{192} + 2^{n} = 2^{192} \cdot [2^8 - (2^5 - 1) + 2^{n-192}] = 2^{192} \cdot [(2^4-1)^2 + 2^{n-192}] = 2^{192} \cdot [15^2 + 2^{n-192}].$

Since we desire this expression to equal a perfect square, consider the $8-15-17$ primitive Pythagorean triplet $\Rightarrow 8^2 + 15^2 = 17^2$ . This gives us:

$2^{(n-192)/2} = 8 = 2^3 \Rightarrow \frac{n-192}{2} = 3 \Rightarrow \boxed{n=198}.$