To solve the problem, we need to determine the value of \( n \) such that the expression \( 2^{200} - 2^{192} \cdot 31 + 2^n \) is a perfect square. We will break down the steps systematically.
### Step 1: Factor out \( 2^{192} \)
We start by factoring out \( 2^{192} \) from the expression:
\[
2^{200} - 2^{192} \cdot 31 + 2^n = 2^{192}(2^8 - 31 + 2^{n-192})
\]
### Step 2: Simplify the expression inside the parentheses
Next, we simplify the expression inside the parentheses:
\[
2^8 = 256
\]
Thus, we have:
\[
2^{192}(256 - 31 + 2^{n-192}) = 2^{192}(225 + 2^{n-192})
\]
### Step 3: Set the expression to be a perfect square
For \( 2^{192}(225 + 2^{n-192}) \) to be a perfect square, \( 225 + 2^{n-192} \) must also be a perfect square, since \( 2^{192} \) is already a perfect square (as \( 2^{192} = (2^{96})^2 \)).
Let:
\[
225 + 2^{n-192} = k^2
\]
for some integer \( k \).
### Step 4: Rearrange the equation
Rearranging gives us:
\[
2^{n-192} = k^2 - 225
\]
### Step 5: Factor the right-hand side
Notice that \( 225 = 15^2 \), so we can factor the right-hand side:
\[
k^2 - 15^2 = (k - 15)(k + 15)
\]
Thus, we have:
\[
2^{n-192} = (k - 15)(k + 15)
\]
### Step 6: Analyze the factors
Since \( 2^{n-192} \) is a power of 2, both \( (k - 15) \) and \( (k + 15) \) must also be powers of 2. Let:
\[
k - 15 = 2^a \quad \text{and} \quad k + 15 = 2^b
\]
where \( b > a \).
### Step 7: Set up the equations
From these equations, we can write:
\[
2^b - 2^a = 30
\]
Factoring out \( 2^a \):
\[
2^a(2^{b-a} - 1) = 30
\]
### Step 8: Determine possible values of \( a \) and \( b \)
The factors of 30 are \( 1, 2, 3, 5, 6, 10, 15, 30 \). We will check for values of \( 2^a \):
- If \( 2^a = 2 \), then \( 2^{b-a} - 1 = 15 \) which gives \( 2^{b-a} = 16 \) or \( b-a = 4 \). Thus, \( a = 1 \) and \( b = 5 \).
- If \( 2^a = 6 \), then \( 2^{b-a} - 1 = 5 \) which gives \( 2^{b-a} = 6 \) which is not a power of 2.
Continuing this way, we find:
- \( a = 1 \) and \( b = 5 \) gives \( k - 15 = 2 \) and \( k + 15 = 32 \), thus \( k = 17 \).
### Step 9: Solve for \( n \)
Now substituting \( k = 17 \):
\[
2^{n-192} = 17^2 - 225 = 289 - 225 = 64
\]
This implies:
\[
2^{n-192} = 2^6 \implies n - 192 = 6 \implies n = 198
\]
### Step 10: Find the sum of the digits of \( n \)
Finally, we find the sum of the digits of \( n = 198 \):
\[
1 + 9 + 8 = 18
\]
### Final Answer
Thus, the sum of the digits of \( n \) is:
\[
\boxed{18}
\]
