The expression `31^(n) + 17^(n)` is divisible by, for every odd positive integer n :

Options

A. 48
B. 14
C. 16
D. both (1) and (3)

Answer

To determine the divisibility of the expression \(31^n + 17^n\) for every odd positive integer \(n\), we can follow these steps:

### Step 1: Identify the Expression
We start with the expression:
\[ 31^n + 17^n \]
where \(n\) is an odd positive integer.

### Step 2: Use the Property of Sums of Powers
For any integers \(a\) and \(b\), if \(n\) is odd, the expression \(a^n + b^n\) can be factored as:
\[ a^n + b^n = (a + b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \ldots + b^{n-1}) \]
In our case, \(a = 31\) and \(b = 17\).

### Step 3: Calculate \(a + b\)
Now, we calculate:
\[ 31 + 17 = 48 \]
This means that \(31^n + 17^n\) is divisible by \(48\) for any odd positive integer \(n\).

### Step 4: Check Divisibility of Options
Now, we need to check which of the given options can divide \(48\):
1. **Option 1: 48** - Yes, \(48\) divides \(48\).
2. **Option 2: 14** - No, \(14\) does not divide \(48\).
3. **Option 3: 16** - Yes, \(16\) divides \(48\) (since \(48 = 16 \times 3\)).
4. **Option 4: Both 1 and 3** - Since both options 1 and 3 are valid, this option is also correct.

### Conclusion
Thus, the expression \(31^n + 17^n\) is divisible by both \(48\) and \(16\) for every odd positive integer \(n\). Therefore, the correct answer is:
\[ \text{Option 4: Both 1 and 3} \]