Consider the following statements. For any integer n,
I. `n^(2)+3` is never divisible by 17.
II. `n^(2)+4` is never divisible by 17.
Then

Options

A. both I and II are true
B. both I and II are false
C. I is false and II is true
D. I is true and II is false

Explanation

To solve the problem, we need to analyze each statement regarding the divisibility of the expressions \( n^2 + 3 \) and \( n^2 + 4 \) by 17 for any integer \( n \).

### Step 1: Analyze Statement I - \( n^2 + 3 \)

We need to determine if \( n^2 + 3 \) is never divisible by 17 for any integer \( n \).

1. **Calculate \( n^2 \mod 17 \)**:
The possible values of \( n^2 \mod 17 \) can be found by squaring the integers from 0 to 16 (since these represent all possible residues modulo 17).

- \( 0^2 \equiv 0 \)
- \( 1^2 \equiv 1 \)
- \( 2^2 \equiv 4 \)
- \( 3^2 \equiv 9 \)
- \( 4^2 \equiv 16 \equiv -1 \)
- \( 5^2 \equiv 8 \)
- \( 6^2 \equiv 2 \)
- \( 7^2 \equiv 15 \equiv -2 \)
- \( 8^2 \equiv 13 \equiv -4 \)
- \( 9^2 \equiv 13 \equiv -4 \)
- \( 10^2 \equiv 15 \equiv -2 \)
- \( 11^2 \equiv 2 \)
- \( 12^2 \equiv 8 \)
- \( 13^2 \equiv -1 \)
- \( 14^2 \equiv 9 \)
- \( 15^2 \equiv 4 \)
- \( 16^2 \equiv 1 \)

The possible values of \( n^2 \mod 17 \) are: \( 0, 1, 2, 4, 8, 9, 13, 15, 16 \).

2. **Add 3 to each residue**:
Now we add 3 to each of these values and check if any of them is congruent to 0 modulo 17.

- \( 0 + 3 \equiv 3 \)
- \( 1 + 3 \equiv 4 \)
- \( 2 + 3 \equiv 5 \)
- \( 4 + 3 \equiv 7 \)
- \( 8 + 3 \equiv 11 \)
- \( 9 + 3 \equiv 12 \)
- \( 13 + 3 \equiv 16 \)
- \( 15 + 3 \equiv 1 \)
- \( 16 + 3 \equiv 2 \)

None of these results is congruent to 0 modulo 17. Therefore, \( n^2 + 3 \) is never divisible by 17.

### Conclusion for Statement I:
**Statement I is true.**

### Step 2: Analyze Statement II - \( n^2 + 4 \)

Now we check if \( n^2 + 4 \) is never divisible by 17 for any integer \( n \).

1. **Add 4 to each residue**:
We take the same residues of \( n^2 \mod 17 \) and add 4 to each.

- \( 0 + 4 \equiv 4 \)
- \( 1 + 4 \equiv 5 \)
- \( 2 + 4 \equiv 6 \)
- \( 4 + 4 \equiv 8 \)
- \( 8 + 4 \equiv 12 \)
- \( 9 + 4 \equiv 13 \)
- \( 13 + 4 \equiv 17 \equiv 0 \) (This is divisible by 17)
- \( 15 + 4 \equiv 19 \equiv 2 \)
- \( 16 + 4 \equiv 20 \equiv 3 \)

We find that when \( n^2 \equiv 13 \mod 17 \) (which occurs when \( n \equiv 4 \) or \( n \equiv 13 \)), \( n^2 + 4 \equiv 0 \mod 17 \). Therefore, \( n^2 + 4 \) can be divisible by 17.

### Conclusion for Statement II:
**Statement II is false.**

### Final Answer:
- Statement I is true.
- Statement II is false.