math-nt-01

# Sharing a Factor

For how many integers $n$ with $1 \le n \le 2025$ is
$$\gcd\!\left(n^{2}+3,\; n+5\right) > 1?$$

Give your final answer as a single integer.

Alright, I have this problem:

**For how many integers \( n \) with \( 1 \le n \le 2025 \) is \(\gcd(n^2 + 3, n + 5) > 1\)?**

First, I need to find all integers \( n \) from 1 to 2025 such that the greatest common divisor (gcd) of \( n^2 + 3 \) and \( n + 5 \) is greater than 1.

### Understanding the Problem
I need to find all \( n \) in the range [1, 2025] where \( n^2 + 3 \) and \( n + 5 \) share a common factor other than 1.

### Approach
The gcd of two numbers \( a \) and \( b \) is the same as the gcd of \( b \) and \( a \mod b \). So, I can use the Euclidean algorithm here.

Let's denote:
- \( a = n^2 + 3 \)
- \( b = n + 5 \)

According to the Euclidean algorithm:
\[ \gcd(a, b) = \gcd(n + 5, n^2 + 3) = \gcd(n + 5, (n^2 + 3) \mod (n + 5)) \]

### Polynomial Division
I need to find \( (n^2 + 3) \mod (n + 5) \). To do this, I can perform polynomial division or use the Remainder Factor Theorem.

The Remainder Factor Theorem says that the remainder of \( n^2 + 3 \) divided by \( n + 5 \) is \( (-5)^2 + 3 = 25 + 3 = 28 \).

So, \( \gcd(n + 5, n^2 + 3) = \gcd(n + 5, 28) \).

### Simplifying the GCD Condition
The condition \( \gcd(n^2 + 3, n + 5) > 1 \) is equivalent to \( \gcd(n + 5, 28) > 1 \).

This means that \( n + 5 \) must share a common factor with 28 other than 1.

### Factorizing 28
First, factorize 28:
\[ 28 = 2^2 \times 7 \]

So, the numbers that share a common factor with 28 are the numbers divisible by 2 or 7.

### Finding Valid \( n \) Values
The condition \( \gcd(n + 5, 28) > 1 \) implies that \( n + 5 \) must be divisible by 2 or 7.

Therefore, \( n + 5 \) must be divisible by 2 or 7.

But \( n \) is from 1 to 2025, so \( n + 5 \) is from 6 to 2030.

#### Case 1: \( n + 5 \) is divisible by 2
This means \( n \) is odd because:
\[ n + 5 \equiv 0 \mod 2 \implies n \equiv -5 \mod 2 \implies n \equiv 1 \mod 2 \]

#### Case 2: \( n + 5 \) is divisible by 7
This means:
\[ n + 5 \equiv 0 \mod 7 \implies n \equiv -5 \mod 7 \implies n \equiv 2 \mod 7 \]

### Counting Numbers in Range
Now, I need to count the numbers \( n \) from 1 to 2025 that are either odd or congruent to 2 modulo 7.

But I need to be careful because some numbers might satisfy both conditions (i.e., numbers that are both odd and \( n \equiv 2 \mod 7 \)).

#### Total Odd Numbers
First, count the number of odd numbers from 1 to 2025.

- The range is 1 to 2025.
- Total numbers: 2025.
- Odd numbers: \( \frac{2025 + 1}{2} = 1013 \) (since 2025 is odd).

#### Numbers \( n \equiv 2 \mod 7 \)
Next, count the numbers \( n \) such that \( n \equiv 2 \mod 7 \).

The smallest \( n \) is 2 (since \( 2 \mod 7 \) is 2).
The largest \( n \) within 2025 is the largest number \( \le 2025 \) such that \( n \equiv 2 \mod 7 \).

Let's find it:
\[ 2025 \div 7 = 289.285... \]
So, \( 7 \times 289 = 2023 \)
Then, \( 2023 + 2 = 2025 \), but \( 2025 \mod 7 = 2025 - 7 \times 289 = 2025 - 2023 = 2 \).

Thus, the largest \( n \) is 2025.

Now, the numbers are \( n = 7k + 2 \) for \( k = 0 \) to \( k = 289 \) (since \( 7 \times 289 + 2 = 2025 \)).

Total numbers: \( 289 - 0 + 1 = 290 \).

#### Numbers Satisfying Both Conditions
Now, I need to find numbers that are both odd and \( n \equiv 2 \mod 7 \).

But \( n \equiv 2 \mod 7 \) implies \( n \) is odd if \( 2 \mod 7 \) is odd, which it is.

But, \( n \equiv 2 \mod 7 \) and \( n \) is odd.

But \( n \equiv 2 \mod 7 \) already implies \( n \) is odd because 2 is odd.

Therefore, all numbers \( n \equiv 2 \mod 7 \) are odd.

### Using the Principle of Inclusion-Exclusion
The problem reduces to counting numbers that are either odd or \( n \equiv 2 \mod 7 \). But since all \( n \equiv 2 \mod 7 \) are odd, the condition simplifies to counting all odd numbers plus the numbers \( n \equiv 2 \mod 7 \) minus the overlap.

But the overlap is exactly the numbers \( n \equiv 2 \mod 7 \), because they are already included in the odd numbers.

Therefore, the total count is just the number of odd numbers, because the numbers \( n \equiv 2 \mod 7