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 \(\gcd\!\left(n^{2}+3,\; n+5\right) > 1\)?**

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:
\[ \gcd(a, b) = \gcd(b, a \mod b) \]

But \( a \mod b \) is the remainder when \( a \) is divided by \( b \). To find \( a \mod b \), I can perform polynomial division or use the remainder theorem.

### Using the Remainder Theorem
The remainder when \( n^2 + 3 \) is divided by \( n + 5 \) is the same as evaluating \( n^2 + 3 \) at \( n = -5 \).

So:
\[ a \mod b = (-5)^2 + 3 = 25 + 3 = 28 \]

Therefore:
\[ \gcd(n^2 + 3, n + 5) = \gcd(n + 5, 28) \]

### 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 common factors of \( n + 5 \) and 28 can be any of the divisors of 28: 2, 4, 7, 14, 28.

### Finding n Values
Thus, \( n + 5 \) must be divisible by at least one of these numbers.

Let's break it down:

1. **Divisible by 2**: \( n + 5 \equiv 0 \mod 2 \) → \( n \equiv -5 \mod 2 \) → \( n \equiv 1 \mod 2 \) (since -5 mod 2 is 1).
   So, \( n \) must be odd.

2. **Divisible by 4**: \( n + 5 \equiv 0 \mod 4 \) → \( n \equiv -5 \mod 4 \) → \( n \equiv 3 \mod 4 \) (since -5 mod 4 is 3).

3. **Divisible by 7**: \( n + 5 \equiv 0 \mod 7 \) → \( n \equiv -5 \mod 7 \) → \( n \equiv 2 \mod 7 \) (since -5 mod 7 is 2).

4. **Divisible by 14**: \( n + 5 \equiv 0 \mod 14 \) → \( n \equiv -5 \mod 14 \) → \( n \equiv 9 \mod 14 \) (since -5 mod 14 is 9).

5. **Divisible by 28**: \( n + 5 \equiv 0 \mod 28 \) → \( n \equiv -5 \mod 28 \) → \( n \equiv 23 \mod 28 \) (since -5 mod 28 is 23).

### Using the Principle of Inclusion-Exclusion
I need to find all \( n \) from 1 to 2025 that satisfy at least one of the above conditions.

But before jumping into inclusion-exclusion, perhaps it's easier to consider the union of all these conditions.

#### Step 1: Count numbers divisible by 2, 4, 7, 14, or 28
But since 4 is a multiple of 2, and 14 and 28 are multiples of 7, I need to be careful.

Actually, the conditions are:
- \( n \equiv 1 \mod 2 \) (odd numbers)
- \( n \equiv 3 \mod 4 \)
- \( n \equiv 2 \mod 7 \)
- \( n \equiv 9 \mod 14 \)
- \( n \equiv 23 \mod 28 \)

But the first condition is about odd numbers, and the second is a subset of the first (since \( n \equiv 3 \mod 4 \) implies \( n \) is odd).

### Using Chinese Remainder Theorem
I need to find numbers that satisfy any of the last three conditions, but the first two are more general.

But perhaps it's better to think in terms of the general condition \(\gcd(n + 5, 28) > 1\).

This is equivalent to \( n + 5 \) not being coprime with 28.

### Total Numbers and Coprime Numbers
Total numbers \( n \) from 1 to 2025: 2025.

Numbers \( n \) such that \( n + 5 \) is coprime with 28: \(\gcd(n + 5, 28) = 1\).

First, find the range of \( n + 5 \): from 6 to 2030.

Now, find numbers \( k \) from 6 to 2030 such that \(\gcd(k, 28) = 1\).

### Euler's Totient Function
The number of integers \( k \) from 1 to \( N \) that are coprime with 28 is given by Euler's totient function \(\phi(28)\).

First, compute \(\phi(28)\):
\[ 28 = 2^2 \times 7 \]
\[ \phi(28) = 28 \times \left(1 - \frac{1}{2}\right) \times \left(1 - \frac{1}{7}\right) = 28 \times \frac{1}{2} \times \frac{6}{7} = 14 \times \frac{6}{7} = 2 \times 6 = 12 \]