关于形如144的数的猜想及寻找使$a_n^2||67$为素数的$a_n$

Hey folks, let's dig into this number theory problem! We've got two parts here: a conjecture about numbers similar to 144, and a task to find all integers (a_n) such that concatenating (a_n^2) with 67 (written as (a_n^2 || 67)) gives a prime number. Let's break this down step by step.

1. Conjecture About Numbers of the Form 144 #

Note: The specific details of the conjecture about numbers resembling 144 weren't provided here, but we'll focus fully on the prime concatenation problem below.

2. Finding All (a_n) Where (a_n^2 || 67) is Prime #

First, let's align on the concatenation rule: (x || y) means sticking the digits of (x) directly before (y). For example, (1 || 2 = 12), just like the problem states.

Known Valid Cases #

From the problem statement, we already have these confirmed primes:

• (a_n = 0): (0^2 || 67 = 67) (prime)
• (a_n = 1): (1^2 || 67 = 167) (prime)
• (a_n = 2): (2^2 || 67 = 467) (prime)
• (a_n = 3): (3^2 || 67 = 967) (prime)
• (a_n = 4): (4^2 || 67 = 1667) (prime)

How to Find All Valid (a_n) #

Let's formalize the concatenated number first: since 67 is a 2-digit number, (a_n^2 || 67) can be written mathematically as:

100 * (a_n²) + 67

This formula makes it way easier to test candidates without dealing with string manipulation.

Step 1: Rule Out Composite Candidates Quickly

We can use basic divisibility rules to skip unnecessary primality checks:

• Divisible by 3?: A number is divisible by 3 if the sum of its digits is. Here, that sum is the digit sum of (a_n²) plus (6+7=13). If that total is a multiple of 3, the number is composite. For example, (a_n=5): digit sum of 25 is 7, 7+13=20 (not divisible by 3), but 2567 is still composite (17×151).
• Divisible by 5?: The number ends with 7, so it's never divisible by 5—we can ignore this check entirely.
• Divisible by 7?: Let's compute modulo 7: (100 ≡ 2 \mod7), (67≡4 \mod7), so the number becomes (2a_n² +4 \mod7). Squares modulo 7 only ever equal 0,1,2,4—so (2a_n² +4) can never be 0 mod7. That means our number is never divisible by 7.

Step 2: Primality Testing for Larger (a_n)

For bigger values of (a_n), we'll need to use a reliable primality test like the Miller-Rabin test (a fast, probabilistic test that's accurate for most practical purposes). Let's test a few more candidates to see the pattern:

• (a_n=5): 2567 = 17×151 (composite)
• (a_n=6): 3667 = 19×193 (composite)
• (a_n=7): 4967 (appears to be prime—confirm with a rigorous test)
• (a_n=8): 6467 = 29×223 (composite)
• (a_n=9): 8167 (confirmed prime)

Is there a finite number of such (a_n)? That's an open question unless we can prove otherwise. For now, the approach is to generate (a_n) values, compute (100a_n²+67), and test for primality.

Key Takeaways #

• Start with small (a_n) values—we already have the first 5 valid ones.
• Use divisibility rules to eliminate composite candidates fast.
• For larger numbers, use a trusted primality test to confirm.

内容的提问来源于stack exchange，提问作者Mr Pie