veprimes

A veprime is a prime number $p$ such that $p^2$ has the same number of digits as $p$ in base 10. For example, the prime number 2 has one digit, and $2^2 = 4$ also has one digit. Similarly, the prime number 3 has one digit, and $3^2 = 9$ also has one digit. However, the prime number 5 has one digit, and $5^2 = 25$ has two digits, so 5 is not a veprime.

The condition for a prime number $p$ to be a veprime can be expressed mathematically. Let $d(n)$ denote the number of digits of a positive integer $n$ in base 10. Then $p$ is a veprime if $d(p^2) = d(p)$. The number of digits of an integer $n$ is given by $\lfloor \log_{10}(n) \rfloor + 1$. Therefore, $p$ is a veprime if $\lfloor \log_{10}(p^2) \rfloor + 1 = \lfloor \log_{10}(p) \rfloor + 1$, which simplifies to $\lfloor 2 \log_{10}(p) \rfloor = \lfloor \log_{10}(p) \rfloor$. This equality holds if and only if $2 \log_{10}(p) < \lfloor \log_{10}(p) \rfloor + 1$.

The first few veprimes are 2, 3, 5, 7. For these single-digit primes, their squares (4, 9, 25, 49) have either one or two digits. The condition for a single-digit prime $p$ to be a veprime is that $p^2$ must also be a single-digit number. This is true for $p=2, 3, 5, 7$.

As prime numbers increase in size, their squares tend to grow much faster, resulting in more digits. Consequently, veprimes become rarer as the primes get larger. Determining all veprimes involves checking this digit-count condition for all prime numbers.