Problem 1529

1. Infinitely many perfect squares are a sum of a perfect square and a prime number:

Consider the perfect square $(n+1)^2$. We can express this as:
$$(n+1)^2 = n^2 + 2n + 1$$
Here, $n^2$ is a perfect square and $2n + 1$ is an odd number. Since there are infinitely many odd prime numbers, we can choose $n$ such that $2n + 1$ is a prime number. For example, if $n = 1$, $2n + 1 = 3$ (which is prime), and $(1+1)^2 = 4 = 1^2 + 3$. Similarly, for $n = 2$, $2n + 1 = 5$ (which is prime), and $(2+1)^2 = 9 = 4 + 5$.

Since there are infinitely many odd primes, we can always find an $n$ such that $2n + 1$ is a prime number. Therefore, there are infinitely many perfect squares that can be expressed as the sum of a perfect square and a prime number.

2. Infinitely many perfect squares are not a sum of a perfect square and a prime number:

Consider the perfect square $(n+1)^2$ again. We can express this as:
$$(n+1)^2 = n^2 + 2n + 1$$
Here, $n^2$ is a perfect square and $2n + 1$ is an odd number. Since there are infinitely many odd composite numbers, we can choose $n$ such that $2n + 1$ is an odd composite number. For example, if $n = 4$, $2n + 1 = 9$ (which is composite), and $(4+1)^2 = 25 = 16 + 9$. Similarly, for $n = 6$, $2n + 1 = 13$ (which is prime), and $(6+1)^2 = 49 = 36 + 13$.

Since there are infinitely many odd composite numbers, we can always find an $n$ such that $2n + 1$ is an odd composite number. Therefore, there are infinitely many perfect squares that cannot be expressed as the sum of a perfect square and a prime number.

To further solidify this, consider the expression $(n+1)^2 - a^2 = (n+a+1)(n-a+1)$. For all $a$ with $a < n$, if $(n+a+1)(n-a+1)$ is an odd composite number, then $(n+1)^2$ cannot be expressed as the sum of a perfect square and a prime number.

This completes the proof for both parts.

$\blacksquare$