I am going to start with an example of two geometric figures. Rectangle must haves:
- quadrilateral
- four right angles
- opposite sides are equal and parallel
- diagonals bisect each other
If we say that every square is a rectangle, a square should possess all properties of a rectangle, plus unique properties on his own, such as:
- all four sides are equal
- diagonals cross at right angles
In comparison to number $2$ and properties of prime numbers, number $2$ is a whole number whose only factors are $1$ and itself. Second one is that every prime cannot be made of rectangle with more than one row. So far, so good, and another property for primes is that every prime can be written as a difference of two squares in a unique way – this is where number $2$ and all evens have non-integral solution:
$(\frac{3}{2})^2 – (\frac{1}{2})^2 = \frac{9}{4} – \frac{1}{4} = \frac{8}{4} = 2$
It is easy to notice all odd composite numbers have two or more integer solutions depending on number of factors. We can see it by intersecting two rectangles of the same size, where both sides of the rectangle are odd or prime numbers:
Clearly number $2$ has got his own properties like parity = $0$, where all prime numbers $>2$ have parity = $1$ inherited from odd numbers. My confusion is that should number $2$ retain all properties of prime numbers "to be qualified" one of them? ( such as in case of square being a rectangle). Or perhaps we can skip this fact and simply call number $2$ the only even prime.
Best Answer
You could raise exactly the same objections if you noted that "all prime numbers are bigger than $4$, except for these two weird exceptions $2$ and $3$". Does that mean you have some kind of existential angst about the primality of $3$, or does it simply mean that numbers do have different properties from each other?
In a ring, "$p$ is prime" means nothing more nor less than "if $p \mid a b$ then $p \mid a$ or $p \mid b$, and $p \not = 0$, and $p$ has no multiplicative inverse". Equivalently, in $\mathbb{Z}$, "$p$ is prime" means "$p$ has no factors other than itself and $1$, and is not $0$ or $\pm 1$". That's all there is to it.