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.


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.

  • $\begingroup$ I just think, that number 2 is a building block of even numbers and number 3 is a building block of prime numbers - and this is where the structure of these two differs. $\endgroup$ – usiro Aug 11 '18 at 20:02
  • $\begingroup$ @usiro Number 2 is the only prime which is even (divisible by 2). Number 3 is the only prime number that's divisible by 3. Number 5 is the only prime number that's divisible by 5. So what? $\endgroup$ – shalop Aug 11 '18 at 20:42
  • $\begingroup$ @Shalop So, if you subtract 2 or 2 squared from any prime number they all get divisibility by 3. $\endgroup$ – usiro Aug 11 '18 at 21:48
  • $\begingroup$ @usiro No that fails for the number 3. Also if you subtract (one of) 2 or 4 from any number not divisible by 3, then it becomes divisible by 3. What is your point? $\endgroup$ – shalop Aug 11 '18 at 23:30
  • $\begingroup$ @Shalop It must fail for 3 since this is the leading coefficient, and I think you meant "any odd number". I was just curious if anyone came across A038509 and notice how consecutive primes get into factorisation. Prime number doesn't have to be seen only as of length, but also as an area formed by intersection of two rectangles not necessarily of the same size . This subject is to long in comments, I would rather raise another question. $\endgroup$ – usiro Aug 12 '18 at 8:21

The number $2$ satisfies lots of properties that other primes don't satisfy, for example, being equal to $2$. There are properties of rectangles that are not satisfied by squares, such as having sides of different lengths. Ellipses are a type of conic section, and yet have the property of being bounded, whereas all other conic sections are unbounded. The number $0$ only has one square root, while all other numbers have two. So is $0$ not a number? There's no problem in saying that $2$ is a prime just because it doesn't satisfy every single property every other prime number happens to satisfy. $2$ is prime because it is.

  • $\begingroup$ The reason I am asking is that this particular property is of greater importance to me in finding arithmetic progressions inside composite numbers than the fact that "all numbers are divisible by 1 and itself". However, I think I can leave it as it is. $\endgroup$ – usiro Aug 11 '18 at 20:22

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