It is well known that there are infinity many primes, the integer $n=1 $ was a prime number and had been deleted from the list of primes because $1$ has one divisor which is $1$ it self, I would like to ask the similar question which states:


Why the integer $n=2$ is considerable to be a prime number however is even integer and is not similar to others primes which are odd?

Note: I would like to know a rigourous proof show that because the definition of prime number is not really enough to me with $n=2$ is different from other primes.

Thank you for any help


closed as unclear what you're asking by Adam Hughes, Leucippus, Alexis Olson, JMP, user91500 Oct 15 '16 at 6:31

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  • $\begingroup$ It helps if you start with a definition. Something which fits with the fundamental theorem of arithmetic might be good too. $\endgroup$ – Henry Oct 14 '16 at 20:59
  • $\begingroup$ somewhat related $\endgroup$ – Lionel Ricci Oct 14 '16 at 21:00
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    $\begingroup$ Even just means multiple of $2$. $2$ is the only prime that is a multiple of $2$. $3$ is the only prime that is a multiple of $3$. $5$ is the only prime that is a multiple of $5$. They’re all the same in this respect: there is nothing special about $2$. $\endgroup$ – Brian M. Scott Oct 14 '16 at 21:01
  • $\begingroup$ ok , why 1 is deleted from primes list however is odd and satisfy the prime property ? Is nothing here about odd or even .... ? $\endgroup$ – zeraoulia rafik Oct 14 '16 at 21:08
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    $\begingroup$ Because a ˆrime must have exactly two divisors, and $1$ has only one. Other than that, if $1$ were considered prime, we wouldn't have the unique factorisation theorem. $\endgroup$ – Bernard Oct 14 '16 at 21:33

For a number to be "even" just means that it is divisible by $2$. In much the same way, I could declare that a number is "threeven" if it is divisble by $3$. Then $3$ is the only threeven prime. Do you object to $3$ being prime? Evenness is only special in the fact that it happens to be an important property of numbers so often.


Primes have the special property that, supposing $p$ is prime, and $p$ divides a positive integer $n$, then no matter how we choose to factorise $n=ab$ into positive integers $a$ and $b$, then $p$ must also divide at least one of $a$ or $b$. This holds equally well for the number $2$ as it does for any other odd prime.

Example: $p=3$, and $n=24$ say. We can write $24=1\cdot 24, 2\cdot 12, 3\cdot 8, 4\cdot 6$, and then reversing the order. You can see the claimed property is true.

Example: consider what happens with $4$, and a number it divides like $12$, then we can write $12=1\cdot 12, 2\cdot 6, 3\cdot 4$ and the reverse order, but we immediately see the problem $12=2\cdot 6$, so four cannot be prime, which we knew, but just as an example that $4$ doesn't have the property.

$1$ is also special in that it also divides both factors in the factorisation, which in generalisations is called a 'unit' and so should be excluded from the definition of a prime.


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