[QUOTE=Mini-Geek;99339]Ok, fine, here's the updated version:

A prime number (or a prime) is a natural number that has exactly (5 - 3) (distinct) natural number divisors, which are (4 - 3) and the prime number itself.
:grin: Hooray for loopholes! :grin:[/QUOTE]

[QUOTE=davar55]Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
[b]or any other specific integer[/b]?[/QUOTE]
I added the emphasis to the original specification of the problem. To the best of my knowledge, 3, 4 and 5 are all specific integers.



Paul

[QUOTE=davar55;99304]Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
or any other specific integer?[/QUOTE]
[spoiler]Let p and q be non-negative integers such that p*p>p and q*q>q.

Then p is prime if and only if the smallest value of q such that q divides p! is q=p.[/spoiler]


Paul

[QUOTE=davar55;99304]Can you define prime numbers over the non-negative
integers without any explicit reference to 0 or 1 or 2
or any other specific integer?[/QUOTE]I have another marvelous definition and the margin of this post is only just wide enough to contain it.

[spoiler]The positive values of this polynomial over the integers are all the primes and nothing but.

(k + (ln(e)+ln(e)))*(ln(e) - (w*z + h + j - q)^(ln(e)+ln(e)) - ((g*k + (ln(e)+ln(e))*g + k + ln(e))*(h + j) + h - z)^(ln(e)+ln(e)) - ((ln(e)+ln(e))*n + p + q + z - e)^(ln(e)+ln(e)) - ((((ln(e)+ln(e))^(ln(e)+ln(e)))^(ln(e)+ln(e)))*(k + ln(e))^(ln(e)+ln(e)+ln(e))*(k + (ln(e)+ln(e)))*(n + ln(e))^(ln(e)+ln(e)) + ln(e) - f^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (e^(ln(e)+ln(e)+ln(e))*(e + (ln(e)+ln(e)))*(a + ln(e))^(ln(e)+ln(e)) + ln(e) - o^(ln(e)+ln(e)))^(ln(e)+ln(e)) - ((a^(ln(e)+ln(e)) - ln(e))*y^(ln(e)+ln(e)) + ln(e) - x^(ln(e)+ln(e)))^(ln(e)+ln(e)) - ((((ln(e)+ln(e))^(ln(e)+ln(e)))^(ln(e)+ln(e)))*r^(ln(e)+ln(e))*y^(ln(e)+ln(e)+ln(e)+ln(e))*(a^(ln(e)+ln(e)) - ln(e)) + ln(e) - u^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (((a + u^(ln(e)+ln(e))*(u^(ln(e)+ln(e)) - a))^(ln(e)+ln(e)) - ln(e))*(n + (ln(e)+ln(e)+ln(e)+ln(e))*d*y)^(ln(e)+ln(e)) + ln(e) - (x + c*u)^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (n + l + v - y)^(ln(e)+ln(e)) - ((a^(ln(e)+ln(e)) - ln(e))*l^(ln(e)+ln(e)) + ln(e) - m^(ln(e)+ln(e)))^(ln(e)+ln(e)) - (a*i + k + ln(e) - l - i)^(ln(e)+ln(e)) - (p + l*(a - n - ln(e)) + b*((ln(e)+ln(e))*a*n + (ln(e)+ln(e))*a - n^(ln(e)+ln(e)) - (ln(e)+ln(e))*n - (ln(e)+ln(e))) - m)^(ln(e)+ln(e)) - (q + y*(a - p - ln(e)) + s*((ln(e)+ln(e))*a*p + (ln(e)+ln(e))*a + p^(ln(e)+ln(e)) - (ln(e)+ln(e))*p - (ln(e)+ln(e))) - x)^(ln(e)+ln(e)) - (z + p*l*(a - p) + t*((ln(e)+ln(e))*a*p - p*(ln(e)+ln(e)) - ln(e)) - p*m)^(ln(e)+ln(e)))
[/spoiler]


Paul

Let n be a non-negative integer, then n is prime if and only if n!*n! isn't divisible by n*n*n. (It's easy to prove this.)

An integer p is prime iff division modulo p is defined, i.e. the residue ring Z/pZ is a field.

A prime number is one where one can't take the equivelent number of objects and place them into group[COLOR="Red"]s[/COLOR] of equal number.
If one is unable have the number of objects or is able to put them into groups with equal number of objects, then it is not prime.

[QUOTE=Uncwilly;99465]group[COLOR="Red"]s[/COLOR][/QUOTE]
Isn't insisting on the plural the same as saying "more than one" and thus an explicit reference to a specific integer ? Or would this be a to restrictive interpretation of the rules ?

[QUOTE=xilman;99334]Counter-example.

3 is a non-negative integer

3 is divisible by itself --- meets your definition

3 is divisible by 1, a unit in the ring of integers --- meets your definition

3 is divisible by -1, a unit in the ring of integers --- meets your definition

-3 is an element of the ring of integers

3 is divisible by -3


Therefore, according to your definition, 3 is not a prime.



Paul[/QUOTE]


I am sure Troels Munkner will be very happy with that derivation.

Mally

What about this one:

p is prime iff there exist integers a and b with:

p is divisible by a and b, and
a <> b and a <> -b

and for each integer c which divides p :
c = a or c = -a or c = b or c = -b

[quote=Andi47;99494]What about this one:

p is prime iff there exist integers a and b with:

p is divisible by a and b, and
a <> b and a <> -b

and for each integer c which divides p :
c = a or c = -a or c = b or c = -b[/quote]

I think it is too obvious that you are referring to
two positive integers. I don't know whether simply
listing the two integers excuses this.

An integer p is prime iff no integer x in the range sqrt(p) <= x < p divides p.