# Create my own PrimeQ function [closed]

I want to create my own functions, for example: Functions like IntegerQ,EvenQ and OddQ.

My code:

meuIntegerQ[_Integer] := True
meuIntegerQ[_] := False

meuEvenQ[n_Integer] /; (Mod[n, 2] == 0) := True
meuEvenQ[_] := False

meuOddQ[n_Integer] /; (Mod[n, 2] != 0) := True
meuOddQ[_] := False


Now I want to create meuPrimeQ. How can I create my own function that verify if the number is a Prime without using PrimeQ obviously (like the examples).

## closed as off-topic by Daniel Lichtblau, Szabolcs, MarcoB, José Antonio Díaz Navas, ÖskåSep 9 '18 at 8:56

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• I'm voting to close this question as off-topic because it is more a math than Mathematica question. – Daniel Lichtblau Aug 31 '18 at 22:24

A simple but by far not the most efficient way is check whether any of the prime numbers $p$ numbers $2 \leq p \leq \lfloor\sqrt{n}\rfloor$ divides $n$ in a recursive way. In order to speed things up, I exploit memoization.

ClearAll[primeQ];
primeQ = False;
primeQ = True;
primeQ[n_] := primeQ[n] = Nor@@Divisible[n,Select[Range[2,Floor[Sqrt[n]]],primeQ]]


A test:

PrimeQ /@ Range == primeQ /@ Range


True

It more efficient, to use short circuiting when the first divisor of n is found, so that non-primes have a chance to be identified quite early:

ClearAll[primeQ2];
primeQ2 = False;
primeQ2 = True;
primeQ2[n_] := primeQ2[n] = Block[{p = 1, b = True},
While[b && p < Floor[Sqrt[N[n]]],
p++;
b = If[primeQ2[p], Not[Divisible[n, p]], True];
];
b
];


A test with timings:

nmax = 100000;
a = PrimeQ /@ Range[nmax]; // AbsoluteTiming // First
b = primeQ /@ Range[nmax]; // AbsoluteTiming // First
c = primeQ2 /@ Range[nmax]; // AbsoluteTiming // First
a == b
a == c


0.039507

12.245

5.72106

True

True

The future runs of primeQ and primeQ2 will be faster since the intermediate results have all been stored.

@Mateus, technically satisfies your requirement! Also, unlike PrimeQ, does not handle n <= 0 and non-integer n.

meuPrimeQ = False;
meuPrimeQ = True;
meuPrimeQ[n_Integer /; n > 2] := Length[Divisors[n]] == 2