这是一个优化版本，仅使用最高sqrt（n）测试并使用Miller-Rabin测试（从Joni的答案开始）：

Here's an optimized version using testing only up to sqrt(n) and using the Miller-Rabin test (as of Joni's answer):

public boolean returnPrime(BigInteger number) {
    //check via BigInteger.isProbablePrime(certainty)
    if (!number.isProbablePrime(5))
        return false;

    //check if even
    BigInteger two = new BigInteger("2");
    if (!two.equals(number) && BigInteger.ZERO.equals(number.mod(two)))
        return false;

    //find divisor if any from 3 to 'number'
    for (BigInteger i = new BigInteger("3"); i.multiply(i).compareTo(number) < 1; i = i.add(two)) { //start from 3, 5, etc. the odd number, and look for a divisor if any
        if (BigInteger.ZERO.equals(number.mod(i))) //check if 'i' is divisor of 'number'
            return false;
    }
    return true;
}