Possibly quit your job with a polyglot
Python 2 / Python 3 - 128

n=int(input())
s="Isf  tthhies ,b oIs sq ufiitn.d"
print(["not "*any([n%a&lt;1for a in range(2,n)])+"prime",s[::2]+s[1::2]][1/2&gt;0])


Identifies prime numbers with Python 2, get's you into trouble with Python 3.



Bonus



__help__ = """In the forloop I'm setting the values i to n.
To be a "notPrimes", it's n % i == 0 if: 1 &lt; i, i &lt; n. (We
do tests i&lt;1/2 i==1, too). Then, prints resulting, prime text.
In case i==n: quit. Just if it aborts: try."""

# read integer from command line
n=int(input())

try:

    # primes = True
    notPrimes = False

    # try each i to n
    for i in range(n):

        # ignore 0 or 1
        if i &lt; 1 / 2 or i == 1:
            continue

        # test divisibility
        if n % i == 0:
            notPrimes = True

    # print result
    if notPrimes:
        print("not prime")
    else:
        print("prime")

except:

    # if program aborts: print help and error code
    print(__help__ [::7])


Try it with Python 2 or Python 3! (In contrast to the golf version above the roles changed: Python 3 is the prime number identifier. Python 2 contains the Easter egg.)

Please, excuse my bad English in the help text! ;)

And I do use the word "quit". But somehow I need to describe when my program terminates. ;)