Goldbach's Conjecture
My solution consists of a single change:


Changed the is_prime function into a prime sieve (from this answer) + lookup


My final code (Also handled odd numbers just for fun, definitely not because I misread the question):

def main():
    import itertools
    import math
    izip = itertools.zip_longest
    chain = itertools.chain.from_iterable
    compress = itertools.compress
    def rwh_primes2_python3(n):
        """ Input n&gt;=6, Returns a list of primes, 2 &lt;= p &lt; n """
        zero = bytearray([False])
        size = n//3 + (n % 6 == 2)
        sieve = bytearray([True]) * size
        sieve[0] = False
        for i in range(int(n**0.5)//3+1):
          if sieve[i]:
            k=3*i+1|1
            start = (k*k+4*k-2*k*(i&amp;1))//3
            sieve[(k*k)//3::2*k]=zero*((size - (k*k)//3 - 1) // (2 * k) + 1)
            sieve[  start ::2*k]=zero*((size -   start  - 1) // (2 * k) + 1)
        ans = [2,3]
        poss = chain(izip(*[range(i, n, 6) for i in (1,5)]))
        ans.extend(compress(poss, sieve))
        return ans

    string2 = "Impossible"
    n = int(input())
    sieve = [t for t in rwh_primes2_python3(n) if t &lt;= math.floor((n // 2) / 2) * 2 + 1][::-1]
    another = [t for t in rwh_primes2_python3(n) if t &gt;= math.floor((n // 2) / 2) * 2 + 1]

    if n &gt; 5 and n % 2 == 0:
        for smaller in sieve:
            if n - smaller in another:
                print (smaller, n - smaller)
                break
    elif n % 2 == 1 and n != 5:
        if n - 2 in another or n-2 in sieve: print (2, n-2)
        else: print ('Impossible')
    elif n == 4:
        print ('2 2')
    else:
        print ('2 3')

main()