The answer in this post provides an excellent functional form to test for primality (AKS primality test):
How to create the most compact mapping n → isprime(n) up to a limit N?
and the test does not require any "memory" of primes found.
The task is then to encode that function without making it an actual function.
n = int(input("Number: "))
M=5
print(f'{M} primes below {n}: ',end='')
itest=n-n%2-1 # number is odd, and is below input number
nab=0
while nab<M:
  if itest == 2:
    print(itest,end=' ')
    nab+=1
    break
  if itest == 3:
    print(itest,end=' ')
    nab+=1
    itest-=1
    continue
  if itest % 2 == 0:
    itest-=2
    continue
  if itest % 3 == 0:
    itest-=2
    continue
  i = 5
  w = 2
  ipass=True
  while i * i <= itest:
    if itest % i == 0:
      ipass=False
      break
    i += w
    w = 6 - w
  if ipass:
    print(itest,end=' ')
    nab+=1
  itest-=2
print('')
print(f'{M} primes above {n}: ',end='')
itest=n+n%2+1 # number is odd, and is above input number
nab=0
while nab<M:
  if itest == 2:
    print(itest,end=' ')
    nab+=1
    itest+=2
    continue
  if itest == 3:
    print(itest,end=' ')
    nab+=1
    itest+=2
    continue
  if itest % 2 == 0:
    itest+=2
    continue
  if itest % 3 == 0:
    itest+=2
    continue
  i = 5
  w = 2
  ipass=True
  while i * i <= itest:
    if itest % i == 0:
      ipass=False
      break
    i += w
    w = 6 - w
  if ipass:
    print(itest,end=' ')
    nab+=1
  itest+=2
The output for 22, for example is
Number: 22
5 primes below 22: 19 17 13 11 7
5 primes above 22: 23 29 31 37 41
and another example:
Number: 131553423669295
5 primes below 131553423669295: 131553423669257 131553423669181 131553423669097 131553423669043 131553423668977
5 primes above 131553423669295: 131553423669299 131553423669347 131553423669413 131553423669419 131553423669439
Time Test
This algorithm is much faster for larger numbers.
For example, a timeit test on this algorithm for the number 5564445, with 1000 executions, took 2.66 seconds. With the naive approach of dividing by every number until a divisor is found, takes 1hr 40 minutes.