Here is a game you can play. It is essentially the simplified version of poker over the telephone from Section 18.2.

There are five cards: ten, jack, queen, king, ace. They are shuffled and disguised by raising their numbers to a random exponent mod the prime 24691313099. You are supposed to guess which one is the ace.

The cards are represented by ten = 200514, etc., because t is the 20th letter, e is the 5th letter, and n is the 14th letter.

Type the following into Sage. The value of $k$ forces the randomization to have 248 as its starting point, so the random choices of $e$ and shuffle do not change when we repeat the process with this value of $k$. Varying $k$ gives different results.

cards=[200514,10010311,1721050514,11091407,10305] 
p=24691313099 
k=248 set_random_seed(k) 
e=randint(10,10^7) 
def pow(list,ex): 
   ret = [] 
   for i in list: 
      ret.append(mod(i,p)^ex) 
   return ret 
s=pow(cards,2*e+1) 
shuffle(s) 
print(s)

Evaluation yields

[10426004161, 16230228497, 12470430058, 3576502017, 2676896936]

These are the five cards. None looks like the ace; that’s because their numbers have been raised to powers mod the prime. Make a guess anyway. Let’s see if you’re correct.

Add the following line to the end of the program:

print(pow(s,mod(2*e+1,p-1)^(-1)))

and evaluate again:

[10426004161, 16230228497, 12470430058, 3576502017, 2676896936] 
[10010311, 200514, 11091407, 10305, 1721050514]

The first line is the shuffled and hidden cards. The second line removed the $k$th power and reveals the cards. The fourth card is the ace. Were you lucky?

If you change the value of $k$, you can play again, but let’s figure out how to cheat. Remove the last line of the program that you just added and replace it with

pow(s,(p-1)/2)

When the program is evaluated, we obtain

[10426004161, 16230228497, 12470430058, 3576502017, 2676896936] 
[1, 1, 1, 24691313098, 1]

Why does this tell us the ace is in the fourth position? Read the part on “How to Cheat” in Section 18.2. Raise the numbers for the cards to the $(p-1)/2$ power mod $p$ (you can put this extra line at the end of the program and ignore the previous output):

print(pow(cards,(p-1)/2))
[1, 1, 1, 1, 24691313098]

We see that the prime $p$ was chosen so that only the ace is a quadratic nonresidue mod $p$.

If you input another value of $k$ and play the game again, you’ll have a similar situation, but with the cards in a different order.