53, 54, 58: Done in class

55:

gap> x:=X(Rationals,"x");
x
gap> a:=x^3+x+1;
x^3+x+1
gap> b:=x^2+x;
x^2+x
gap> ShowGcd(a,b);
x^3+x+1=(x-1)*(x^2+x) + 2*x+1
x^2+x=(1/2*x+1/4)*(2*x+1) + -1/4
2*x+1=(-8*x-4)*(-1/4) + 0
The Gcd is -1/4
 = 1*(x^2+x) + (-1/2*x-1/4)*(2*x+1)
 = (-1/2*x-1/4)*(x^3+x+1) + (1/2*x^2-1/4*x+3/4)*(x^2+x)
-1/4


56:

gap> a:=Filtered([1..23],x->Gcd(x,24)=1);
[ 1, 5, 7, 11, 13, 17, 19, 23 ]
gap> List(a,x->OrderMod(x,24));
[ 1, 2, 2, 2, 2, 2, 2, 2 ]

No primitive element. For a prime there would be only one element of order
2, here are many.



57:

Define A and B

gap> A:=[[0,0,0,1],[1,0,0,0],[0,1,0,0],[0,0,1,2]]*One(GF(3));
[ [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3)^0 ], [ Z(3)^0, 0*Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), Z(3)^0, 0*Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), Z(3)^0, Z(3) ] ]
gap> Display(A);
 . . . 1
 1 . . .
 . 1 . .
 . . 1 2
gap> B:=[[0,2,1,2],[0,0,2,1],[0,0,0,2],[2,1,2,1]]*One(GF(3));
[ [ 0*Z(3), Z(3), Z(3)^0, Z(3) ], [ 0*Z(3), 0*Z(3), Z(3), Z(3)^0 ], 
  [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ], [ Z(3), Z(3)^0, Z(3), Z(3)^0 ] ]

# order 5 (16th power)
gap> a:=A^16;
[ [ Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ Z(3), Z(3), 0*Z(3), Z(3) ], 
  [ Z(3), Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), Z(3) ] ]
gap> b:=B^16;
[ [ Z(3)^0, 0*Z(3), Z(3), Z(3) ], [ Z(3)^0, Z(3)^0, 0*Z(3), Z(3) ], 
  [ Z(3), Z(3)^0, Z(3)^0, 0*Z(3) ], [ 0*Z(3), Z(3), Z(3), Z(3) ] ]
# naive discrete log
gap> First([0..4],x->a^x=b);
3

# order 16:

gap> a:=A^5;
[ [ Z(3), Z(3)^0, Z(3), Z(3) ], [ Z(3)^0, Z(3), Z(3)^0, Z(3) ], 
  [ 0*Z(3), Z(3)^0, Z(3), Z(3)^0 ], [ Z(3)^0, Z(3), Z(3), 0*Z(3) ] ]
gap> b:=B^5;
[ [ Z(3)^0, Z(3)^0, 0*Z(3), Z(3)^0 ], [ Z(3)^0, Z(3)^0, Z(3)^0, 0*Z(3) ], 
  [ Z(3)^0, Z(3)^0, Z(3)^0, Z(3)^0 ], [ Z(3)^0, 0*Z(3), Z(3)^0, 0*Z(3) ] ]

Hensel lifting, Order 16
Step 1, take 8th powers:
gap> a^8;
[ [ Z(3), 0*Z(3), 0*Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ] ]
gap> b^8;
[ [ Z(3), 0*Z(3), 0*Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ] ]

# as p=2 we can determine the exponent simply by testing for equality
gap> a^8=b^8;
true
gap> e0:=1; # yes, exponent is 1
1

# now change b to divide off a^e0. The result will be a power of a^2.
gap> b1:=b/a;
[ [ Z(3), Z(3), Z(3), Z(3)^0 ], [ 0*Z(3), Z(3), Z(3), Z(3) ], 
  [ Z(3)^0, 0*Z(3), Z(3), Z(3) ], [ Z(3), Z(3), Z(3)^0, Z(3)^0 ] ]
gap> a1:=a^2;
[ [ Z(3)^0, Z(3)^0, Z(3)^0, Z(3) ], [ 0*Z(3), Z(3)^0, Z(3)^0, Z(3)^0 ], 
  [ Z(3), 0*Z(3), Z(3)^0, Z(3)^0 ], [ Z(3)^0, Z(3)^0, Z(3), Z(3) ] ]

# repeat the process, but a1 now has order 8, so 4th powers:

gap> a1^4=b1^4;
true
gap> e1:=1;
1

# again change b, again p-th power of a

gap> b2:=b1/a1;
[ [ Z(3), 0*Z(3), 0*Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ] ]
gap> a2:=a1^2;
[ [ Z(3), Z(3)^0, Z(3)^0, Z(3) ], [ 0*Z(3), Z(3), Z(3)^0, Z(3)^0 ], 
  [ Z(3), 0*Z(3), Z(3), Z(3)^0 ], [ Z(3)^0, Z(3)^0, Z(3), 0*Z(3) ] ]

# now order 4

gap> a1^2=b1^2;
false
gap> e2:=0;
0

gap> b3:=b2/a2^0;
[ [ Z(3), 0*Z(3), 0*Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ] ]
gap> a3:=a2^2;
[ [ Z(3), 0*Z(3), 0*Z(3), 0*Z(3) ], [ 0*Z(3), Z(3), 0*Z(3), 0*Z(3) ], 
  [ 0*Z(3), 0*Z(3), Z(3), 0*Z(3) ], [ 0*Z(3), 0*Z(3), 0*Z(3), Z(3) ] ]

gap> a3^4=b3^4;
true
gap> e3:=1;
1

# we've found the digits of the exponent
gap> e:=e0+2*e1+4*e2+8*e3;
11
gap> a^e=b; # test
true

Combine with Chinese remainder
gap> ChineseRem([16,5],[11,3]);
43
gap> A^43=B; # test
true