MATH 115A SOLUTION SET VIFEBRUARY 24, 2005

(1) Find the index of 5 relative to each of the primitive roots of 13. [Hint: Recall that2 is a primitive root modulo 13. To find the other prinitive roots, use the table that waswritten down today in class.]

Solution:

Since 2 is a primitive root modulo 13, the φ(12) = 4 primitive roots modulo 13 are,

21 ≡ 2 (mod 13), 25 ≡ 6 (mod 13), 27 ≡ 11 (mod 13), 211 ≡ 7 (mod 13).

Now the congruences

29 ≡ 5 (mod 13), 69 ≡ 5 (mod 13), 73 ≡ 5 (mod 13), 113 ≡ 5 (mod 13)

imply that

ind2(5) = 9, ind6(5) = 9, ind7(5) = 3, ind11(5) = 3.

(2) Find a primitive root modulo 11, and construct a table of indices relative to thisprimitive root. Use your table to solve the following congruences:

(a) 7x3 ≡ 3 (mod 11).(b) 3x4 ≡ 5 (mod 11).(c) x8 ≡ 10 (mod 11).

Solution:

(a) Consider the congruence 7x3 ≡ 3 (mod 11). Taking indices relative to the primitiveroot 2 leads to the congruence

ind(7) + 3 ind(x) ≡ ind(3) (mod 10),

and this in turn implies that

3 ind(x) ≡ 1 (mod 10).

This has the solution ind(x) ≡ 7 (mod 10), whence it follows that x ≡ 7 (mod 11).

(b) Taking indices yields the congruence

ind(3) + 4 ind(x) ≡ ind(5) (mod 10),

and this in turn implies that

4 ind(x) ≡ 6 (mod 10).

This last congruence has solutions ind(x) ≡ 4 (mod 10) and ind(x) ≡ 9 (mod 10). Thisgives x ≡ 5 (mod 11) and x ≡ 6 (mod 11) respectively.

(c) If x8 ≡ 10 (mod 11), then 8 ind(x) ≡ ind(10) (mod 10), that is, 8 ind(x) ≡ 5 (mod 10).This linear congruence is not soluble, and so the congruence x8 ≡ 10 (mod 11) has no solu-tions.

1

2 MATH 115A SOLUTION SET VI FEBRUARY 24, 2005

(3) The following is a table of indices for integers modulo 17 relative to the primitive root3:

a 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16ind3(a) 16 14 1 12 5 15 11 10 2 3 7 13 4 9 6 8

Use this table to solve the following congruences:(a) x12 ≡ 13 (mod 17).(b) 8x15 ≡ 10 (mod 17).(c) 9x8 ≡ 8 (mod 17).(d) 7x ≡ 7 (mod 17).

Solution:

(a) Consider the congruence x12 ≡ 13 (mod 17). Taking indices with respect to theprimitive root 3 yields

12 ind(x) ≡ ind(13) (mod 16),

or12 ind(x) ≡ 4 (mod 16).

This last congruence has solutions ind(x) ≡ 3, 7, 11 and 15 (mod 16), which leads to x ≡10, 11, 7, and 6 (mod 17).

(b) For the congruence 8x5 ≡ 10 (mod 17), we have

ind(8) + 5 ind(x) ≡ ind(10) (mod 16),

or10 + 5 ind(x) ≡ 3 (mod 16).

Thus 5 ind(x) ≡ 9 (mod 16), and this has solution ind(x) ≡ 5 (mod 16). It follows thatx ≡ 5 (mod 17).

(c) If 9x8 ≡ 8 (mod 17), then

ind(9) + 8 ind(x) ≡ ind(8) (mod 16).

It follows that 2 + 8 ind(x) ≡ 10 (mod 16), or

8 ind(x) ≡ 8 (mod 16).

This last congruence has solutions ind(x) ≡ 1, 3, 5, 7, 9, 11 and 15 (mod 16), and so x ≡3, 10, 5, 11, 14, 7, 12 and 6 (mod 17).

(d) Let 7x ≡ 7 (mod 17). Then x ind(7) ≡ ind(7) (mod 16), and so x ≡ 1 (mod 16).

(4) Find the remainder when 324 · 513 is divided by 17. [Hint: Use the theory of indices!]

Solution:

Given the congruence x ≡ 324 · 513 (mod 17), take indices with respect to the primitiveroot 3. Then

ind(x) ≡ 24 ind(3) + 13 ind(5) (mod 16)

≡ 24 · 1 + 13 · 5 (mod 16)

≡ 9 (mod 16),

MATH 115A SOLUTION SET VI FEBRUARY 24, 2005 3

which means that x ≡ 14 (mod 17).

(5) Show that the congruence x3 ≡ 3 (mod 19) has no solutions, while the congruencex3 ≡ 11 (mod 19) has three distinct solutions.

Solution:

Since (3, 18) = 3 and 36 6≡ 1 (mod 19), the congruence x3 ≡ 3 (mod 19) has no solutions.On the other hand, 116 ≡ 1 (mod 19), and so x3 ≡ 11 (mod 19) has three solutions.