Math 5248 Lec 2 Homework #4 Due April 2.
As always, you must show your work.
10 POINTS EACH PROBLEM FOR A TOTAL OF 130 POINTS
9.1.5: 21000000 %17 = (262500 )16 %17 =1
9.5.7 The MI of 23 is 23519%521 = 68. (3pts)
Why this works: Mod 521, 23 • 23519 = 23520 = 1, so 23519 is the MI of 23.(7pts)
9.6.6: 2260%1039=630. So 630 is a square root of 2, provided 2 has a square root. So we
must check: 6302%1039=2, so indeed 630 is a square root of 2.
For comparison, look at exercise 1 in this section, which is not part of the homework:
25%19=13, so 13 is a square root of 2, provided 2 has a square root. Bu in this case,
132%19≠2, so 2 does not have a square root. Indeed, using Euler,
29%19≡─1, not +1, mod19, so 2does not have a square root.
9.7.7: p=199 and n =11.Note that 199≡1 mod 11 and gcd(11,198/11)=gcd(11,18)=1. So
then we find a MI of 11 mod 18, which is 5.Then 585mod199=43. So 43 is an 11th root of
58,provide 58 has an 11th root. We check 4311%199=58.
14.4: Using the simple procedure described on page 186, and not the complicated
procedure described (?) on page 187, suppose n = 437 and Vera chooses x =100. What
value does Vera send to Peter? What value does Peter send to Vera? Answer the second
question using Sun Ze’s theorem.
Vera sends 1004 %437 = 416.
Peter knows that 437=(19)(23).
First he computes a square root of 416 mod 19. Now 416 mod 19=17, [you could omit
this simplification], so we want a square root of 17 mod 19. We compute 175 %19 = 6.
Then Peter computes a square root of 416 mod 23. Now 416 mod 23 =2, And to compute a
square root of 2 mod 23 we calculate 26 %23 = 18.
Now we combine these two solutions using CRT. We must find a and b such that
a23+b19=1. We get (5)(23)─(6)(19)=1, or 115─114=1. Then we get the solution
(6)(115)─(18)(114)%437=386,so Peter sends 386 to Vera.
NOTE: There is a misprint on page 187: (p+1)/2 should be (p+1)/4. I am sorry I did not
notice this when I assigned the homework. However if z is a square mod p and you raise it
to the power (p+1)/2, you get z again, so it is obvious the formula in the book is wrong.
10.8.6: 6936 %109 = 63, so 69 is not a cube mod 109.
10.8.8: 6836 %109 = 1, so 68 is a cube mod 109. [We do not know what a cube root of
68 is, but in fact 233 %109 = 68.]
23.1.3
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23.1.7: (gh)(h-1g-1)= g(hh-1)g-1=geg-1=gg-1=e, so the inverse of gh is h-1g-1. This is not the
only way to do this exercise.
23.2.3: We must check three things:
𝑒 ∈ 𝐻 and 𝑒 ∈ 𝐾, so 𝑒 ∈ 𝐻 ∩ 𝐾
If 𝑥 ∈ 𝐻 ∩ 𝐾, then 𝑥 ∈ 𝐻 and 𝑥 ∈ 𝐾. Since H is a subgroup, 𝑥 −1 ∈ 𝐻 .
Similarly 𝑥 −1 ∈ 𝐾. Hence 𝑥 −1 ∈ 𝐻 ∩ 𝐾.
Finally, if a and b are in H∩K, then a and b are in H so ab is in H; and a and b are in K, so
ab is in K. Hence ab is in H∩K.
23.2.5: Using notation we learned in section 23.5, the 5 subgroups are
<0>={0}, <1> = Z/16, <2>, <4>, and <8>. You were not expected to show that these are
all the subgroups, although with the knowledge gained from 23.6 this is not hard to do.
23.3.1 (You can use 23.2.3 in your explanation.) H∩K is a subgroup of both H and K, so
the order of H∩K must divide both 16 and 35. Hence the order of H∩K is one and so
H∩K consists solely of the identity element.
23.6.1 Recall that we mentioned several times in class that the last bullet on page 309 is
incorrect. There will be φ(4)=2 element of order 4 in Z/8. To find them, first note that the cyclic
subgroup of order 4 is generated by 8/4=2. Then 2 is an element of order 4, and all the elements
of order 4 are given by k2, where 1<k<4 and k is relatively prime to 4; that is k = 1,3. So the two
elements of order 4 are 2 and 6.
There will be φ(6)=2 elements of order 6 in Z/72. Note that the cyclic group of order 6 is
generated by 72/6 = 12. Then 12 has order 6,and all the elements of order 6 are given k12,
where 1<k<6 and k is relatively prime to 6; that is k =1,5. So the elements of order 6 are 12 and
60.