nt-assignment1 soln.pdf

8/9/2019 NT-Assignment1 soln.pdf

1/11

MTH 4436 Section 5.2 Homework Solutions

Fall 2009

Pat Rossi Name

1. Use Fermat’s Theorem to verify that 17 divides 11104 + 1

Observe: 17 is prime and 17 - 11. Hence, by Fermat’s Theorem, 1117−

1 ≡ 1(mod17) .

i.e., 1116 ≡ 1(mod17)

Next, note that 104 = (6) (16) + 8.

Finally, note that since 1116 ≡ 1(mod17) , we have:

1116 − 1 ≡ 0(mod17) ⇒ (118 + 1) (118 − 1) ≡ 0(mod17) ⇒ either:

(118 + 1) ≡ 0(mod17) or (118 − 1) ≡ 0(mod17) .(because there are no zero divisors

mod 17, since 17 is prime)

If (118 − 1) ≡ 0(mod17) , then (114 + 1) (114 − 1) ≡ 0(mod17) ⇒ either:

(114 + 1) ≡ 0(mod17) or (114 − 1) ≡ 0(mod17) .

But neither (114 + 1) ≡ 0(mod17) nor (114 − 1) ≡ 0(mod17) .

Therefore, (118 + 1) ≡ 0(mod17) .

⇒ 118 ≡ −1(mod17)

Hence,

11104 = 11(6)(16)+8 = 11(6)(16) · 118 = (1116)6· 118 ≡ (1)6 · (−1)(mod17) ≡ −1(mod17)

i.e., 11104 = −1(mod17) ⇒ 11104 + 1 = 0 (mod 17)

Hence, 17 divides 11104 + 1.Proof.


2/11

2. ~

(a) If gcd (a, 35) = 1, show that a12 ≡ 1(mod35)

Proof. Let the hypothesis be given (i.e., suppose that gcd (a, 35) = 1.)

Since 35 = 5 · 7, it follows that 5 - a nor does 7 - a.

(If 5|a, then gcd(a, 35) ≥ 5 and if 7|a, then gcd(a, 35) ≥ 7, contrary to ourhypothesis.)

Hence, by Fermat’s Theorem, we have: a5 ≡ a (mod 5) and a7 ≡ a (mod7) , orequivalently, a4 ≡ 1(mod5) and a6 ≡ a (mod 7) .

The latter pair of congruences yield: a12 = (a4)3≡ 13 (mod5) ≡ 1(mod5)

and: a12 = (a6)2≡ 12 (mod 7) ≡ 1(mod7)

i.e., a12 ≡ 1(mod5) and a12 ≡ 1(mod7)

⇒ 5| (a12 − 1) and 7| (a12 − 1) .

Since gcd (5, 7) = 1, it follows that (5 · 7) | (a12 − 1)

i.e., a12 ≡ 1(mod35)

Remark 1 Our proof made use of the fact that if a|c and b|c and if gcd (a, b) = 1,then ab|c.

(b) If gcd (a, 42) = 1, show that 168 = 3 · 7 · 8 divides a6 − 1.

Proof. Let the hypothesis be given (i.e., suppose that gcd (a, 42) = 1.)

Since 42 = 2 · 3 · 7, it follows that neither 2, 3, nor 7 divide a.

(If 2|a, then gcd (a, 42) ≥ 2, if 3|a, then gcd(a, 42) ≥ 3, and if 7|a, then gcd (a, 42) ≥7, contrary to our hypothesis.)

Hence, by Fermat’s Theorem, we have: a2 ≡ a (mod 2) , a3 ≡ a (mod 3) , and a7 ≡a (mod 7) , or equivalently, a ≡ 1(mod2) , a2 ≡ 1(mod3) , and a6 ≡ 1(mod7) .

The three congruences yield: a6 = 16 (mod 2) ≡ 1(mod2) ,

a6 = (a2)3 ≡ 13 (mod 3) ≡ 1(mod3) ,

and: a6 = 1 (mod 7)

Our concern here is that our proposed divisor 168 has factorization 23 · 3 · 7.Thus, it will probably not be enough for us to have that a6 = 1(mod2) . Wewill probably need to show that a6 ≡ 1(mod23) . (i.e., we will need to show thata6 ≡ 1(mod8) .)

2


3/11

To get this result, note that a ≡ 1(mod2) ⇒ a − 1 ≡ 0(mod2)

⇒ a + 1 = (a − 1) + 2 ≡ 2 + 2 (mod 2) ≡ 0(mod2)

i.e., a + 1 ≡ 0(mod2) also.

Claim: Either (a− 1) ≡ 0(mod4) and (a + 1) ≡ 2(mod4) OR (a − 1) ≡

2(mod4) and (a + 1) ≡ 0(mod4)

To see this, note that since both (a − 1) and (a + 1) are even, neither can becongruent to 1 or 3 modulo 4.

We observe that it is impossible for both (a − 1) and (a + 1) to be congruent to 2modulo 4, for if (a − 1) ≡ 2(mod4) , then (a + 1) = (a − 1)+2 ≡ 2+2(mod4) ≡0(mod4) .

A similar argument shows that it is impossible for both (a − 1) and (a + 1) to becongruent to 0 modulo 4.

Claim: (a2 − 1) ≡ 0(mod8) .

From the previous claim, we can assume, without loss of generality, that (a − 1) ≡0(mod4) and (a + 1) ≡ 2(mod4) .

Hence, (a2 − 1) = (a − 1) | {z }

divis by 4

· (a + 1) | {z }

divis by 2

Therefore, (a2 − 1) is divisible by 8 and is congruent to 0 modulo 8.

Claim: a6 ≡ 1(mod8)

Observe: a6

− 1 = (a2

− 1) (a4

+ a2

+ 1) ≡ 0 · (a4

+ a2

+ 1) (mod 8) ≡ 0(mod8)

⇒ a6 ≡ 1(mod8)

Thus, we finally have the congruences: a6 ≡ 1(mod8) ,

a6 ≡ 1(mod3) ,

and: a6 ≡ 1(mod7)

i.e., a6 ≡ 1(mod2) , a6 ≡ 1(mod3) , and a6 ≡ 1(mod7) .

⇒ 8| (a6 − 1) , 3| (a6 − 1) , and 7| (a6 − 1) .

Since gcd (3, 7) = gcd (3, 8) = gcd (7, 8) = 1, it follows that (3 · 7 · 8) | (a6 − 1)

i.e., a6 ≡ 1(mod168)

3


4/11


5/11

3. From Fermat’s Theorem, deduce that for any integer n ≥ 0, 13| (1112n+6 + 1) .

Proof. Since 13 - 11, Fermat’s Theorem tells us that 1112 ≡ 1(mod13) .

⇒ (1112)n

≡ 1n (mod 13) ⇒ 1112n ≡ 1(mod13)

Hence, (1112n − 1) ≡ 0(mod13) .

When n = 1, we have (1112 − 1) ≡ 0(mod13) .

⇒ (116 + 1) (116 − 1) ≡ 0(mod13)

Since 13 is prime, there are no zero divisors mod 13, and hence:

either (116 + 1) ≡ 0(mod13) or (116 − 1) ≡ 0(mod13)

⇒ either 116 ≡ −1(mod13) or 116 ≡ 1(mod13).

Since, given k ∈ {2, 3, . . . , p − 1} , p − 1 is the smallest positive power of k that iscongruent to 1 modulo p, we have:

116 ≡ −1(mod13)

Observe: (1112n − 1) ≡ 0(mod13) ⇒ (1112n − 1) · 116 ≡ 0 · 116 (mod 13)

⇒ (1112n+6 − 116) ≡ 0(mod13) ⇒ (1112n+6 − (−1)) ≡ 0(mod13)

⇒ (1112n+6 + 1) ≡ 0(mod13)

Hence, 13| (1112n+6

+ 1)

5


6/11

4. Derive each of the following congruences:

(a) a21 ≡ a (mod 15) ∀a.

Proof. By Fermat’s Theorem, a5 ≡ a (mod5) for all a

⇒ (a5)4≡ a4 (mod 5) ⇒ a20 ≡ a4 (mod 5)

⇒ a21 = a20 · a ≡ a4 · a (mod 5) ≡ a5 (mod5) ≡ a (mod 5)

i.e., a21 ≡ a (mod 5) ∀a.

Next, observe that a3 ≡ a (mod3)

⇒ a21 = (a3)7≡ a7 (mod 3) ≡ a6 · a (mod 3) ≡ (a3)

2· a (mod3) ≡ a2 · a (mod3)

≡ a3 (mod 3) ≡ a (mod 3) a ≡ 0(mod3) ⇒ a21 ≡ 021 (mod 3) ≡ 0(mod3)

≡ a (mod 3)

i.e., a21 ≡ a (mod 3) ∀a.

Thus, we have: a21 − a ≡ 0(mod5) and a21 − a ≡ 0(mod3)

⇒ 5| (a21 − a) and 3| (a21 − a) .

Since gcd (3, 5) = 0, we have: (5 · 3) | (a21 − a) .

i.e., (a21 − a) ≡ 0(mod15) ⇒ a21 ≡ a (mod15)

(b) a7 ≡ a (mod 42) ∀a

Proof. By Fermat’s Theorem, a7 ≡ a (mod7) ∀a

Also, a3 ≡ a (mod3) ⇒ (a3)2≡ a2 (mod 3) ⇒ a6 ≡ a2 (mod 3) ⇒

a6 · a ≡ a2 · a (mod 3) ⇒ a7 ≡ a3 (mod 3) ≡ a (mod 3)

i.e., a7 ≡ a (mod 3) ∀a

Finally, a2 ≡ a (mod2) ⇒ (a2)3≡ a3 (mod 2) ⇒ a6 ≡ a3 (mod2)

⇒ a6 · a ≡ a3 · a (mod 2) ⇒ a7 ≡ a4 (mod 2) ≡ (a2)2

(mod 2) ≡ a2 (mod2)

≡ a (mod2)i.e., a7 ≡ a (mod 2) ∀a

Hence, 2| (a7 − a) ; 3| (a7 − a) ; and 7| (a7 − a) .

Since gcd (2, 3) = gcd (2, 7) = gcd (3, 7) = 1, it follows that (2 · 3 · 7) | (a7 − a)

i.e., (a7 − a) ≡ 0(mod42) ⇒ a7 ≡ a (mod42)

6


7/11

(c) a13 ≡ a (mod3 · 7 · 13) ∀a

Proof. By Fermat’s Theorem, a13 ≡ a (mod13)

a3 ≡ a (mod 3)

and a7 ≡ a (mod7)

Observe: a3 ≡ a (mod3) ⇒ a12 = (a3)4 ≡ a4 (mod 3) ≡ a3 · a (mod 3)

≡ a · a (mod3) ≡ a2 (mod3)

i.e., a12 ≡ a2 (mod 3)

⇒ a13 = a12 · a ≡ a2 · a (mod 3) ≡ a3 (mod3) ≡ a (mod 3)

i.e., a13 ≡ a (mod 3)

Also, if 7 - a, then a6 ≡ 1(mod7)

Since a7 ≡ a (mod 7) , we have a13 = a7 · a6 ≡ a · 1 (mod 7)≡ a (mod7)

i.e., a13 ≡ a (mod 7) if 7 - a

If 7|a, then a ≡ 0(mod7) ⇒ a13 ≡ 013 (mod 7) ≡ 0(mod7) ≡ a (mod7)

i.e., a13 ≡ a (mod 7) if 7|a.

Thus, a13 ≡ a (mod7) ∀a.

Thus, we have: a13 ≡ a (mod3); a13 ≡ a (mod7); and a13 ≡ a (mod13)

⇒ 3| (a13 − a) ; 7 (a13 − a) ; and 13 (a13 − a) .

Since 3, 7, 13 are pairwise relatively prime, it follows that (3 · 7 · 13) | (a13 − a) ,

or equivalently: a13 ≡ a (mod3 · 7 · 13)

7


8/11

(d) a9 ≡ a (mod 30) ∀a.

Proof. By Fermat’s Theorem, a2 ≡ a (mod2)

a3 ≡ a (mod 3)

and a5 ≡ a (mod5)

Observe: a2 ≡ a (mod2) ⇒ a8 = (a2)4 ≡ a4 (mod2)

≡ (a2)2

(mod2) ≡ a2 (mod2) ≡ a (mod 2)

i.e., a8 ≡ a (mod 2) .

Thus, a9 = a8 · a ≡ a · a (mod2) ≡ a2 (mod 2) ≡ a (mod 2)

i.e., a9 ≡ a (mod 2)

Also, a3 ≡ a (mod3) ⇒ a9 = (a3)3≡ a3 (mod 3) ≡ a (mod 3)

i.e., a9 ≡ a (mod 3) .

Finally, if 5 - a, then a4 ≡ 1(mod5)

This combined with the fact that a5 ≡ a (mod 5) yields:

a9 = a5 · a4 ≡ a · 1 (mod 5) ≡ a (mod 5) .

i.e., a9 ≡ a (mod 5) if 5 - a.

If 5|a, then a ≡ 0(mod5) ⇒ a9 ≡ 09 (mod 5) ≡ 0(mod5) ≡ a (mod5)

i.e., If 5|a, then a9 ≡≡ a (mod5) .

Thus, a9 ≡ a (mod 5) ∀a.

So we have: a9 ≡ a (mod2); a9 ≡ a (mod3); and a9 ≡≡ a (mod 5) , or equivalently:

2| (a9 − a) ; 3| (a9 − a) ; and 5| (a9 − a) .

Since 2, 3, 5 are pairwise relatively prime, it follows that (2 · 3 · 5) | (a9 − a) .

i.e., a9 ≡ a (mod 30) ∀a.

8


9/11


10/11

6. ~

(a) Find the units digit of 3100 by the use of Fermat’s Theorem.

First, let’s recognize that the units digit of 3100 is the same as 3100 (mod 10) .

Next, note that since neither 2 nor 5 divide 3, that by Fermat’s Theorem:

3 ≡ 1(mod2) , and hence, 3100 ≡ 1100 (mod 2) ≡ 1(mod2) .

i.e., 3100 ≡ 1(mod2) .

Hence, 2| (3100 − 1) .

Also: 32 ≡ −1(mod5) , and hence, 3100 = (32)50≡ (−1)50 (mod5) ≡ 1(mod5) .

i.e., 3100 ≡ 1(mod5) .

Hence, 5| (3100 − 1) .

Since 2 and 5 are relatively prime, (2 · 5) | (3100 − 1) .

i.e., 3100 ≡ 1(mod10) .

(b) For any integer a, verify that a5 and a have the same digits unit.

Proof. From part 6.a, it suffices to show that a5 ≡ a (mod10) .

By Fermat’s Theorem, a5 ≡ a (mod5) ∀a ∈ Z

⇒ 5| (a5 − a)

Also: a2

≡ a (mod 2) ∀a ∈ Z

⇒ a4 = (a2)2≡ a2 (mod2) ≡ a (mod2)

i.e., a4 ≡ a (mod 2)

⇒ a5 = a4 · a ≡ a · a (mod 2) ≡ a2 (mod2) ≡ a (mod2)

i.e., a5 ≡ a (mod 2) ∀a ∈ Z

⇒ 2| (a5 − a)

Since 2 and 5 are relatively prime, we have:

(2 · 5) | (a5 − a) .

i.e., a5 ≡ a (mod 10) ∀a ∈ Z

10


11/11

nt-assignment1 soln.pdf

Documents