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MAT 320 Elementary Number Theory

Homework 7 Solution

Due Nov. 12th in class

1. Find the last digit of the decimal expansion of 31000.

Solution. This is equivalent to finding the least positive residue of 31000 mod 10. By

Euler’s theorem, we have φ(10) = 10 · (1− 12)(1− 1

5) = 4. Since 3 is coprime to 10, thus

we have

34 ≡ 1 (mod 10).

Hence 31000 ≡ 1 (mod 10).

2. Use Euler’s theorem to find the least positive residue of 3100,000 modulo 35.

Solution. Same approach as problem (1), we compute φ(35) = 35 · (1− 15)(1− 1

7) = 24.

Thus by Euler’s theorem, we obtain

3100,000 = 324·4166+16 ≡ 316 ≡ 814 ≡ 114 ≡ 162 ≡ 11 (mod 35).

Hence the least positive residue is 11 modulo 35.

3. Show that if a is an integer such that a is not divisible by 3 or such that a is divisible by

9, then a7 ≡ a (mod 63).

Proof. We first observe that 7 is coprime to 9. Thus to show that statement, it is

equivalent to showing that a7 ≡ a (mod 7) and a7 ≡ a (mod 9). Since 7 is a prime

number, thus by Fermat’s little theorem, we obtain, for any a,

a7 ≡ a (mod 7).

To show the second congruence, let’s first assume that a is divisible by 9, then a ≡ 0

(mod 9), thus a7 ≡ a ≡ 0 (mod 9). So the congruence is really trivial. Now we assume

that a is not divisible by 3, this is same as gcd(a, 9) = 1. Then we use Euler’s theorem

and obtain

a6 ≡ 1 (mod 9).

Hence a7 ≡ a (mod 9).

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4. Show that if a is an integer coprime to 32,760, then a12 ≡ 1 (mod 32, 760).

Proof. We first factor 32760 = 23 · 32 · 5 · 7 · 13. Thus the statement is equivalent

to showing a12 ≡ 1 (mod 8), a12 ≡ 1 (mod 9), a12 ≡ 1 (mod 5), a12 ≡ 1 (mod 7), and

a12 ≡ 1 (mod 13). We compute φ(8) = 4, φ(9) = 6, φ(5) = 4, φ(7) = 6 and φ(13) = 12.

Hence these congruences are true.

5. Show that aφ(b) + bφ(a) ≡ 1 (mod ab), if a and b are coprime to each other.

Proof. We use the condition that a and b are coprime to each other. So it suffices to

prove the following two congruences:

aφ(b) + bφ(a) ≡ 1 (mod a)

and

aφ(b) + bφ(a) ≡ 1 (mod b).

Since a and b are symmetric, we only need to show one of these two. Let’s show the first

one: aφ(b) + bφ(a) ≡ 1 (mod a). By Euler’s theorem, we have

bφ(a) ≡ 1 (mod a).

It’s easy to see that

aφ(b) ≡ 0 (mod a).

Hence adding these two congruences up, we obtain aφ(b) + bφ(a) ≡ 1 (mod a).

6. Suppose that n = p1p2 · · · pk where p1, · · · , pk are distinct odd primes. Show that

aφ(n)+1 ≡ a (mod n).

Proof. Using Euler’s formula we obtain φ(n) = (p1− 1)(p2− 1) · · · (pk − 1). The desired

congruence is equivalent to the following ones:

aφ(n)+1 ≡ a (mod pi), ∀i = 1, 2, · · · , k.

Now let’s show aφ(n)+1 ≡ a (mod pi). First we look at the trivial case, i.e., a ≡ 0

(mod pi). It’s easy to see that the congruence is true. Now we assume a 6≡ 0 (mod pi),

Since pi is prime, by Fermat’s little theorem, we obtain

api−1 ≡ a (mod pi), ,∀a 6≡ 0 (mod pi).

Thus

aφ(n) = a(p1−1)(p2−1)···(pk−1) ≡ 1 (mod pi).

Hence

aφ(n)+1 ≡ a (mod pi), ∀a.

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7. Show that if m is a positive integer, m > 1, then am ≡ am−φ(m) (mod m) for all positive

integers a (optional).

Proof. Write m = pe11 pe2

2 · · · pekk . Then it is easy to see that the statement is equivalent

to the following congruences:

am ≡ am−φ(m) (mod peii ), ∀i.

Now write a = pfii a1, where gcd(a1, p

eii ) = 1. By Euler’s theorem, we have

aφ(m)1 ≡ 1 (mod pei

i ).

Thus am1 ≡ a

m−φ(m)1 (mod pei

i ). So it suffices to show

(pfii )m ≡ (pfi

i )m−φ(m) (mod peii ).

Equivalently we need to show

(pfii )m ≡ 0 (mod pei

i )

and

(pfii )m−φ(m) ≡ 0 (mod pei

i ).

In fact, it suffices to show the latter since (pfii )m is a power of (pfi

i )m−φ(m). To show the

latter, we just need to show that m − φ(m) ≥ ei. In fact we may assume that m = peii .

Thus we have

m− φ(m) = pei−1i ≥ 2ei−1 = (1 + 1)ei−1 = 1 + (ei − 1) + · · · ) > ei.

Hence the statement.