Number Theory 2018 Midterm Solutions

Number

Theory 2018 Midterm Solutions

1. (a) Define what it means for an integer p to be prime.

An integer is prime if (it is positive and) it only positive divisors are 1 and itself.

Solution

(b) The primes 3, 5, and 7 are a ‘prime triplet’ ? Determine how many other prime triplets primes there are. (How many primes p 6= 3 are there such that p + 2 and p + 4 are also prime.)

The given on is the only one. For any triple,p, p+ 2, p+ 4 at least one of the numbers is divisible by 3. So they can only be a triplet primes if one of them is 3.

Solution

2. (a) What does it mean that (a, b, c) is a pythagorean triple?

a²+b²=c²

Solution

(b) What does it mean that it is a primitive pythagorean triple?

a, bandchave no common factors.

Solution

(c) For consecutive positive odd integers t and t + 2, show that 1

t + 1 t + 2 = a

b

where (a, b, c) is a pythagorean triple.

If tands=t+ 2 are consecutive odds, we have that (st,^s2−t₂ ²,^s2+t₂ ²) is a primitive pythagorean triple. Usings=t+ 2 we have

s²−t²

2 =(t+ 2)²−t²

2 = 2t+ 2.

So 1

t+ 1

t+ 2=2t+ 2 st = a

b

wherea=standb= 2t+ 2, and (a, b, c) is a pythagorean triple.

Solution

3. (a) Use the Euclidean algorithm to compute gcd(899, 493).

Step a = b · q + r

1 899 = 493 · 1 + 406

2 493 = 406 · 1 + 87

3 406 = 87 · 4 + 58

4 87 = 58 · 1 + 29

5 58 = 29 · 2 + 0

So gcd(899,493) = 29

Solution

(b) Use the extended Euclidean algorithm to write gcd(899, 493) as a linear combination of 899 and 493.

Step r 899 493

1 406 1 −1

2 87 −1 1−(−1) = 2

3 58 1−4(−1) = 5 −1−4(2) =−9 4 29 −1−1(5) =−6 2−1(−9) = 11 So 29 =−6(899) + 11(493)

Solution

4. (a) What does it mean that a ≡

m

b?

a≡mbifmdividesa−b.

Solution

(b) Show that if a ≡

m

b and a

⁰

≡

m

b

⁰

then aa

⁰

≡

m

bb

⁰

.

Ifa≡mbanda⁰≡mb⁰thenmdividesa−banda⁰−b⁰soa=cm+bfor some integercanda⁰=c⁰m+bfor some integersc⁰. Thusaa⁰= (cm+b)(c⁰m+b⁰) = m(cc⁰m+bc⁰m+b⁰cm) +bb⁰. Takingbb⁰from both sides shows thatmdivides aa⁰−bb⁰, and soaa⁰≡mbb⁰.

(c) Prove that n is divisible by 3 if and only if the the sum of its digits is divisible by 3.

Where di is thei^th digit ofn(counting from the right) we have thatn= d1+ 10¹(d2) + 10²(d3) +· · ·+ 10^r−1dr for somer. Observing that 10≡31 and so that 10ⁱ≡₃1ⁱ= 1, this gives, modulo 3, that

n≡₃d1+ 1(d2) +· · ·+ 1(dr) =d1+d2+· · ·+dr.

Sonis divisible by 3, or congruent to 0 mod 3, if and only if the sum of its digits is.

Solution

5. Let p ≥ 3 be a prime number.

(a) Assume that a

²

≡

p

b

²

for integers a and b {1, 2, . . . , p − 1}. Show that b ≡

p

±a.

(b) What are the numbers a such that a

²

≡

p

1.

(c) Show that (p − 1)! ≡

_p

p − 1 for any prime p.

(a) If a² ≡p b² thenp|(a²−b²) = (a−b)(a+b). Asp has no zero di- visors this means one of these is a multiple ofp, and so thatb≡_paorb≡_p−a.

(b) Clearly 1 and −1 ≡p (p−1) are such numbers. By (a) there are not more.

(c) Aspis prime, every integera∈ {1,2, . . . , p−1}has a unique multiplicative inverse: an integer 1/ain{1,2, . . . , p−1}such thata·1/a= 1. As by part (b) the element 1 has only two square roots: 1 and−1. So when we take the product (p−1)! every number cancels with its multiplicative inverse except 1 and−1. So (p−1)! = 1·1· · · · ·1· −1 =−1≡p(p−1).

Solution

6. (a) What does the euler function φ count? (I.e.: What is φ(n)?)

φ(n) is the number of integersamodulonfor which gcd(a, n) = 1.

Solution

(b) What is φ(525)?

φ(525) =φ(3)·φ(25)·φ(7) = 2·(25−5)·6 = 240

Solution

(c) Find a number n such that n ≡

₅

3 and n ≡

₂₃

5

We use the Chinese Remainder Theorem Algorithm. Asn≡₅3 we have that n= 3 +c·5. Plugging this inton≡235 we get that

3 +c·5≡235⇒c≡232/5.

As 9∗5≡23−1 we see that−9 is the multiplicative inverse of 5 mod 23. (One could also use the extended euclidean algorithm to find this.) So c=−18.

Thusn= 3 + (−18)5 =−87≡11528 is such ann.

Solution

7. What is the highest power d of 2 such that 2

^d

divides (2

ⁿ

)!?

2ⁿ⁻¹ of the numbers from 1 to 2ⁿhave a factor of 2;

2ⁿ⁻² of the numbers have a second factor of 2;

2ⁿ⁻ⁱhave ani^thfactor of 2;

upto one having ann^thfactor of 2.

So the product (2ⁿ)! of these numbers has

d= 2ⁿ⁻¹+ 2ⁿ⁻²+· · ·+ 2¹+ 1 = 2ⁿ−1

factors of 2. Sod= 2ⁿ−1 is the largest power such that 2^ddivides (2ⁿ)!.

Solution

8. (a) What is a Mersenne prime?

A primepof the formp= 2^q−1 for a primeq.

Solution

(b) Show that if p is a prime of the form a − 1 then it is a Mersenne prime.

Using the geometric series expansion

aⁿ−1 = (a−1)(aⁿ⁻¹+aⁿ⁻²+· · ·+a¹+ 1)

we see that such a prime p =aⁿ−1 must havea−1 = 1, so a= 2. If p= 2ⁿ−1 andnisn’t prime, then wheren=stwe have that

2ⁿ−1 = (2^s)^t−1 = (2^s−1)((2^s)^t−1+ (2^s)^t−2+· · ·+ (2^s)¹+ 1) and sopisn’t prime. So whenp=aⁿ−1 we have thata= 2 andnis prime;

thuspis a Mersenne prime.

Solution

9. (a) Let σ(n) be the sum of the divisors of n. What is σ(10)?

σ(10) = 1 + 2 + 5 + 10 = 18.

Solution

(b) If p = 2

^q

− 1 is a Mersenne prime, what is σ(p · (2

^q−1

))?

If p = 2^q−1 is a Mersenne prime, then p·(2^q−1) is a perfect number, so σ(p·2^q−1) = 2p·2^q−1=p2^q.

Solution

(c) Show that if gcd(a, b) = 1 then σ(ab) = σ(a)σ(b).

Observe that any divisor d of ab is product d = a⁰b⁰ of a divisor a⁰ of a and a divisor b⁰ of b. Indeed, because gcd(a, b) = 1 we have that d = gcd(d, a) gcd(d, b). On the other hand for any divisora⁰ of a and b⁰ ofb,d=a⁰b⁰is clearly a divisor ofab. So the set of divisors ofabis exactly

{a⁰b⁰:a⁰|aandb⁰|b}.

Thus

σ(ab) =X

d|ab

d=X

a⁰|a

X

b⁰|b

a⁰b⁰= (X

a⁰|a

a⁰)(X

b⁰|b

b⁰) =σ(a)σ(b).

as needed.

Solution

10. (a) Explain how you would use the method of successive squaring to compute 45

¹¹²

mod 211.

Writing 112 as 64 + 32 + 16 = 2⁶+ 2⁵+ 2⁴ we would compute 45,45²,45²²,45²³,45²⁴,45²⁵,45²⁶ mod 211 by successive squaring they we would compute

45²⁶·45²⁵·45²⁴ mod 211.

Solution

(b) Compute 47

¹¹⁹

mod 11.

Modulo 11 we do:

47¹¹⁹≡₁₁3⁹= 3·3²⁸

Computing 3²≡119, 3²² = 9² ≡114, 3²⁴ = 4² ≡11 5 and 3²⁸ = 5² ≡113, we get

47¹¹⁹≡113·3 = 9.

Solution