Bachelor of Ecole Polytechnique

Bachelor of Ecole Polytechnique

Algorithms for Discrete Mathematics, year 2, semester 1

Arithmetic 2: Modulos and Fermat

Table of contents

Arithmetic of square root and continued fractions Arithmetic of modulos

The little Fermat theorem

(Bonus) The Goldbach conjecture

Arithmetic with matrices

The aim of this Section is to use linear algebra and python to compute exact expressions in Arithmetic.

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## loading python libraries

# necessary to display plots inline:

%matplotlib inline

# load the libraries

import matplotlib.pyplot as plt # 2D plotting library

import numpy as np # package for scientific computing

from math import * # package for mathematics (pi, arctan, sqrt, factorial ...)

Do it yourself.(Theory)

Prove by induction that there exist integers such that for every 1.

Find a matrix such that 2.

,

a

b

n ≥ 1

(1 + √ ) 2 ⎯⎯

= a

+ b

√ 2 ⎯⎯ .

2 × 2 A

( a

) = A × ( ) . b

a

b

Answers.

For this is true with . Assume this holds for some .

Finally, one can write where

which are easily proved to be integers by induction.

1.

By the above computation we have

which can be written as:

By induction this implies that 2.

n = 1 a

= b

= 1 n ≥ 1

(1 + 2 √ ) ⎯⎯

= (1 + √ 2 ⎯⎯ ) × (1 + √ ) 2 ⎯⎯

= (1 + √ 2 ⎯⎯ ) × ( + a

b

√ 2 ⎯⎯ )

= a

+ b

√ 2 ⎯⎯ + a

√ 2 ⎯⎯ + 2 b

= ( + 2 ) + ( + ) a

b

a

b

√ 2 ⎯⎯ . (1 + √ ) 2 ⎯⎯

= a

+ b

√ 2 ⎯⎯ ,

= + 2 , = + ,

a

a

b

b

a

b

{ a

b

= a

+ 2 , b

= a

+ , b

( a

) = ( ) × ( ) . b

1 1 2

1

a

b

( ) a

= × ( ) = × ( ) . b

( 1 )

1 2 1

n−1

a

b

( 1 ) 1

2 1

n−1

1

1

Do it yourself. Using the powers of matrix , write a small script which returns . (Recall that a matrix can be de�ned by np.matrix([[a,b],[c,d]]) .) Your script should return something like

(1+Root(2))^100 is 5608832401712713489 + 630483757982286570 8 * SquareRoot(2)

A a

, b

2 × 2

Do it yourself. It can be proved that

for some constants .

Estimate numerically with a plot. Can you guess the exact value?

1.

Estimate numerically.

2.

c a

∼ R

c , R

R c

Answers.

With the script below we plot , which should converge to . We obtain approximately . We can guess (?) that

1.

We then plot , which seems to converge to . 2.

n ↦ a

R

2.3803 R = 1 + √ 2 ⎯⎯ ≈ 2.414 n ↦ /(1 + a

√ ) 2 ⎯⎯

1/2

(1+Root(2))^100 is 5608832401712713489 + 6304837579822865708 * SquareRo ot(2)

Final value: 0.5 n=100

MatrixA=np.matrix([[1,2],[1,1]])

PowerofMatrix=(MatrixA**(n-1))*np.matrix([[1],[1]]) a=np.asscalar(PowerofMatrix[0])

b=np.asscalar(PowerofMatrix[1])

print('(1+Root(2))^'+str(n)+' is '+str(a)+' + '+str(b)+' * SquareRoot(2)')

def Coefficient_a(n):

MatrixA=np.matrix([[1,2],[1,1]])

PowerofMatrix=(MatrixA**(n-1))*np.matrix([[1],[1]]) return np.asscalar(PowerofMatrix[0])

N=50# Estimation of R:

RenormalizedValues=[Coefficient_a(n)**(1/(n+0.0)) for n in range(1,N)]

print('Final value: ' +str(RenormalizedValues[-1]) )

# Estimation of c:

#RenormalizedValues=[Coefficient_a(n)/((1+np.sqrt(2))**n) for n in range(1,N)]

#plt.plot(RenormalizedValues) plt.show()

print('Final value: '+str(RenormalizedValues[-1]))

Do it yourself.

We set

(Theory) Let us write as an integer ratio (for example , you can check ). Find a matrix such that

(Hint: Find a simple relation between and .) 1.

(Application) Compute the exact value (as a fraction) of 2.

(Side question) Compute a numerical evaluation of . Does it look familiar?

3.

= , = , = , = , …

u

1

1 + 1 u

1

1 +

u

1 1 +

1+1

u

1

1 +

u

u

= / a

b

a

= 1, b

= 2

= 3, = 5

a

b

2 × 2 B

( a

) = B × ( ) . b

a

b

u

u

r = 1 .

1 +

1+ 1

1+ 1

1+ 1

1+ 1

1+ 1

1+ 1 1+ 11+1

1 + r

Answers.

We have that

Therefore we obtain

i.e.

1.

We have that

See the script below for the answer: . 2.

According to the script below, is surprisingly close to the golden ratio . This is no coincidence, it can be proved that .

3.

= .

u

1 1 + u

= = .

a

b

1 1 +

b

+ b

a

( a

) = ( ) × ( ) . b

0 1

1 1

a

b

r = u

= ( 0 ) ( ) = ( ) . 1 1

1

9

u

b

( 0 ) 1 1

1

9

1 r = 89/144 2

1 + r ϕ = (1 + √ 5 ⎯⎯ )/2 ≈ 1.61803398875...

lim u

= ϕ

Arithmetic of modulos

Do it yourself. Write a script which computes . (Explain in the cell below the successive steps.)

38911

(mod 188)

Answers. The above script shows that . Therefore the �rst simpli�cation is

Now, we observe that the small script gives So �nally

38911 ≡ 183 (mod 188)

≡ (mod 188).

38911

183

≡ 1 (mod 188).

183

= = ( × ≡ 1 × (mod 188) = 7.

183

183

183

)

183

183

--- Question 2:

r =89 / 144 --- Question 3:

1+r = 1.6180555555555556

# Script for Question 2.

MatrixB=np.matrix([[0,1],[1,1]])

PowerofMatrix=(MatrixB**(9))*np.matrix([[1],[2]]) a=np.asscalar(PowerofMatrix[0])

b=np.asscalar(PowerofMatrix[1]) print('---')

print('Question 2:')

print('r =' +str(a)+' / '+str(b))

# Script for Question 3.

print('---') print('Question 3:')

print('1+r = ' +str(1+(a+0.0)/b))

# Too big, the kernel explodes!

#print((38911**21025413)%188) k=38911%188

print('38911 = '+str(k)+ ' mod(188) =:k') FoundAPeriod=False

i=0while FoundAPeriod == False:

i=i+1

remainder = (k**i)%188

print('for i='+str(i)+' we have k**i mod(188) = '+str(remainder)) if remainder == 1:

FoundAPeriod = True

print('We have that 21025413%'+str(i)+' = '+str(21025413%i))

print('The final answer is '+str(k)+'**'+str(21025413%i)+' (mod 188) ='+str((183**9)%

The little Fermat theorem

The "little" Fermat Theorem states the following:

Theorem

Let be a prime number. For every integer , we have

(In the sequel we will use the above formulation rather than "for every , we have ".)

p 1 ≤ a < p

≡ 1(mod p ).

a

a≥ 0 a^p ≡a(modp)

Do it yourself. Write a script which checks that the little Fermat Theorem is true for

p = 17

for a = 1, a^{p-1} mod p =1 for a = 2, a^{p-1} mod p =1 for a = 3, a^{p-1} mod p =1 for a = 4, a^{p-1} mod p =1 for a = 5, a^{p-1} mod p =1 for a = 6, a^{p-1} mod p =1 for a = 7, a^{p-1} mod p =1 for a = 8, a^{p-1} mod p =1 for a = 9, a^{p-1} mod p =1 for a = 10, a^{p-1} mod p =1 for a = 11, a^{p-1} mod p =1 for a = 12, a^{p-1} mod p =1 for a = 13, a^{p-1} mod p =1 for a = 14, a^{p-1} mod p =1 for a = 15, a^{p-1} mod p =1 for a = 16, a^{p-1} mod p =1 Fermat theorem is true for n = 17 p=17

Check=True

for a in range(1,p):

Value= a**(p-1)%p

print('for a = '+str(a)+', a^{p-1} mod p ='+str(Value)) if Value != 1:

Check=False if Check==True:

print('Fermat theorem is true for n = '+str(p)) else:

print('Fermat theorem is not true for n = '+str(p))

We say that is composite if is not prime. The contraposition of the little Fermat Theorem is very useful: it says that

In this case, we say that is a Fermat witness for (the non-primeness of) . For example, you can check that

(and of course is composite). Therefore is a Fermat witness for , and this is a (somehow convoluted) proof of the fact that is composite.

n n

( there exists  a < p  such that  a

≢ 1 (mod n ) ) ⇒ p  is composite.

a p

= 16384 ≡ 4 (mod 15) ≢ 1(mod 15) 2

15 a = 2 p = 15

15

Do it yourself.

Check that for every composite then is a Fermat witness for . The output should look like

n = 2 is prime n = 3 is prime

n = 4 is composite and 2 is a Fermat witness n = 5 is prime

n = 6 is composite and 2 is a Fermat witness ...

1.

Find the smallest composite such that is not a Fermat witness for . 2.

Same question with . 3.

n ≤ 60 a = 2 n

n a = 2 n

a = 3

To save you time we have copy/pasted the function IsPrime() from Notebook 1:

True False

def IsPrime(n):

# input: integer n

# output: True or False depending on whether n is prime or not if n==1:

return False if n==2:

return True elif n%2==0:

return False factor=3

while factor**2 < n+1:

if n%factor == 0:

return False factor=factor+2 return True

# Tests

print(IsPrime(2)) print(IsPrime(108))

Do it yourself. Find the smallest Fermat witness which proves that

1105

---Question 1--- n = 2 is prime

n = 3 is prime

n = 4 is composite and 2 is a Fermat witness n = 5 is prime

n = 6 is composite and 2 is a Fermat witness n = 7 is prime

n = 8 is composite and 2 is a Fermat witness n = 9 is composite and 2 is a Fermat witness n = 10 is composite and 2 is a Fermat witness n = 11 is prime

n = 12 is composite and 2 is a Fermat witness n = 13 is prime

n = 14 is composite and 2 is a Fermat witness n = 15 is composite and 2 is a Fermat witness n = 16 is composite and 2 is a Fermat witness n = 17 is prime

n = 18 is composite and 2 is a Fermat witness n = 19 is prime

n = 20 is composite and 2 is a Fermat witness n = 21 is composite and 2 is a Fermat witness n = 22 is composite and 2 is a Fermat witness n = 23 is prime

n = 24 is composite and 2 is a Fermat witness n = 25 is composite and 2 is a Fermat witness n = 26 is composite and 2 is a Fermat witness n = 27 is composite and 2 is a Fermat witness n = 28 is composite and 2 is a Fermat witness n = 29 is prime

n = 30 is composite and 2 is a Fermat witness ---Questions 2-3---

n = 341 is composite and a = 2 is NOT a Fermat witness n = 91 is composite and a = 3 is NOT a Fermat witness

# Question 1

print('---Question 1---') for n in range(2,31):

if not IsPrime(n):

if (2**(n-1)-1)%n!=0:

print('n = '+str(n)+' is composite and 2 is a Fermat witness') else:

print('n = '+str(n)+' is composite and 2 is NOT a Fermat witness') else:

print('n = '+str(n)+' is prime')

# Question 2 print('')

print('---Questions 2-3---') for a in [2,3]:

Check=True n=a+1

while Check==True:

if IsPrime(n)==False:

if (a**(n-1)-1)%n==0:

print('n = '+str(n)+' is composite and a = '+str(a)+' is NOT a Fermat witness' Check=False

n=n+1 print('')

Fermat prime numbers

A Fermat number is an integer of the form for some . First Fermat

numbers are given by

F

= 2

+ 1 n ≥ 0

= + 1 = 3, = + 1 = 5, = + 1 = 17, , = + 1 = 257.

F

2

F

2

F

2

F

2

Do it yourself. Fermat conjectured that every Fermat number is prime. Can you test his conjecture up to

n = 8

(Bonus) The Goldbach conjecture

Goldbach's conjecture is one of the most famous unsolved problems in number theory. It states that

Every even integer can be written as the sum of two primes.

We will try to check the conjecture.

≥ 2

p=1105 a=2Test=False

while a<p and Test==False:

if (a**(p-1)-1)%p!=0:

Test=True a=a+1

print('a = '+str(a-1)+' is the smallest Fermat witness of p = '+str(p))

def Fermat(n):

return 2**(2**n)+1

# With my code it works only up to n=6 !!!

for n in range(7):

F=Fermat(n) if IsPrime(F):

print('For n = '+str(n)+', F_n = '+str(F)+' => prime') else:

print('For n = '+str(n)+', F_n = '+str(F)+' => not prime')

Do it yourself. Write a script which checks the conjecture for every (and returns a decomposition for every ).

For you should get something like

For n = 60, we have 60 = 7 + 53, Goldbach is true For n = 62, we have 62 = 3 + 59, Goldbach is true For n = 64, we have 64 = 3 + 61, Goldbach is true For n = 66, we have 66 = 5 + 61, Goldbach is true For n = 68, we have 68 = 7 + 61, Goldbach is true For n = 70, we have 70 = 3 + 67, Goldbach is true

M ≤ n ≤ N n

M = 60, N = 70

For n = 1100, we have 1100 = 3 + 1097, Goldbach is true For n = 1102, we have 1102 = 5 + 1097, Goldbach is true For n = 1104, we have 1104 = 7 + 1097, Goldbach is true For n = 1106, we have 1106 = 3 + 1103, Goldbach is true For n = 1108, we have 1108 = 5 + 1103, Goldbach is true For n = 1110, we have 1110 = 7 + 1103, Goldbach is true For n = 1112, we have 1112 = 3 + 1109, Goldbach is true For n = 1114, we have 1114 = 5 + 1109, Goldbach is true For n = 1116, we have 1116 = 7 + 1109, Goldbach is true For n = 1118, we have 1118 = 31 + 1087, Goldbach is true For n = 1120, we have 1120 = 3 + 1117, Goldbach is true For n = 1122, we have 1122 = 5 + 1117, Goldbach is true For n = 1124, we have 1124 = 7 + 1117, Goldbach is true For n = 1126, we have 1126 = 3 + 1123, Goldbach is true For n = 1128, we have 1128 = 5 + 1123, Goldbach is true For n = 1130, we have 1130 = 7 + 1123, Goldbach is true For n = 1132, we have 1132 = 3 + 1129, Goldbach is true For n = 1134, we have 1134 = 5 + 1129, Goldbach is true For n = 1136, we have 1136 = 7 + 1129, Goldbach is true For n = 1138, we have 1138 = 29 + 1109, Goldbach is true def IsPrime(n):

# input: integer n

# output: True or False depending on whether n is prime or not if n==1:

return False if n==2:

return True elif n%2==0:

return False factor=3

while factor**2 < n+1:

if n%factor == 0:

return False factor=factor+2 return True

M=1100 N=1140

for n in range(M,N,2): # we test the conjecture for even integers GoldbachTrueFor_n = False # so far Goldbach is False

p=2

while GoldbachTrueFor_n == False and p <= n:

if IsPrime(p) and IsPrime(n-p):

print('For n = '+str(n)+', we have '+str(n)+' = '+str(p)+' + '+str(n-p)+

GoldbachTrueFor_n = True # We have an example: Goldbach holds p = p+1