Chapter - introduction,


I have started reading a book - How to Prove it (author: Daniel J. Velleman). I shall be posting its solutions from this post onwards, each titled with corresponding chapter. This post contains solutions for the introduction section of the book.

Note: In this book natural numbers include also. But it seems like in some places in the proofs, I messed up with this convention. In some placed I might have considered includes while in other places vice versa. Please point out to me, I will correct to use the book’s version of in all the places.

Soln.1(a)

Using theorem:

\( 2^n - 1 \) , is not prime, if \( n \) is not prime. Also with \( n = a \cdot b \), \( 2^n - 1 \) can be decomposed into product \( x.y \) where \( x = 2^b - 1 \) and \( y = 1 + 2^{1 \cdot b} + 2^{2 \cdot b} + ... 2^{(a-1) \cdot b} \)

We have \( n = 3 \times 5 \), Taking \( a = 3 \) and \( b = 5 \)

\( \Rightarrow x = 2^5 - 1 = 31 \) and \( y = 1 + 2^{1 \cdot 5} + 2^{2 \cdot 5} = 1057 \)

\( \Rightarrow x = 31, y = 1057 \)

Soln.1(b)

We can use the same theorem again. Now we have \( n = 32767 = 31 \times 1057 \). Taking \( b = 31 \) gives \( x = 2^b - 1 = 2^{31} - 1 = 2147483647 \)


Soln.2

Table for \( 3^n - 1\) :

\( n \) Is \( n \) prime? \( 3^n - 1 \) Is \( 3^n - 1 \) is prime?
1 no 2 yes
2 yes 8 no
3 yes 26 no
4 no 80 no
5 yes 242 no
6 no 728 no
7 yes 2186 no
8 no 6560 no
9 no 19682 no
10 no 59048 no
11 yes 177146 no
12 no 531440 no
13 yes 1594322 no
14 no 4782968 no

It can be observed from above table that \( 3^n - 1 \) is always an even number and it is never prime except for \( n = 1 \).

Table for \( 3^n - 2^n \) :

\( n \) Is \( n \) prime? \( 3^n - 2^n \) Is \( 3^n - 2^n \) is prime?
1 no 1 no
2 yes 5 yes
3 yes 19 yes
4 no 65 no
5 yes 211 yes
6 no 665 no
7 yes 2059 no
8 no 6305 no
9 no 19171 no
10 no 58025 no
11 yes 175099 no
12 no 527345 no
13 yes 1586131 no
14 no 4766585 no
15 no 14316139 no
16 no 42981185 no
17 yes 129009091 yes

It can be observed that if \( n \) is not prime than \( 3^n - 2^n \) is also not prime. Also it is observed that \( 3^n - 2^n \) is prime only if \( n \) is prime.


Soln3

Theorem 3 says:

There are infinite prime numbers

Proof of this theorem is done by contradiction. It was shown if \( m = p_1 \cdot p_2 \cdot p_3 … p_n + 1 \) where \( p_1, p_2, p_3 … p_n \) are finite(assumption) list of prime numbers. Now it can be seen that \( m \) is a prime number which contradicts the assumption of finite prime numbers.

(a) Use this method to find a prime different from 2, 3, 5, and 7.

Lets find \( m = 2 \times 3 \times 5 \times 7 + 1 = 210 + 1 = 211 \)

Now, if \( 2, 3, 5, 7 \) is the list of all prime numbers known(except m), than from the proof \( m = 211 \) can be the new prime number.

But this is not the case here as there are infinite prime numbers. Thus \( m \) may not result in a prime number.

If we use all prime numbers less than a given number(say \( g \) ), for computing \( m \), than from the steps of the proof we are sure that \( m \) is not divisible by any of the numbers less than \( g \). Thus if m is not prime than all of its prime-factors must be greater than \( g \).

As, \( 211 \) is a big number to manually check from all prime-number greater than 7 for prime-factors of \( 211 \), we may take a smaller set say \( 2, 3, 5 \) to find \( m = 2 \times 3 \times 5 + 1 = 31 \) and check for prime-factors of m greater than \( 5 \).

After checking, \( 31 \) is a prime number.

(b) Use this method to find a prime different from 2, 5, and 11.

Using the method from above, taking only \( 2 \) results \( m = 2 + 1 = 3 \) which is a prime number.

Now taking only \( 2, 3 \) results \( m = 2 \times 3 + 1 = 7 \) which is also a prime number.


Soln4

We may use following theorem here:

For every positive integer n, there is a sequence of n consecutive positive integers containing no primes.

Its proof shows that a set of such consecutive integers can be

Putting \( n = 5 \Rightarrow x + i = 6! + 2 + i = 722 + i \) which gives us 722, 723, 724, 725, 726


Soln5

Table:

\( n \) Is \( n \) prime? \( 2^n - 1 \) Is \( 2^n - 1 \) is prime?
1 no 1 no
2 yes 3 yes
3 yes 7 yes
4 no 15 no
5 yes 31 yes
6 no 63 no
7 yes 127 yes
8 no 255 no
9 no 511 no
10 no 1023 no
11 yes 2047 no
12 no 4095 no

The discussion on p.5 says that if \( 2^n - 1 \) is prime than \( 2^{n-1} \cdot (2^n - 1) \) is a perfect number,

For \( n = 3 \Rightarrow 2^{3-1} \cdot (2^3 - 1) \Rightarrow 4 \times 7 = 28 \).

Similarly, for \( n = 5 \Rightarrow 2^{5-1} \cdot (2^5 - 1) \Rightarrow 16 \times 31 = 496 \).


Soln6

There are no more such prime triplets. From wiki:

In mathematics, a prime triplet is a set of three prime numbers of the form (p, p + 2, p + 6) or (p, p + 4, p + 6). With the exceptions of (2, 3, 5) and (3, 5, 7), this is the closest possible grouping of three prime numbers, since one of every three sequential odd numbers is a multiple of three, and hence not prime. wiki

which is true, as all prime numbers except \( 2 \) are odd and no three consecutive odd numbers can be prime as at-least one of them must be a multiple of three.