Chapter 1
Elementary Properties of Integers

1.1 Fundamental Notions and Laws

In the present chapter we are concerned primarily with certain elementary properties of the positive integers 1, 2, 3, 4, …It will sometimes be convenient, when no confusion can arise, to employ the word integer or the word number in the sense of positive integer.

We shall suppose that the integers are already defined, either by the process of counting or otherwise. We assume further that the meaning of the terms greater, less, equal, sum, difference, product is known.

From the ideas and definitions thus assumed to be known follow immediately the theorems:

Other fundamental theorems, which we take without proof, are embodied in the following formulas:

Here $a$, $b$, $c$ denote any positive integers.

These formulas are equivalent in order to the following five theorems: addition is commutative; multiplication is commutative; addition is associative; multiplication is associative; multiplication is distributive with respect to addition.

1.

Prove the following relations:

2.

Find the sum of each of the following series:

3.

Discover and establish the law suggested by the equations $1^2=0+1$, $2^2=1+3$, $3^2=3+6$, $4^2=6+10$, $\dots$; by the equations $1=1^3$, $3+5=2^3$, $7+9+11=3^3$, $13+15+17+19=4^3$, $\dots$.

1.2 Definition of Divisibility. The Unit

DEFINITIONS. An integer $a$ is said to be divisible by an integer $b$ if there exists an integer $c$ such that $a=bc$. It is clear from this definition that $a$ is also divisible by $c$. The integers $b$ and $c$ are said to be divisors or factors of $a$; and $a$ is said to be a multiple of $b$ or of $c$. The process of finding two integers $b$ and $c$ such that $bc$ is equal to a given integer $a$ is called the process of resolving $a$ into factors or of factoring $a$; and $a$ is said to be resolved into factors or to be factored.

We have the following fundamental theorems:

I. If $b$ is a divisor of $a$ and $c$ is a divisor of $b$, then $c$ is a divisor of $a$.

Since $b$ is a divisor of a there exists an integer $\beta$ such that $a=b\beta$. Since $c$ is a divisor of $b$ there exists an integer $\gamma$ such that $b=c\gamma$. Substituting this value of $b$ in the equation $a=b\gamma$ we have $a=c\gamma \beta$. But from theorem III of § 1.1 it follows that $\gamma \beta$ is an integer; hence, $c$ is a divisor of $a$, as was to be proved.

II. If $c$ is a divisor of both $a$ and $b$, then $c$ is a divisor of the sum of $a$ and $b$.

From the hypothesis of the theorem it follows that integers $\alpha$ and $\beta$ exist such that

Adding, we have

where $\delta$ is an integer. Hence, $c$ is a divisor of $a+b$.

III. If $c$ is a divisor of both $a$ and $b$, then $c$ is a divisor of the difference of $a$ and $b$.

The proof is analogous to that of the preceding theorem.

DEFINITIONS. If $a$ and $b$ are both divisible by $c$, then $c$ is said to be a common divisor or a common factor of $a$ and $b$. Every two integers have the common factor 1. The greatest integer which divides both $a$ and $b$ is called the greatest common divisor of $a$ and $b$. More generally, we define in a similar way a common divisor and the greatest common divisor of $n$ integers $a_1$, $a_2$, $\dots$, $a_n$.

DEFINITIONS. If an integer $a$ is a multiple of each of two or more integers it is called a common multiple of these integers. The product of any set of integers is a common multiple of the set. The least integer which is a multiple of each of two or more integers is called their least common multiple.

It is evident that the integer $1$ is a divisor of every integer and that it is the only integer which has this property. It is called the unit.

DEFINITION. Two or more integers which have no common factor except $1$ are said to be prime to each other or to be relatively prime.

DEFINITION. If a set of integers is such that no two of them have a common divisor besides $1$ they are said to be prime each to each.

1.

Prove that $n^3-n$ is divisible by $6$ for every positive integer $n$.

2.

If the product of four consecutive integers is increased by $1$ the result is a square number.

3.

Show that $2^{4n+2}+1$ has a factor different from itself and $1$ when $n$ is a positive integer.

1.3 Prime Numbers. The Sieve of Eratosthenes

DEFINITION. If an integer $p$ is different from 1 and has no divisor except itself and 1 it is said to be a prime number or to be a prime.

DEFINITION. An integer which has at least one divisor other than itself and 1 is said to be a composite number or to be composite.

All integers are thus divided into three classes:

1.

The unit;

2.

Prime numbers;

3.

Composite numbers.

We have seen that the first class contains only a single number. The third class evidently contains an infinitude of numbers; for, it contains all the numbers $2^2,2^3,2^4,\dots$ In the next section we shall show that the second class also contains an infinitude of numbers. We shall now show that every number of the third class contains one of the second class as a factor, by proving the following theorem:

I. Every integer greater than 1 has a prime factor.

Let $m$ be any integer which is greater than 1. We have to show that it has a prime factor. If $m$ is prime there is the prime factor $m$ itself. If $m$ is not prime we have

where $m_1$ and $m_2$ are positive integers both of which are less than $m$. If either $m_1$ or $m_2$ is prime we have thus obtained a prime factor of $m$. If neither of these numbers is prime, then write

Both $m{\prime }_1$ and $m{\prime }_2$ are factors of $m$ and each of them is less than $m_1$. Either we have not found in $m{\prime }_1$ or $m{\prime }_2$ a prime factor of $m$ or the process can be continued by separating one of these numbers into factors. Since for any given $m$ there is evidently only a finite number of such steps possible, it is clear that we must finally arrive at a prime factor of $m$. From this conclusion, the theorem follows immediately.

Eratosthenes has given a useful means of finding the prime numbers which are less than any given integer $m$. It may be described as follows:

Every prime except 2 is odd. Hence if we write down every odd number from 3 up to $m$ we shall have it the list every prime less than $m$ except 2. Now 3 is prime. Leave it in the list; but beginning to count from 3 strike out every third number in the list. Thus every number divisible by 3, except 3 itself, is cancelled. Then begin from 5 and cancel every fifth number. Then begin from from the next uncancelled number, namely 7, and strike out every seventh number. Then begin from the next uncancelled number, namely 11, and strike out every eleventh number. Proceed in this way up to $m$. The uncancelled numbers remaining will be the odd primes not greater than $m$.

It is obvious that this process of cancellation need not be carried altogether so far as indicated; for if $p$ is a prime greater than $\sqrt{m}$, the cancellation of any $p^{\text{ th}}$ number from $p$ will be merely a repetition of cancellations effected by means of another factor smaller than $p$, as one my see by the use of the following theorem.

II. An integer $m$ is prime if it has no prime factor equal or less than $I$, where $I$ is the greatest integer whose square is equal to or less than $m$.

Since $m$ has no prime factor less than $I$, it follows from theorem I that is has no factor but unity less than $I$. Hence, if $m$ is not prime it must be the product of two numbers each greater than $I$; and hence it must be equal to or greater than . This contradicts the hypothesis on $I$; and hence we conclude that $m$ is prime.

By means of the method of Eratosthenes determine the primes less than 200.

1.4 The Number of Primes is Infinite

I. The number of primes is infinite.

We shall prove this theorem by supposing that the number of primes is not infinite and showing that this leads to a contradiction. If the number of primes is not infinite there is a greatest prime number, which we shall denote by $p$. Then form the number

Now by theorem 1 of § 1.3 $N$ has a prime divisor $q$. But every non-unit divisor of $N$ is obviously greater than $p$. Hence $q$ is greater than $p$, in contradiction to the conclusion that $p$ is the greatest prime. Thus the proof of the theorem is complete.

In a similar way we may prove the following theorem:

II. Among the integers of the arithmetic progression $5$, $11$, $17$, $23$, $\dots$, there is an infinite number of primes.

If the number of primes in this sequence is not infinite there is a greatest prime number in the sequence; supposing that this greatest prime number exists we shall denote it by $p$. Then the number $N$,

is not divisible by any number less than or equal to $p$. This number $N$, which is of the form $6n-1$, has a prime factor. If this factor is of the form $6k-1$ we have already reached a contradiction, and our theorem is proved. If the prime is of the form $6k_1+1$ the complementary factor is of the form $6k_2-1$. Every prime factor of $6k_2-1$ is greater than $p$. Hence we may treat $6k_2-1$ as we did $6n-1$, and with a like result. Hence we must ultimately reach a prime factor of the form $6k_3-1$; for, otherwise, we should have $6n-1$ expressed as a product of prime factors all of the form $6t+1$—a result which is clearly impossible. Hence we must in any case reach a contradiction of the hypothesis. Thus the theorem is proved.

The preceding results are special cases of the following more general theorem:

III. Among the integers of the arithmetic progression $a$, $a+d$, $a+2d$, $a+3d$, $\dots$, there is an infinite number of primes, provided that $a$ and $b$ are relatively prime.

For the special case given in theorem II we have an elementary proof; but for the general theorem the proof is difficult. We shall not give it here.

1.

Prove that there is an infinite number of primes of the form $4n-1$.

2.

Show that an odd prime number can be represented as the difference of two squares in one and in only one way.

3.

The expression $m^p-n^p$, in which $m$ and $n$ are integers and $p$ is a prime, is either prime to $p$ or is divisible by $p^2$.

4.

Prove that any prime number except $2$ and $3$ is of one of the forms $6n+1$, $6n-1$.

1.5 The Fundamental Theorem of Euclid

If $a$ and $b$ are any two positive integers there exist integers $q$ and $r$, $q\stackrel{=}{>}0,0\leqq r<b$, such that

If $a$ is a multiple of $b$ the theorem is at once verified, $r$ being in this case $0$. If $a$ is not a multiple of $b$ it must lie between two consecutive multiples of $b$; that is, there exists a $q$ such that

Hence there is an integer $r$, $0<r<b$, such that $a=qb+r$. In case $b$ is greater than $a$ it is evident that $q=0$ and $r=a$. Thus the proof of the theorem is complete.

1.6 Divisibility by a Prime Number

I. If $p$ is a prime number and $m$ is any integer, then $m$ either is divisible by $p$ or is prime to $p$.

This theorem follows at once from the fact that the only divisors of $p$ are $1$ and $p$.

II. The product of two integers each less than a given prime number $p$ is not divisible by $p$.

Let $a$ be a number which is less than $p$ and suppose that $b$ is a number less than $p$ such that $ab$ is divisible by $p$, and let $b$ be the least number for which $ab$ is so divisible. Evidently there exists an integer $m$ such that

Then $p-mb<b$. Since $ab$ is divisible by $p$ it is clear that $mab$ is divisible by $p$; so is $ap$ also; and hence their difference $ap-mab$, , is divisible by $p$. That is, the product of $a$ by an integer less than $b$ is divisible by $p$, contrary to the assumption that $b$ is the least integer such that $ab$ is divisible by $p$. The assumption that the theorem is not true has thus led to a contradiction; and thus the theorem is proved.

III. If neither of two integers is divisible by a given prime number $p$ their product is not divisible by $p$.

Let $a$ and $b$ be two integers neither of which is divisible by the prime $p$. According to the fundamental theorem of Euclid there exist integers $m$, $n$, $\alpha$, $\beta$ such that

Then

If now we suppose $ab$ to be divisible by $p$ we have $\alpha \beta$ divisible by $p$. This contradicts II, since $\alpha$ and $\beta$ are less than $p$. Hence $ab$ is not divisible by $p$.

By an application of this theorem to the continued product of several factors, the following result is readily obtained:

IV. If no one of several integers is divisible by a given prime $p$ their product is not divisible by $p$.

1.7 The Unique Factorization Theorem

I. Every integer greater than unity can be represented in one and in only one way as a product of prime numbers.

In the first place we shall show that it is always possible to resolve a given integer $m$ greater than unity into prime factors by a finite number of operations. In the proof of theorem I, § 1.3, we showed how to find a prime factor $p_1$ of $m$ by a finite number of operations. Let us write

If $m_1$ is not unity we may now find a prime factor $p_2$ of $m_1$. Then we may write

If $m_2$ is not unity we may apply to it the same process as that applied to $m_1$ and thus obtain a third prime factor of $m$. Since $m_1>m_2>m_3>\dots$ it is clear that after a finite number of operations we shall arrive at a decomposition of $m$ into prime factors. Thus we shall have

where $p_1$, $p_2$, $\dots$, $p_r$ are prime numbers. We have thus proved the first part of our theorem, which says that the decomposition of an integer (greater than unity) into prime factors is always possible.

Let us now suppose that we have also a decomposition of $m$ into prime factors as follows:

Then we have

Now $p_1$ divides the first member of this equation. Hence it also divides the second member of the equation. But $p_1$ is prime; and therefore by theorem IV of the preceding section we see that $p_1$ divides some one of the factors $q$; we suppose that $p_1$ is a factor of $q_1$. It must then be equal to $q_1$. Hence we have

By the same argument we prove that $p_2$ is equal to some $q$, say $q_2$. Then we have

Evidently the process may be continued until one side of the equation is reduced to $1$. The other side must also be reduced to $1$ at the same time. Hence it follows that the two decompositions of $m$ are in fact identical.

This completes the proof of the theorem.

The result which we have thus demonstrated is easily the most important theorem in the theory of integers. It can also be stated in a different form more convenient for some purposes:

II. Every non-unit positive integer $m$ can be represented in one and in only one way in the form

where $p_1$, $p_2$, $\dots$, $p_n$ are different primes and ${\alpha }_1$, ${\alpha }_2$, $\dots$, ${\alpha }_n$ are positive integers.

This comes immediately from the preceding representation of $m$ in the form $m=p_1{p}_2\dots p_r$ by combining into a power of $p_1$ all the primes which are equal to $p_1$.

COROLLARY 1. If $a$ and $b$ are relatively prime integers and $c$ is divisible by both $a$ and $b$, then $c$ is divisible by $ab$.

COROLLARY 2. If $a$ and $b$ are each prime to $c$ then $ab$ is prime to $c$.

COROLLARY 3. If $a$ is prime to $c$ and $ab$ is divisible by $c$, then $b$ is divisible by $c$.

1.8 The Divisors of an Integer

The following theorem is an immediate corollary of the results in the preceding section:

I. All the divisors of $m$,

and every such number is a divisor of $m$.

From this it is clear that every divisor of $m$ is included once and only once among the terms of the product

when this product is expanded by multiplication. It is obvious that the number of terms in the expansion is . Hence we have the theorem:

II. The number of divisors of $m$ is .

Again we have

Hence,

III. The sum of the divisors of $m$ is

In a similar manner we may prove the following theorem:

IV. The sum of the $h^{th}$ powers of the divisors of $m$ is

1.

Find numbers $x$ such that the sum of the divisors of $x$ is a perfect square.

2.

Show that the sum of the divisors of each of the following integers is twice the integer itself: 6, 28, 496, 8128, 33550336. Find other integers $x$ such that the sum of the divisors of $x$ is a multiple of $x$.

3.

Prove that the sum of two odd squares cannot be a square.

4.

Prove that the cube of any integer is the difference of the squares of two integers.

5.

In order that a number shall be the sum of consecutive integers, it is necessary and sufficient that it shall not be a power of 2.

6.

Show that there exist no integers $x$ and $y$ (zero excluded) such that $y^2=2x^2$. Hence, show that there does not exist a rational fraction whose square is 2.

7.

The number $m=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_n^{{\alpha }_n}$, where the $p$’s are different primes and the $\alpha$’s are positive integers, may be separated into relatively prime factors in $2^{n-1}$ different ways.

8.

The product of the divisors of $m$ is $\sqrt{m^v}$ where $v$ is the number of divisors of $m$.

1.9 The Greatest Common Factor of Two or More Integers

Let $m$ and $n$ be two positive integers such that $m$ is greater than $n$. Then, according to the fundamental theorem of Euclid, we can form the set of equations

If $m$ is a multiple of $n$ we write $n=n_0$, $k=0$, in the above equations.

DEFINITION. The process of reckoning involved in determining the above set of equations is called the Euclidian Algorithm.

I. The number $n_k$ to which the Euclidian algorithm leads is the greatest common divisor of $m$ and $n$.

In order to prove this theorem we have to show two things:

1) That $n_k$ is a divisor of both $m$ and $n$;

2) That the greatest common divisor $d$ of $m$ and $n$ is a divisor of $n_k$.

To prove the first statement we examine the above set of equations, working from the last to the first. From the last equation we see that $n_k$ is a divisor of $n_{k-1}$. Using this result we see that the second member of next to the last equation is divisible by $n_k$ Hence its first member $n_{k-2}$ must be divisible by $n_k$. Proceeding in this way step by step we show that $n_2$ and $n_1$, and finally that $n$ and $m$, are divisible by $n_k$.

For the second part of the proof we employ the same set of equations and work from the first one to the last one. Let $d$ be any common divisor of $m$ and $n$. From the first equation we see that $d$ is a divisor of $n_1$. Then from the second equation it follows that $d$ is a divisor of $n_2$. Proceeding in this way we show finally that $d$ is a divisor of $n_k$. Hence any common divisor, and in particular the greatest common divisor, of $m$ and $n$ is a factor of $n_k$.

This completes the proof of the theorem.

COROLLARY. Every common divisor of $m$ and $n$ is a factor of their greatest common divisor.

II. Any number $n_i$ in the above set of equations is the difference of multiples of $m$ and $n$.

From the first equation we have

so that the theorem is true for $i=1$. We shall suppose that the theorem is true for every subscript up to $i-1$ and prove it true for the subscript $i$. Thus by hypothesis we have¹

Substituting in the equation

we have a result of the form

From this we conclude at once to the truth of the theorem.

Since $n_k$ is the greatest common divisor of $m$ and $n$, we have as a corollary the following important theorem:

III. If $d$ is the greatest common divisor of the positive integers $m$ and $n$, then there exist positive integers $\alpha$ and $\beta$ such that

If we consider the particular case in which $m$ and $n$ are relatively prime, so that $d=1$, we see that there exist positive integers $\alpha$ and $\beta$ such that $\alpha m-\beta n=\pm 1$. Obviously, if $m$ and $n$ have a common divisor $d$, greater than $1$, there do not exist integers $\alpha$ and $\beta$ satisfying this relation; for, if so, $d$ would be a divisor of the first member of the equation and not of the second. Thus we have the following theorem:

IV. A necessary and sufficient condition that $m$ and $n$ are relatively prime is that there exist integers $\alpha$ and $\beta$ such that $\alpha m-\beta n=\pm 1$.

The theory of the greatest common divisor of three or more numbers is based directly on that of the greatest common divisor of two numbers; consequently it does not require to be developed in detail.

1.

If $d$ is the greatest common divisor of $m$ and $n$, then $m/d$ and $n/d$ are relatively prime.

2.

If $d$ is the greatest common divisor of $m$ and $n$ and $k$ is prime to $n$, then $d$ is the greatest common divisor of $km$ and $n$.

3.

The number of multiplies of $6$ in the sequence $a,2a,3a,\cdots ,ba$ is equal to the greatest common divisor of $a$ and $b$.

4.

If the sum or the difference of two irreducible fractions is an integer, the denominators of the fractions are equal.

5.

The algebraic sum of any number of irreducible fractions, whose denominators are prime each to each, cannot be an integer.

6*.

The number of divisions to be effected in finding the greatest common divisor of two numbers by the Euclidian algorithm does not exceed five times the number of digits in the smaller number (when this number is written in the usual scale of 10).

1.10 The Least Common Multiple of Two or More Integers

I. The common multiples of two or more numbers are the multiples of their least common multiple.

This may be readily proved by means of the unique factorization theorem. The method is obvious. We shall, however, give a proof independent of this theorem.

Consider first the case of two numbers; denote them by $m$ and $n$ and their greatest common divisor by $d$. Then we have

where $\mu$ and $\nu$ are relatively prime integers. The common multiples sought are multiples of $m$ and are all comprised in the numbers $am=ad\mu$, where $a$ is any integer whatever. In order that these numbers shall be multiples of $n$ it is necessary and sufficient that $ad\mu$ shall be a multiple of $d\nu$; that is, that $a\mu$ shall be a multiple of $\nu$; that is, that $a$ shall be a multiple of $\nu$, since $\mu$ and $\nu$ are relatively prime. Writing $a=\delta \nu$ we have as the multiples in question the set $\delta d\mu \nu$ where $\delta$ is an arbitrary integer. This proves the theorem for the case of two numbers; for $d\mu \nu$ is evidently the least common multiple of $m$ and $n$.

We shall now extend the proposition to any number of integers $m,n,p,q,\dots$. The multiples in question must be common multiples of $m$ and $n$ and hence of their least common multiple $\mu$. Then the multiples must be multiples of $\mu$ and $p$ and hence of their least common multiple ${\mu }_1$. But ${\mu }_1$ is evidently the least common multiple of $m,n,p$. Continuing in a similar manner we may show that every multiple in question is a multiple of $\mu$, the least common multiple of $m,n,p,q,\dots$. And evidently every such number is a multiple of each of the numbers $m,n,p,q,\dots$.

Thus the proof of the theorem is complete.

When the two integers $m$ and $n$ are relatively prime their greatest common divisor is $1$ and their least common multiple is their product. Again if $p$ is prime to both $m$ and $n$ it is prime to their product $mn$; and hence the least common multiple of $m,n,p$ is in this case $mnp$. Continuing in a similar manner we have the theorem:

II. The least common multiple of several integers, prime each to each, is equal to their product.

1.

In order that a common multiple of $n$ numbers shall be the least, it is necessary and sufficient that the quotients obtained by dividing it successively by the numbers shall be relatively prime.

2.

The product of $n$ numbers is equal to the product of their least common multiple by the greatest common divisor of their products $n-1$ at a time.

3.

The least common multiple of $n$ numbers is equal to any common multiple $M$ divided by the greatest common divisor of the quotients obtained on dividing this common multiple by each of the numbers.

4.

The product of $n$ numbers is equal to the product of their greatest common divisor by the least common multiple of the products of the numbers taken $n-1$ at a time.

1.11 Scales of Notation

I. If $m$ and $n$ are positive integers and $n>1$, then $m$ can be represented in terms of $n$ in one and in only one way in the form

where

That such a representation of $m$ exists is readily proved by means of the fundamental theorem of Euclid. For we have

If the value of $n_{h-1}$ given in the last of these equations is substituted in the second last we have

This with the preceding gives

Substituting from this in the preceding and continuing the process we have finally

a representation of $m$ in the form specified in the theorem.

To prove that this representation is unique, we shall suppose that $m$ has the representation

where

and show that the two representations are identical. We have

Then

The first member is divisible by $n$. Hence the second is also. But the second member is less than $n$ in absolute value; and hence, in order to be divisible by $n$, it must be zero. That is, $b_k=a_h$. Dividing the equation through by $n$ and transposing we have

It may now be seen that $b_{k-1}=a_{h-1}$. It is evident that this process may be continued until either the $a$’s are all eliminated from the equation or the $b$’s are all eliminated. But it is obvious that when one of these sets is eliminated the other is also. Hence, $h=k$. Also, every $a$ equals the $b$ which multiplies the same power of $n$ as the corresponding $a$. That is, the two representations of $m$ are identical. Hence the representation in the theorem is unique.

From this theorem it follows as a special case that any positive integer can be represented in one and in only one way in the scale of 10; that is, in the familiar Hindoo notation. It can also be represented in one and in only one way in any other scale. Thus

Or, using a subscript to denote the scale of notation, this may be written

For the case in which $n$ (of theorem I) is equal to 2, the only possible values for the $a$’s are 0 and 1. Hence we have at once the following theorem:

II. Any positive integer can be represented in one and in only one way as a sum of different powers of 2.

1.

Any positive integer can be represented as an aggregate of different powers of $3$, the terms in the aggregate being combined by the signs $+$ and $-$ appropriately chosen.

2.

Let $m$ and $n$ be two positive integers of which $n$ is the smaller and suppose that $2^k\le n<2^{k+1}$. By means of the representation of $m$ and $n$ in the scale of 2 prove that the number of divisions to be effected in finding the greatest common divisor of $m$ and $n$ by the Euclidian algorithm does not exceed $2k$.

1.12 Highest Power of a Prime $p$ Contained in $n!$.

Let $n$ be any positive integer and $p$ any prime number not greater than $n$. We inquire as to what is the highest power $p^{\nu }$ of the prime $p$ contained in $n!$.

In solving this problem we shall find it convenient to employ the notation

to denote the greatest integer $\alpha$ such that $\alpha s\le r$. With this notation it is evident that we have

and more generally

If now we use to denote the index of the highest power of $p$ contained in an integer $x$, it is clear that we have

since only multiples of $p$ contain the factor $p$. Hence

Applying the same process to the $H$-function in the second member and remembering relation (1) it is easy to see that we have

the series on the right containing evidently only a finite number of terms different from zero. Thus we have the theorem:

I. The index of the highest power of a prime $p$ contained in $n!$ is

The theorem just obtained may be written in a different form, more convenient for certain of its applications. Let $n$ be expressed in the scale of $p$ in the form

where

Then evidently

Adding these equations member by member and combining the second members in columns as written, we have

Comparing this result with theorem I we have the following theorem:

II. If $n$ is represented in the scale of $p$ in the form

where $p$ is prime and

then the index of the highest power of $p$ contained in $n!$ is

Note the simple form of the theorem for the case $p=2$; in this case the denominator $p-1$ is unity.

We shall make a single application of these theorems by proving the following theorem:

III. If $n$, $\alpha$, $\beta$, $\dots$, $\lambda$ are any positive integers such that $n=\alpha +\beta +\dots +\lambda$, then

is an integer.

Let $p$ be any prime factor of the denominator of the fraction (A). To prove the theorem it is sufficient to show that the index of the highest power of $p$ contained in the numerator is at least as great as the index of the highest power of $p$ contained in the denominator. This index for the denominator is the sum of the expressions

The corresponding index for the numerator is

But, since $n=\alpha +\beta +\dots +\lambda$, it is evident that

From this and the expressions in (B) and (C) it follows that the index of the highest power of any prime $p$ in the numerator of (A) is equal to or greater than the index of the highest power of p contained in its denominator. The theorem now follows at once.

COROLLARY. The product of $n$ consecutive integers is divisible by $n!$.

1.

Show that the highest power of 2 contained in 1000! is $2^{994}$; in 1900! is $2^{1893}$. Show that the highest power of 7 contained in 10000! is $7^{1665}$.

2.

Find the highest power of 72 contained in 1000!

3.

Show that 1000! ends with 249 zeros.

4.

Show that there is no number $n$ such that $3^7$ is the highest power of 3 contained in $n!$.

5.

Find the smallest number $n$ such that the highest power of 5 contained in $n!$ is $5^{31}$. What other numbers have the same property?

6.

If $n=rs$, $r$ and $s$ being positive integers, show that $n!$ is divisible by by ; by the least common multiple of and .

7.

If $n=\alpha +\beta +pq+rs$, where $\alpha ,\beta ,p,q,r,s$, are positive integers, then $n!$ is divisible by

8.

When $m$ and $n$ are two relatively prime positive integers the quotient

as an integer.

9*.

If $m$ and $n$ are positive integers, then each of the quotients

is an integer. Generalize to $k$ integers $m,n,p,\dots$.

10*.

If $n=\alpha +\beta +pq+rs$ where $\alpha ,\beta ,p,q,r,s$ are positive integers, then $n!$ is divisible by

11*.

Show that

is an integer ($r,s,t$ being positive integers). Generalize to the case of $n$ integers $r,s,t,u,\dots$.

1.13 Remarks Concerning Prime Numbers

We have seen that the number of primes is infinite. But the integers which have actually been identified as prime are finite in number. Moreover, the question as to whether a large number, as for instance $2^{257}-1$, is prime is in general very difficult to answer. Among the large primes actually identified as such are the following:

No analytical expression for the representation of prime numbers has yet been discovered. Fermat believed, though he confessed that he was unable to prove, that he had found such an analytical expression in

Euler showed the error of this opinion by finding that $641$ is a factor of this number for the case when $n=5$.

The subject of prime numbers is in general one of exceeding difficulty. In fact it is an easy matter to propose problems about prime numbers which no one has been able to solve. Some of the simplest of these are the following:

1.

Is there an infinite number of pairs of primes differing by 2?

2.

Is every even number (other than 2) the sum of two primes or the sum of a prime and the unit?

3.

Is every even number the difference of two primes or the difference of 1 and a prime number?

4.

To find a prime number greater than a given prime.

5.

To find the prime number which follows a given prime.

6.

To find the number of primes not greater than a given number.

7.

To compute directly the $n^{\text{ th}}$ prime number, when $n$ is given.