Miscellaneous Problems

Let $q$ denote an odd prime number and let $r_1,r_2,\ldots ,r_q$ and $s_1,s_2,\ldots ,s_q$ be two complete systems of residues mod $q$. Then the $q$ numbers $r_1s_1,r_2s_2,\ldots ,r_qs_q$ do not form a complete system of residues mod $q$.

Let the highest power of the odd prime number $p$ that divides the number $n$ be $p^\alpha$, $\alpha \geq 1$. Then we have

according as $\lambda$ is or is not divisible by $p-1$, $\lambda =1,2,3,\ldots$.

Let $p$ be the smallest prime number that divides $n$. Then there exist two complete systems of residues mod $n$

such that each of the $p-2$ lines

represents a complete system of residues mod $n$. However then

is not a complete system of residues mod $n$.

If $p$ is a prime number, then

are divisible by $p$. (For the definitions of $s_k^n$ and $S_k^n$ see the introductions to Problem 197 and Problem 186 of Part I, respectively.)

If $p$ is an odd prime, the coefficients of the Tchebychev polynomial $T_p(x)$ (defined at the beginning of Part Six) are all divisible by $p$, with the exception of the leading coefficient which is equal to $2^{p-1}$.

Every integral power of a number $n$ can be represented as the sum of $n$ consecutive odd integers.

A number in the sequence $2,3,4,\ldots ,n$ ($n>2$) is relatively prime to all the other numbers if and only if it is a prime number that exceeds $n/2$. (That such a prime number always exists for $n>2$ was proved by Tchebychev. Cf. Oeuvres, Vol. 1, p. 63, St. Petersbourg 1899.)

The partial sums of the harmonic series

for $n>1$ are not equal to an integer. This follows directly from the Tchebychev theorem [249], but is to be proved without using that theorem.

The sum of two or more consecutive terms of the harmonic series, i.e. a sum of the form

cannot be an integer. If it is written in the form of a fraction in its lowest terms, then its denominator is even and its numerator is odd.

If the positive integer $n$ is divisible by all numbers that are $\leq \sqrt n$, then $n$ is either 24 or a divisor of 24.

More generally show by an elementary argument the following: If $0<\alpha <1$, then there exist only a finite number of positive integers $n$ such that $n$ is divisible by $1,2,3,\ldots ,[n^\alpha ]$.

Let $Q$ be a prime number and relatively prime to $10P$. The arithmetic mean of the digits in a period of the expression of $P/Q$ as a decimal fraction is 4.5 if and only if the length of the period is an even number.

The number $n=2^h+1$, $h\geq 2$ is a prime number if and only if

The following theorem given by Euler as a conjecture is true: The Diophantine equation

has no solution in positive integers $x,y,z,t$.

If the prime number $q\geq 11$, then there always exist positive odd prime numbers $p_1,p_2,p_3,p_4$, not necessarily distinct, that are all smaller than $q$, and satisfy the equations

respectively. $\left[\left(\dfrac {p_1}{q}\right)\right.$ etc., are Legendre symbols defined in Problem 45 of Part V.]

The decimal fraction

(formed by writing consecutively all the prime numbers) is irrational. [249, 110.]

The number

is irrational.

The number $e$ is not only irrational, but it is also not an algebraic number of the second degree, i.e. it cannot satisfy an equation of the form

where $a,b,c$ are integers and are not all $=0$.

If the Euler (Mascheroni) constant $C=-\Gamma '(1)$ were a rational number, then $\Gamma '(n+1)$ would have to be an integer for all integers $n$ from some point onward.

The number

is irrational.

The numbers $a$ and $b$ are positive; $a$ may be rational or irrational and the same holds for $b$, and so there are four possible cases. Show by means of examples that $a^b$ can be rational or irrational in each of these four cases (eight examples are required).

The question whether the number $\pi$ has the property that the arithmetic mean of its first $n$ decimal places converges to 4.5 must be left unresolved. If however the number $\pi$ has this property then the number $4-\pi$ has the same property.

Denote by $q_n$ the denominator of the $n$th Bernoulli number $B_n$ [182]. Then we have

If $\zeta (s)$ denotes the Riemann zeta-function defined in the introduction to 39,

Let the arithmetic function $f(n)$ be multiplicative [introduction to Chapter 1, § 6] and converge to 0 as $n$ tends to infinity through the prime numbers and the powers of prime numbers. Then $f(n)$ also satisfies the stronger requirement that

as $n$ tends to infinity through all positive integers.

For every given positive number $\delta$,

Let the quadratic polynomial $ax^2+bx+c$ with integral coefficients be represented by the lattice point $a,b,c$ of a three-dimensional cubic lattice. Let the number of those lattice points that lie in the cube

and correspond to a reducible polynomial be $r_n$. Prove that

(It is in a certain sense the ``normal case'' for a quadratic polynomial to be irreducible.)

The probability that a polynomial with integral coefficients of given degree is reducible is equal to 0. More precisely: Let $h$ be an integer, $h\geq 2$, and let $r_n$ denote the number of those lattice points of $(h+1)$-dimensional space that lie in the cube

and correspond to a reducible polynomial $a_0x^h+a_1x^{h-1}+\cdots +a_h$. Then we have

(This result generalizes and improves 265.) [Problem 46 of Part II.]