# Richert Theorem

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In 1852, Chebyshev proved the Bertrand’s postulate:

For any integer $n>1$, there always exists at least one prime number $p$ with $n < p < 2n$.

You can find Erdős’ elementary proof here. In this post I would like to discuss an application of this fantastic result, discovered by Hans-Egon Richert in 1948:

Every integer $n\geq 7$ can be expressed as a sum of distinct primes.

There are several proofs available in literature, but we will follow the short proof given by Richert himself (english translation has been taken from here and here):

Consider the set of prime integers $\{p_1, p_2,p_3,\ldots\}$ where $p_1=2, p_2=3, p_3=5,\ldots$. By Bertrand’s postulate we know that $\boxed{p_i < p_{i+1} < 2p_i }$.

Next, we observe that, any integer between 7 and 19 can be written as a sum of distinct first 5 prime integers $\{2,3,5,7,11\}$:

7 = 5+2; 8 = 5+3; 9 = 7+2; 10 = 5+3+2; 11 = 11; 12 = 7+5; 13 = 11+2; 14 = 7+5+2; 15 = 7+5+3; 16 = 11+5; 17 = 7+5+3+2; 18 = 11+7; 19 = 11+5+3

Hence we fix $a=7-1=6$, $b=19-a=13$, and $k=5$ to conclude that $\boxed{b \geq p_{k+1}}$.

Let, $S_i = \{a+1,a+2,\ldots, a+p_{i+1}\}=\{7,8,\ldots, 6+p_{i+1}\}$. Then by the above observation we know that the elements of $S_k = S_5 = \{7,8,\ldots, 19\}$ are the sum of distinct first $k=5$ prime integers.
Moreover, if the elements of $S_i$ can be written as the sum of distinct first $i$ prime integers, then the elements of $S_{i+1}$ can also be written as the sum of distinct first $i+1$ prime integers since

$S_{i+1}\subset S_i \cup \{m + p_{i+1}: m\in S_{i}\}$

as a consequence of $2p_{i+1} \geq p_{i+2}$.

Hence inductively the result follows by considering $\bigcup_{i\ge k} S_i$, which contains all integers greater than $a=6$, and contains only elements which are distinct sums of primes.

Exercise: Use Bertrand’s postulate to generalize the statement proved earlier: If $n>1$ and $k$  are natural numbers , then the sum

$\displaystyle{\frac{1}{n}+ \frac{1}{n+1} + \ldots + \frac{1}{n+k}}$

cannot be an integer.

[HINT: Look at the comment by Dan in the earlier post.]

References:
[0] Turner, C. (2015) A theorem of Richert. Math.SE profile: https://math.stackexchange.com/users/37268/sharkos

[1] Richert, H. E. (1950). Über Zerfällungen in ungleiche Primzahlen. (German). Mathematische Zeitschrift 52, pp. 342-343.  https://doi.org/10.1007/BF02230699

[2] Sierpiński,W. (1988). Elementary theory of numbers. North-Holland Mathematical Library 31, pp. 137-153.

# Conway’s Prime Producing Machine

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Primes are not randomly arranged (since their position is predetermined) but we can’t find an equation which directly gives us nth prime number. However, we can ask for a function (which surely can’t be a polynomial) which will give only the prime numbers as output. For example, the following one is used for MRDP theorem:

But it’s useless to use this to find bigger primes because the computations are much more difficult than the primality tests.

Conway’s PRIMEGAME takes whole numbers as inputs and outputs $2^k$ if and only if $k$ is prime.

Source: https://cstheory.stackexchange.com/a/14727 [Richard Guy, © 1983 Mathematical Association of America]

PRIMEGAME is based on a Turing-complete esoteric programming language called FRACTRAN, invented by John Conway. A FRACTRAN program is an ordered list of positive fractions together with an initial positive integer input n. The program is run by updating the integer $n$ as follows:

1. for the first fraction $f$ in the list for which $nf$ is an integer, replace $n$ by $nf$;
2. repeat this rule until no fraction in the list produces an integer when multiplied by n, then halt.

PRIMEGAME is an algorithm devised to generate primes using a sequence of 14 rational numbers:

$\displaystyle{\left( \frac{17}{91}, \frac{78}{85}, \frac{19}{51}, \frac{23}{38}, \frac{29}{33}, \frac{77}{29}, \frac{95}{23}, \frac{77}{19}, \frac{1}{17}, \frac{11}{13}, \frac{13}{11}, \frac{15}{2}, \frac{1}{7}, \frac{55}{1} \right)}$

Starting with 2, one finds the first number in the machine that multiplied by 2 gives an integer; then for that integer we find the first number in the machine that generates another integer. Except for the initial 2, each number output have an integer for a binary logarithm is a prime number, which is to say that powers of 2 with composite exponents don’t show up.

If you have some knowledge of computability and unsolvability theory, you can try to understand the working of this Turing machine. There is a nice exposition on OeisWiki  to begin with.

“Hilbert’s 10th Problem” by Martin Davis and Reuben Hersh [© 1973 Scientific American, doi: 10.1038/scientificamerican1173-84] Illustrating the basic idea of machines from unsoilvability theory.

Following is an online program by Prof. Andrew Granville illustrating the working of PRIMEGAME:

Motivation for this post came from Andrew Granville’s Math Mornings at Yale.

# Prime Number Problem

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Following is a problem about prime factorization of the sum of consecutive odd primes. (source: problem 80 from The Green Book of Mathematical Problems)

Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeated) prime factors.

The first thing to observe is that sum of odd numbers is even, hence the sum of two consecutive odd primes will be divisible by 2. Let’s see factorization of some of the examples:

$3 + 5 = 2\times 2 \times 2$
$5 + 7 = 2 \times 2\times 3$
$7+11 = 2 \times 3\times 3$
$11+13 = 2 \times 2 \times 2 \times 3$
$13+17 = 2 \times 3 \times 5$
$17+19 = 2\times 2\times 3 \times 3$
$19+23 = 42 = 2\times 3\times 7$
$23+29 = 52 = 2\times 2 \times 13$

Now let $p_n$ and $p_{n+1}$ be the consecutive odd primes, then from above observations we can conjecture that either $p_n+p_{n+1}$ is product of at least three distinct primes or $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ such that $k+\ell \geq 3$.

To prove our conjecture, let’s assume that $p_n+p_{n+1}$ is NOT a product of three (or more) distinct primes (otherwise we are done). Now we will have to show that if $p_n+p_{n+1}= 2^k p^\ell$ for some odd prime $p$ then $k+\ell \geq 3$.

If $\ell = 0$ then we should have $k\geq 3$. This is true since $3+5=8$.

Now let $\ell > 0$. Since $k\geq 1$ (sum of odd numbers is even), we just need to show that $k=1, \ell=1$ is not possible. On the contrary, let’s assume that $k=1,\ell = 1$. Then $p_n+p_{n+1} = 2p$. By arithmetic mean property, we have

$\displaystyle{p_n < \frac{p_n+p_{n+1}}{2}} = p

But, this contradicts the fact that $p_n,p_{n+1}$ are consecutive primes. Hence completing the proof of our conjecture.

This is a nice problem where we are equating the sum of prime numbers to product of prime numbers. Please let me know the flaws in my solution (if any) in the comments.

# Finite Sum & Divisibility – 2

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Earlier this year I discussed a finite analogue of the harmonic sum. Today I wish to discuss a simple fact about finite harmonic sums.

If $p$ is a prime integer, the numerator of the fraction $1+\frac{1}{2}+\frac{1}{3}+\ldots + \frac{1}{p-1}$ is divisible by $p$.

We wish to treat the given finite sum modulo $p$, hence we can’t just add up fractions. We will have to consider each fraction as inverse of an integer modulo p. Observe that for $0, we have inverse of each element $i$ in the multiplicative group $\left(\mathbb{Z}/p\mathbb{Z}\right)^\times$, i.e. there exist an $i^{-1}$ such that $i\cdot i^{-1}\equiv 1 \pmod p$.

For example, for $p=5$, we have $1^{-1}=1, 2^{-1}=3, 3^{-1}=2$ and $4^{-1}=4$.

Hence we have
$\displaystyle{i\cdot \frac{1}{i}\equiv 1 \pmod p ,\qquad (p-i)\cdot \frac{1}{p-i} \equiv 1 \pmod p }$
for all $0.

Hence we have:
$\displaystyle{i\left(\frac{1}{i}+\frac{1}{p-i}\right)\equiv i\cdot \frac{1}{i} - (p-i)\cdot \frac{1}{p-i} \equiv 0 \pmod p}$

Thus we have:
$\displaystyle{\frac{1}{i}+\frac{1}{p-i}\equiv 0 \pmod p}$

The desired result follows by summation.

We can, in fact, prove that the above harmonic sum is divisible by $p^2$, see section 7.8 of G. H. Hardy and E. M. Wright’s An Introduction to the Theory of Numbers for the proof.

# Ulam Spiral

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Some of you may know what Ulam’s spiral is (I am not describing what it is because the present Wikipedia entry is awesome, though I mentioned it earlier also). When I first read about it, I thought that it is just a coincidence and is a useless observation. But a few days ago while reading an article by Yuri Matiyasevich, I came to know about the importance of this observation. (Though just now I realised that Wikipedia article describes is clearly, so in this post I just want to re-write that idea.)

It’s an open problem in number theory to find a non-linear, non-constant polynomial which can take prime values infinitely many times. There are some conjectures about the conditions to be satisfied by such polynomials but very little progress has been made in this direction. This is a place where Ulam’s spiral raises some hope. In Ulam spiral, the prime numbers tend to create longish chain formations along the diagonals. And the numbers on some diagonals represent the values of some quadratic polynomial with integer coefficients.

Ulam spiral consists of the numbers between 1 and 400, in a square spiral. All the prime numbers are highlighted. ( Ulam Spiral by SplatBang)

Surprisingly, this pattern continues for large numbers. A point to be noted is that this pattern is a feature of spirals not necessarily begin with 1. For examples, the values of the polynomial $x^2+x+41$ form a diagonal pattern on a spiral beginning with 41.

# Repelling Numbers

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An important fact in the theory of prime numbers is the Deuring-Heilbronn phenomenon, which roughly says that:

The zeros of L-functions repel each other.

Interestingly, Andrew Granville in his article for The Princeton Companion to Mathematics remarks that:

This phenomenon is akin to the fact that different algebraic numbers repel one another, part of the basis of the subject of Diophantine approximation.

I am amazed by this repelling relation between two different aspects of arithmetic (a.k.a. number theory). Since I have already discussed the post Colourful Complex Functions, wanted to share this picture of the algebraic numbers in the complex plane, made by David Moore based on earlier work by Stephen J. Brooks:

In this picture, the colour of a point indicates the degree of the polynomial of which it’s a root, where red represents the roots of linear polynomials, i.e. rational numbers,  green represents the roots of quadratic polynomials, blue represents the roots of cubic polynomials, yellow represents the roots of quartic polynomials, and so on.  Also, the size of a point decreases exponentially with the complexity of the simplest polynomial with integer coefficient of which it’s a root, where the complexity is the sum of the absolute values of the coefficients of that polynomial.

Moreover,  John Baez comments in his blog post that:

There are many patterns in this picture that call for an explanation! For example, look near the point $i$. Can you describe some of these patterns, formulate some conjectures about them, and prove some theorems? Maybe you can dream up a stronger version of Roth’s theorem, which says roughly that algebraic numbers tend to ‘repel’ rational numbers of low complexity.

To read more about complex plane plots of families of polynomials, see this write-up by John Baez. I will end this post with the following GIF from Reddit (click on it for details):

# Prime Consequences

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Most of us are aware of the following consequence of Fundamental Theorem of Arithmetic:

There are infinitely many prime numbers.

The classic proof by Euclid is easy to follow. But I wanted to share the following two analytic equivalents (infinite series and infinite products) of the above purely arithmetical statement:

• $\displaystyle{\sum_{p}\frac{1}{p}}$   diverges.

For proof, refer to this discussion: https://math.stackexchange.com/q/361308/214604

• $\displaystyle{\sum_{n=1}^\infty \frac{1}{n^{s}} = \prod_p\left(1-\frac{1}{p^s}\right)^{-1}}$, where $s$ is any complex number with $\text{Re}(s)>1$.

The outline of proof,   when $s$ is a real number, has been discussed here: http://mathworld.wolfram.com/EulerProduct.html