The next issue of the Carnival of Mathematics, rounding up blog posts from the month of November 2023, is now online at John D Cook’s blog.
The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.
In the last Finite Group livestream, Katie told us about emirps. If a number p is prime, and reversing its digits is also prime, the reversal is an emirp (‘prime’ backwards, geddit?).
For example, 13, 3541 and 9999713 are prime. Reversing their digits we get the primes 31, 1453 and 3179999, so these are all emirps. It doesn’t work for all primes – for example, 19 is prime, but 91 is \(7 \times 13 \).
In the livestream chat the concept of primemirp emerged. This would be a concatenation of a prime with its emirp. There’s a niggle here: just like in the word ‘primemirp’ the ‘e’ is both the end of ‘prime’ and the start of ’emirp’, so too in the number the middle digit is end of the prime and the start of its emirp.
Why? Say the digits of a prime number are \( a_1 a_2 \dots a_n \), and its reversal \( a_n \dots a_2 a_1 \) is also a prime. Then the straight concatenation would be \( a_1 a_2 \dots a_n a_n \dots a_2 a_1 \). Each number \(a_i\) is in an even numbered place and an odd numbered place. Now, since
\[ 10^k \pmod{11} = \begin{cases} 10, & \text{if } k \text{ is even;}\\ 1, & \text{otherwise,} \end{cases} \]
it follows that each \(a_i \) contributes a multiple of eleven to the concatenation. A mismatched central digit breaks this pattern, allowing for the possibility of a prime.
I wrote some code to search for primemirps by finding primes, reversing them and checking whether they were emirps, then concatenating them and checking the concatenation. I found a few! Then I did what is perfectly natural to do when a sequence of integers appears in front of you – I put it into the OEIS search box.
Imagine my surprise to learn that the concept exists and is already included in the OEIS! It was added by Patrick De Geest in February 2000, based on an idea from G. L. Honaker, Jr. But there was no program code to find these primes and only the first 32 examples were given. I edited the entry to include a Python program to search for primemirps and added entries up to the 8,668th, which I believe is all primemirps where the underlying prime is less than ten million. My edits to the entry just went live at A054218: Palindromic primes of the form ‘primemirp’.
At this year’s MathsJam UK Gathering, I had the pleasure of running one of the Saturday Night Tables – a chance to invite attendees at the Gathering to drop by and play with something. Together with fellow Manchester MathsJam regular Andrew Taylor, I ran a table of Mathematical Drawing Hacks – ways to make drawing complex mathematical objects and shapes easier.
We had over 30 entries in the competition, of which 20 achieved correct answers, and have picked a random set of winners to pass on to Naylor Games, who should be in touch with them by email in the next few days.
For anyone interested in seeing the answers, here’s what they were. As a reminder, the challenge here is to find a value for \(x\), given that \(n\) represents the number of cards, to get the total of all the card values closest to 21.
The next issue of the Carnival of Mathematics, rounding up blog posts from the month of October 2023, is now online at Beauty of Mathematics.
The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.
Reminder: I’m occasionally working to (sort of) recreate Martin Gardner’s cover images from Scientific American, the so-called Gardner’s Dozen.
This time I’m looking at the cover image from the July 1965 issue, accompanying the column on ‘op art’ (which became chapter 24 in Martin Gardner’s Sixth Book of Mathematical Diversions from Scientific American).
In this guest post, David Benjamin shares a cornucopia of concepts and stories relating to Pythagoras and his famous theorem.
I admit to mild irritation when I’m told that Pythagoras’ theorem is $a^2+b^2=c^2$. The theorem is based on area – in particular, that of squares. There are many proofs of the theorem and in this post we present a miscellany of Pythagorean Theorem curiosities, including some of my favourite proofs, the theorem’s links to algebra, geometry and number theory, an assassination of a president of the USA, an alleged murder in Greece, an infinite spiral of surds, Gauss and coordinate geometry – plus another connection between Pascal and Fibonacci.
The theorem
A square is added to each side of a right-angled triangle as shown in the above image. The sum of the the areas of the two smaller squares is equal to the area of the largest square. If the hypotenuse of the triangle has a length of $c$ and the other two sides are of length $a$ and $b$ then $a^2+b^2=c^2$
When the length of each side of the triangle is a positive integer, the three numbers make a Pythagorean triple. $(3, 4, 5)$ is the smallest triple with $3^2+4^2=5^2$. The Chinese text Chou Pei Suan Ching – original title Zhoubi – (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives this visual proof for the $(3, 4, 5)$ triple.
A visual proof for the Pythagorean triple $(3, 4, 5)$ from the Chinese text Chou Pei Suan Ching
$(3, 4, 5)$ is a primitive triple since $3, 4$ and $5$ are coprime – their only common divisor is $1$. $(n\times3, n\times4, n\times5), n = 2, 3, 4,…$ are part of the same ‘family’ and clearly not primitive triples. Another primitive triple is $(5, 12, 13)$ and an ordered sequence of hypotenuses for such triples are listed here. In the sequence I was surprised to see $185$ appearing twice. In fact there are exactly four distinct triples with hypotenuse $185$. With the aid of a spreadsheet, I was able to find them: $(57, 176, 185), (60, 175, 185),(104, 153, 185)$ and $(111, 148, 185)$
Generating Pythagorean triples
The sequence $\frac{1}{1},\frac{3}{2},\frac{7}{5},\frac{17}{12},\frac{41}{29},\frac{99}{70},\frac{239}{169},\frac{577}{408},\frac{1393}{985},…$ produces a Pythagorean triple from every other term beginning with $\frac{7}{5}$:
As an added bonus, the decimal equivalent of each term of the sequence converges to $\sqrt2$, in a similar way the Fibonacci sequence converges to the golden ratio $\psi=\frac{1+\sqrt5}{2}$
Another method to find Pythagorean triples uses consecutive even numbers and the sum of their reciprocals as shown below.
Euclid of Alexandria (325BC – 265BC) was a Greek mathematician who wrote a treatise, The Elements – a collection of 13 books. Books 1 to 6 are on plane geometry and books 7 to 9 on number theory. Euclid created a formula for generating Pythagorean triples from any pair of positive integers $m$ and $n$, where $m>n$:
($m^2-n^2, 2mn, m^2+n^2$) is the triple.
If $m=7$ and $n=3$, the triple is ($40, 42, 58$) $\rightarrow 40^2 + 42^2 = 3364 = 58^2$
If $m=89$ and $n=11$, the triple is ($7800, 1958, 8042$) $\rightarrow 7800^2 + 1958^2 = 64673764 = 8042^2$
A lovely link between Pascal, Fibonacci, Euclid and Pythagoras comes via any four consecutive Fibonacci numbers
$F_{1}$
$F_{2}$
$F_{3}$
$F_{4}$
$F_{5}$
$F_{6}$
$F_{7}$
$F_{8}$
$F_{9}$
$F_{10}$
$1$
$1$
$2$
$3$
$5$
$8$
$13$
$21$
$34$
$55$
The first 10 Fibonacci numbers
Using $3, 5, 8, 13$
Multiply the first and the last numbers $\rightarrow 3 \times{13}=39$
Multiply then double the middle two numbers $\rightarrow 5 \times{8}\times{2}=80$
Sum the squares of the middle two numbers $\rightarrow 5^{2}+8^{2}=89$
The Pythagorean triple is $(39, 80, 89) \rightarrow 39^{2}+80^{2}=7921=89^{2}$
$F_{1}$ to $F_{4}$ gives the primitive triple $(3, 4, 5)$
$F_{2}$ to $F_{5}$ gives the primitive triple $(5, 12, 13)$
$F_{3}$ to $F_{6}$ gives the triple $(16, 30, 34)$
$F_{7}$ to $F_{10}$ gives the primitive triple $(715, 1428, 1597)$
In addition, for any set of four consecutive Fibonacci numbers, $F_{7}$ to $F_{10}$ for example, the following connection is true
$7+10=17$ and the $17^{th}$ Fibonacci number is $1597$, the third member of the triple, the hypotenuse of the triangle!
Amazingly, if we use $m$ and $n$ as consecutive Fibonacci numbers when using Euclid’s method, then the last number of the triple is again a Fibonacci number
$m$
$n$
Triple
Fibonacci number
$2$
$1$
$(3, 4, 5)$
$5^{th}$
$3$
$2$
$(5, 12, 13)$
$7^{th}$
$5$
$3$
$(16, 30, 34)$
$9^{th}$
$8$
$5$
$(39, 80, 89)$
$11^{th}$
$13$
$8$
$(105, 208, 233)$
$13^{th}$
Another pattern is created with the the sequence of odd numbers $5, 7, 9, 11, 13,…$ The $15^{th}$ Fibonacci number is $610$, the hypotenuse when $m=21$ and $n=13$
A visual proof of the theorem
I first came across a visual proof of Pythagoras’ theorem for all right-angled triangles in Roger B. Nelsen’s wonderful book Proofs Without Words, Exercises in Visual Thinking. Nelsen noted the proof (author unknown, circa B.C. 200?) is adapted from the Chou Pei Suan Ching. The two images below combine to show the proof:
A visual proof of Pythagoras’ theorem for all right-angled triangles
A proof by trapezium area and an untimely painful death
James Garfield (November 19, 1831 – September 19, 1881) was elected as the United States’ 20th President in 1880. He was assassinated after just 200 days in office after being shot on July 2, 1881, in a Washington railroad station. Garfield remained mortally wounded in the White House for many weeks where Alexander Graham Bell, inventor of the telephone, attempted to locate the bullet with an induction-balance electrical device which he had designed. Bell and physicians were unsuccessful in their attempts and Garfield died from an infection and an internal haemorrhage on September 19, 1881.
In 1876, Garfield had an elegant proof of Pythagoras’ theorem published. The proof makes use of the formulae for the areas of a triangle and a trapezium. The proof is demonstrated in the image below
A proof using the Shoelace formula
The brilliant German mathematician Johann Carl Friedrich Gauss (30 April 1777 – 23 February 1855) developed a formula to calculate the area of a polygon if every vertex of the polygon lies on a known Cartesian coordinate. The formula is widely known as the Shoelace formula and makes use of the calculation to find the determinant of a 2 by 2 matrix.
John Molokach observed that the Pythagorean theorem follows from Gauss’ Shoelace Formula, as shown below
It’s not just squares
If the same regular $n$-gon, $n$ = 3, 4, 5,.. is drawn on each side of a right-angled triangle, then the sum of the areas of the two smaller $n$–gons equals the area of the $n$–gon on the hypotenuse. Semicircles also produce the same result. As $n\rightarrow \infty$, a regular $n$-gon approaches a circle and so circles, where the sides of the triangle act as tangents to the circles can be said to satisfy Pythagoras’ theorem. The same result can be obtained by rotating the semicircles through $180^ \circ$ and adding matching semicircles.
Three regular hexagons drawn on a $3, 4, 5$ triangle with $\frac{27\sqrt3}{2}+\frac{48\sqrt3}{2}=\frac{75\sqrt3}{2}$
The spiral of Theodorus of Cyrene
When $n$ is not a square number, $\sqrt{n}$ is called a surd
Pythagoras and his followers, the Pythagoreans, believed that the universe can be explained by whole numbers and the ratio of whole numbers. Their moto – “All is number” – was carved above the entrance of their meeting place. However, a spanner was thrown in the works when Hippasus of Metapontum, one of the Pythagoreans suggested that $\sqrt2$, the length of the hypotenuse of the right-angled triangle with sides $1, 1$ and $\sqrt2$, could not be written as the ratio of two whole numbers. Such numbers are now called irrational and it was approximately 200 years before Euclid of Alexandria proved $\sqrt2$ was irrational. The Pythagoreans were sworn to secrecy and one legend suggests that Hippasus was thrown off a boat and drowned for revealing his discovery to non-Pythagoreans. Then again, some believe Hippasus drowned because he revealed how to construct a dodecahedron inside a sphere. The ($1, 1$,$\sqrt2$) triangle is the start of the spiral of Theodorus. Further right-angled triangles, each containing a side of length $1$ unit, are added sequentially to produce the sequence $\sqrt2, \sqrt3, \sqrt4, \sqrt5,…$. The sequence is the length of each new hypotenuse, as shown in the diagram below. Theodorus probably stopped at $\sqrt17$ as it the length of the hypotenuse of the triangle before the triangles begin to overlap. The spiral can be expanded here.
The Theodorus spiral – also known as the Einstein spiral, Pythagorean spiral, square root spiral