3,628,800

This post is inspired by a recent puzzle published by Alex Bellos in the Guardian: Did you solve it? Puzzles you can do in the pub | Mathematics | The Guardian

The title of the post is 10! = 1x2x3x4x5x6x7x8x9x10 = 3,628,800. Interestingly, this value is the same number of seconds in six weeks. The challenge in Bellos's puzzle was to make the comparison without using a calculator.

One can confirm this without a calculator. Write out the number of seconds in six weeks as:

6 (weeks) x 7 (days) x 24 (hours) x (60 minutes) x (60 seconds)

Now one can cancel factors in the above equation by matching up with terms in the equation for 10!.

Cancel the 6 in both equations.

Cancel the 7 in both equations.

Cancel the 24 by matching with the 3 and 8 in the 10! equation

Reduce the 60 to 6 in the seconds equation by cancelling the 10 in the 10! equation.

Reduce the other 60 to 6 in the seconds equation by canceling the the 2 and the 5 in the 10! equation.

These canceling operations leaves us with 6x6=36 in the seconds equations and 4x9=36 in the factorial equation, therefore the number of seconds in six weeks equals 10!.

Alex Bellos is also the author of one of my favorite math books: Math Vacation: My Favorite Math Websites (jamesmacmath.blogspot.com)

Georg Cantor and Infinity

Image: https://photos.aip.org/history-programs/niels-bohr-library/photos/cantor-georg-a1

Georg Cantor was a German mathematician whose work on set theory changed the way mathematicians viewed infinity. Before Cantor, infinity was seen as a vague and imprecise concept. Cantor's ideas gave infinity a rigorous mathematical framework and led to the development of new branches of mathematics.

Cantor was born in 1845 in Russia and studied mathematics in Germany. He became interested in the concept of infinity and began to study it in depth. One of Cantor's key insights was that there are different levels of infinity. He showed that some infinities are "larger" than others, and that this could be rigorously proven using mathematical concepts such as sets and cardinality.

Cantor's work on infinity was not initially well received by his peers. Many mathematicians at the time were uncomfortable with the idea of infinity and saw it as a source of paradoxes and contradictions. They believed that mathematics should only deal with finite quantities and that infinity was not a legitimate mathematical concept.

Cantor's work challenged this view and showed that infinity could be studied rigorously and systematically. He introduced the concept of transfinite numbers, which are numbers that are greater than any finite number but less than infinity. He also developed the idea of cardinality, which measures the size of a set by comparing it to another set.

Cantor's ideas led to the development of new branches of mathematics such as set theory and topology. They also had important implications for other areas of mathematics, such as analysis and algebra. Cantor's work was a major contribution to the development of modern mathematics and has had a profound impact on the field.

Despite his contributions to mathematics, Cantor faced considerable opposition during his lifetime. He was often criticized by his peers and suffered from depression and mental illness. However, his ideas eventually gained acceptance and his work has had a lasting impact on the field of mathematics.

One of Cantor's most important contributions to the study of infinity was his development of the concept of cardinality, which measures the "size" of a set. Cantor introduced a notation to represent the cardinality of a set: aleph-null (ℵ₀) for the cardinality of the set of natural numbers, and aleph-one (ℵ₁) for the next smallest infinity.

Cantor's work showed that not all infinite sets have the same cardinality, and that there are different levels of infinity. For example, the set of real numbers is larger than the set of natural numbers, and Cantor showed that there is no way to match up every real number with a unique natural number. This led Cantor to propose the continuum hypothesis, which states that there is no set with cardinality between that of the natural numbers and that of the real numbers.

Cantor's work on cardinality and infinity led to the development of a hierarchy of aleph numbers, denoted by ℵ₀, ℵ₁, ℵ₂, and so on. Each aleph number represents the cardinality of a set that is larger than the previous one. For example, ℵ₁ is the cardinality of the set of all countable ordinals, which is larger than the cardinality of the set of natural numbers.

The concept of aleph levels has been used in many areas of mathematics, including set theory, topology, and analysis. It has also had important implications for the foundations of mathematics, as it showed that the notion of infinity can be studied rigorously and systematically.

In conclusion, Georg Cantor's work on infinity changed the way mathematicians viewed this concept. His ideas gave infinity a rigorous mathematical framework and led to the development of new branches of mathematics. Cantor's work was initially met with opposition, but it has had a lasting impact on the field of mathematics and has led to new insights and discoveries.

Contributor: This post was written by ChatGPT, a language model developed by OpenAI. For more information on language models, visit the OpenAI website Edited by: J. McMahon

A Chat with A.I.

 

                                                        Image: Howcolour on Iconfinder

I recently tested ChatGPT Feb 13 Version for researching topics for this blog site. The first entry was a recent post on the topic of zero: Math Vacation: Zero (jamesmacmath.blogspot.com). The short article written by the bot was the appropriate length and I haven't found errors.

To test the bot's capabilities a little more, I challenged it to write another article on the topic of how Fibonacci found that every other number, starting with 5, in the Fibonacci sequence is the length of the hypotenuse of a right-angled triangle defined by Pythagorean triple (see: Pythagorean Triples, Part II, Part III).

The article produced was interesting; however, I found a few errors. I'm posting the bot's product unedited with errors, which I'll correct in a future post:

A.I. request: Can you write how it is that every other number in the Fibonacci sequence is the length of a Pythagorean triple hypotenuse?

Response: Certainly! The relationship between the Fibonacci sequence and Pythagorean triples is a fascinating one. Let me explain how it works.

First, let's define what we mean by Pythagorean triples. A Pythagorean triple is a set of three positive integers (a, b, c) that satisfy the equation a^2 + b^2 = c^2, where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides.

Now, let's consider the Fibonacci sequence. The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding numbers, starting with 0 and 1. So the sequence goes like this: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on.

Now, here's the interesting part: every third number in the Fibonacci sequence is the length of the hypotenuse of a Pythagorean triple! To see why this is the case, let's consider a general Pythagorean triple (a, b, c) where c is odd. We can then write c as the sum of two consecutive integers: c = n + (n+1), where n is a positive integer.

Now, using the Pythagorean theorem, we can express a and b in terms of n as follows:

a = n^2 - (n+1)^2 = -2n-1 b = 2n^2 + 2n

Notice that a and b are both odd integers. We can then express the ratio of a and b as follows:

a/b = -(2n+1)/(2n^2+2n) = -(2n+1)/(2n(n+1))

Now, if we let n be a Fibonacci number, then we have:

n = F_k for some positive integer k.

Therefore, we can express the ratio of a and b in terms of Fibonacci numbers:

a/b = -(2F_k+1)/(2F_k(F_k+1)) = -1/F_{k+1}

This means that the ratio of a and b is a negative reciprocal of a Fibonacci number. And since every third Fibonacci number is even, it follows that every other number in the Fibonacci sequence is the length of a Pythagorean triple hypotenuse.

In summary, every third number in the Fibonacci sequence is the length of a Pythagorean triple hypotenuse because the ratio of the other two sides of the corresponding right triangle can be expressed as a negative reciprocal of a Fibonacci number.

A Million Dollar Prize for Solving a Math Problem

 

(Yitang Zhang, Photo credit: http://www.voachinese.com/media/video/i-america-math-zhang-yitang-20131204/1803128.html)

The Riemann hypothesis, first proposed by Bernhard Riemann in 1859, deals with the distribution of prime numbers. At the beginning of the 20th century, it was one of the top 23 problems identified by David Hilbert for the mathematical community to work on for the upcoming century. Math Vacation: David Hilbert Problems (jamesmacmath.blogspot.com) Riemann hypothesis - Wikipedia

The Riemann hypothesis and five others remain unsolved. The Clay Mathematics Institute has offered a US$1 million prize to anyone who can solve the Riemann hypothesis. Millennium Problems | Clay Mathematics Institute

The list of unsolved Hilbert problems may be closer to being shorten by University of California – Santa Barbara professor Yitang Zhang. https://www.scmp.com/news/china/science/article/3198779/has-chinese-born-professor-discovered-big-piece-150-year-old-maths-puzzle

If successful, this would be remarkable for a 67-year-old mathematician. However, it won’t be the first time for Zhang to solve a difficult problem later in life. Nine years ago, he shocked the world with his work of the twin-prime conjecture. While he didn’t prove the twin prime conjecture, he did prove there exists a limit below which (at the time it was 70 million) there must be an infinite number of primes separated by that a specific gap size, N. Since then, using Zhang’s techniques, the limit for that minimum gap size has been dropped from 70 million to a few hundred. If mathematicians could bring this limit down to 2, it would prove the twin prime conjecture.

August 20, 2023 Update: New York Post Article - Riemann hypothesis: Unsolved math problem worth $1 million (nypost.com).

Geometry for dummies - A book review

 

In high school, I remember enjoying geometry class. It was in this class where most students first learn about proofs, and I liked the satisfaction of putting together solid proofs for theorems based a collection of given facts. Recently, I searched out for books to re-learn these skills. My first book selected from the local library was too advanced. I felt hugely disappointed in not being able to follow the first proof presented. I returned that book and found Mark Ryan’s Geometry for dummies in the same section.

Ryan’s approach is to introduce a concept and show a proof based on it. Then, he gives the reader a chance to apply the concept to another proof. He then walks the reader through longer proofs using multiple concepts earlier introduced. The author has a very clear way of explaining the ideas that makes this book easier to read (especially compared to the first book I checked out).

The book gives the basics and some more advance concepts on angles, triangles, polygons, circles and has an introduction to 3-dimensional geometry.

The author's biography says Mark Ryan practiced law for four years before deciding he should do something he enjoys. To me, this isn't surprising. In another post, I introduced a famous lawyer who improved his debating skill by studying Euclid: Math Vacation: Abraham Lincoln - A President Trained by Euclid (jamesmacmath.blogspot.com)

Book Website: Geometry For Dummies, 3rd Edition | Wiley

David Hilbert Problems

 

By Unknown author - Possibly Reid, Constance (1970) Hilbert, Berlin, Heidelberg: Springer Berlin Heidelberg Imprint Springer, p. 230 ISBN: 978-3-662-27132-2., Public Domain, https://commons.wikimedia.org/w/index.php?curid=36302

In 1900, a renowned German mathematician, David Hilbert, published 23 problems for the mathematical community to work on for the 20th century. At the time of the publication, the problems had not been solved. A link to the English translation of the problems: S0002-9904-1902-00923-3.pdf (ams.org)

To date, 8 of the problems have been resolved, meaning the problem was solved or a proposed proposition was proven or disproven. Another 9 have been partially resolved, meaning there exists some controversy as to whether the proposed solution fully resolves the question. Six problems remain unsolved.

A breakdown of the problems' status is given here: Hilbert's problems - Wikipedia.

Problem 2 was written about in a prior post about Kurt Gödel's incompleteness theorem.

Problem 18 sought a proof for the densest sphere packing which wasn't resolved until 1998. Also see this blog's post on sphere packing in higher dimensions.

Abraham Lincoln - A President Trained by Euclid

In studying law, Abraham Lincoln (16th United States of America President) came across the word demonstrate frequently. He searched for further meaning. Consulting Websters, he found “certain proof” and “proof beyond the possibility of doubt.” 

So, what did Lincoln do? He studied the works of Euclid – six volumes of geometric proofs and he mastered all of them. Proofs where a proposition is demonstrated beyond doubt based on a foundation of facts and logic. We don’t know Lincoln as a mathematician, but we do know him as a famous debater and orator.

Geometry in high school is the point at which students either really start to understand the mathematical process or really despise math.

It’s in geometry where we learn how to set-up a proof. We start with given axioms and step, by step, we establish a proof. At the end of the proof, it is the practice to write “Q.E.D.” Which is the Latin acronym for Q.E.D., Quod erat demonstrandum -

– That which is to be demonstrated.

 Some of this chain of logic is seen in the 1^st paragraph of the Gettysburg Address (note the term proposition):

             "Four score and seven years ago our fathers brought forth upon this continent, a new nation, conceived in Liberty, and dedicated to the proposition that all men are created equal."

Another example is from AL’s private notes in 1854 (perhaps formulating an opposing viewpoint to those who tried to justify slavery):

“If A can prove, however conclusively, that he may, of right, enslave B – why may not B snatch the same argument, and prove equally, that he may enslave A.” Source: Abraham Lincoln: A Life by Michael Burlingame, Johns Hopkins University Press, 2013.

Also, in the ending paragraph of AL 1^st inaugural address (more of a music analogy but linked by chords and harmony to math):

The mystic chords which, proceeding from so many battlefields and so many patriot graves, pass through all the hearts and all hearths in this broad continent of ours, will yet again harmonize in their ancient music when breathed upon by the guardian angel of the nation.

So, Lincoln learned from the discipline of geometry how to make a solid, logical argument and how make a proof.

1.   Carpenter, F.B. The Inner Life of Abraham Lincoln: Six Months at the White House Hurd & Houghton, New York, NY (1874).

Article: From Euclid to Abraham Lincoln, Logical Minds Think Alike (nautil.us)

A Simple Pythagorean Theorem Proof

 

I recently came across a proof for the Pythagorean Theorem that is simple and contains a minimum amount of equations. It may be one of the most direct and understandable proofs of the many that exist. A nice animation of this proof is found at MathAdam.

 

Start with your standard right-angle triangle of sides A, B and C with C being the hypotenuse.

 

Draw a line perpendicular to C to the opposite vertex.

Now, in addition to the original triangle, there are two additional right-angle triangles formed. One has the side A as its hypotenuse (well name triangle A) and the other has side B has its hypotenuse (triangle B). The original triangle with sides ABC will now be called triangle C. The three triangles, A, B, and C are all similar A ~ B ~ C. (This can be established by the fact all three has the same interior angles.) Another key fact that we can see is the area of A + B =  area of C

The next step is to flip over each of the three triangles.

The next step is to the draw out squares on each of the three sides.

The relative sizes of the squares have the same relative ratios as the triangles A:B:C.

Since the areas of triangles has the relationship: A + B = C, then the same applies for the squares, therefore A² + B² = C².

                

Incompleteness

For thousands of years, mathematicians searched for a set of axioms from which all mathematical truths could be derived. Their dreams were shattered when Kurt Gödel published his incompleteness theorem. He proved that there can be no set of axioms from which all other mathematical facts can be derived. The result is there will always be truths that cannot be proven. We are then left with questions that may or may not have answers such as the Collatz Conjecture, Goldbach Conjecture and many other unproven mathematical conjectures. We may be disappointed that we'll never have a single set of axioms for mathematics, but Gödel's proof leaves us with the extra challenge of proving these conjectures knowing such proof may or may not exist.

Also see recent article: 

Natalie Wolchover

https://bigthink.com/surprising-science/kurt-godel-foundations-mathematics-unproven

Trigonometric Proof of Pythagorean Theorem

This proof is similar to Bhaskara's second proof, but at the end we also derive an important trigonometric identity. In the diagram, A, B, C are the lengths of the sides of a right triangle. The lower case, a, represents the angle opposite of side A.

From the definitions of the common trigonometric functions of sin(a) and cos(a), sin(a) = A/C and cos(a) = B/C. These two equations can be rewritten as A = C sin(a) and B = C cos(a).

On the hypotenuse side of the triangle, C can be broken into two sub-lengths, A’ and B’, which represent the projection of sides A and B, respectively, onto side C.

From the similar triangles that are formed, sin(a) = A’/A and cos(b) = B’/B. These two equations can be rewritten as A’ = A sin(a) and B’ = B cos(a).

Substituting for sin(a) and cos(b), A’= A (A/C) and B’ = B (B/C) or A’ = A²/C and B’ = B²/C.

Given that side C = A’ + B’, we get C = A²/C and B’ = B²/C. Multiplying both sides by C, we now derive the Pythagorean Theorem:

C² = A² + B².

The above proof of the Pythagorean Theorem used the trigonometric functions to derive the terms of A² and B². Alternatively, we can keep the sin(a) and cos(a) functions to show another important identity. Returning to the terms for A’ and B’, we can also express these two lengths as A’ = A sin(a) and B’ = B cos(a). Substituting for A and B in these two equations, A’ = C sin(a) sin(a) and B’ = C cos(a) cos(a). Trigonometric convention allows these terms to be stated as A’ = C sin²(a) and B’ = C cos²(a).

Substituting these last two equations into the relationship of side C being the sum of A’ and B’,

C = A’ + B’

C = C sin²(a) + C cos²(a)

Next divide both sides of the equation by C, we now have an important trigonometric identity:

1 = sin²(a) + cos²(a)

Four Color Map

I first heard about the four-color map theorem in the early 1980's when I was in grad school at the University of Illinois at Urbana-Champaign. When I mentioned to a friend the name of my professor for a math class I was taking, he reminded me of Wolfgang Haken's fame. Just a few years earlier, Haken and a colleague, Kenneth Appel, proved the four-color map theorem. The theorem states for a planar surface, divided into any number and shape of regions, four colors suffice for coloring the regions so no two adjacent regions have the same color. 

In the first, simple checkerboard pattern below, only two colors are needed. However, in the second example, which already has used three colors, the middle region must use a fourth color to satisfy the condition that no adjacent regions have the same color.

Cartographers had known of the restriction for years but no one had been successful in proving the theorem until Haken and Appel in 1976 - published in 1977 in the Illinois Journal of Mathematics.

Haken and Appel's proof was ground-breaking and controversial as it was the first mathematical proof that relied on computer assistance. In their proof, they used mathematical rules to reduce the infinitude of possible map configurations to 1,834 configurations which were then checked one by one by computer.

7/25/2023 Update: Scientific American recently published a history of the four-color map problem: How a Doodler's Problem Sparked a Controversy in Math - Scientific American.

2/6/2024 Update: Quanta Magazine has a video explaining the four-color theorem and the use of computers to establish the proof: How Math’s Famous Map Theorem Was Solved With Computers.

Thomas Bayes - an Attempt to Prove the Existence of God

Thomas Bayes was an eighteen-century Presbyterian minister and gifted statistician. Two of his publications, Divine Benevolence, or an Attempt to Prove That the Principal End of the Divine Providence and Government is the Happiness of His Creatures (1731), and An Essay towards solving a Problem in the Doctrine ofChances (1763), are examples of how he wanted to use his intellect to prove the existence of God. His attempts were unsuccessful, otherwise Thomas Bayes would be a known to all and not just to mathematicians. However, his devotion is honored as his approach was so rigorous that it established a new field of mathematics, now named Bayesian Statistics.

Our current world-wide Covid-19 crisis provides an example of how Bayesian Statistics is applied.

First, here is the nomenclature for revised probabilities:

P(A)       Probability of Event A

P(B)       Probability of Event B

P(A|B)  Probability of A, given B is true

Fictitious Covid-19 problem

Suppose a test for Covid-19 is 95% effective in detecting the disease when present in a patient. The test gives a false positive result 10% of time when the disease is not present. Approximately 4% of the population has the disease. Given a positive test result, what is the probability that a patient actually has the disease.

The three facts stated above can be expressed as:

A.     P(D) = .04               4% of the population has the disease, and therefore P(N)=.96 or 96% of population doesn’t have the disease.

B.       P(Pos|D) = .95       The test gives a positive result 95% of the time when the disease is present

C.       P(Pos|N) = .10       Given no disease, 10% of those tested will falsely test positive

Now, a typical question is: given a positive test result, what is the probability a patient actually has the disease? Using our nomenclature, we are looking for P(D|Pos)  or probability of D, given Pos. We did have P(Pos|D) = .95, but that is different than P(D|Pos). In Bayesian terms, this question is called an inverse probability problem.

Bayes Theorem provides a formula for determining this revised probability, but I think it easier to express our known and unknown probabilities in a table. Our first given facts are pre-filled in the table. 

	Disease Present (D)	No Disease (N)	Total
Positive Test result (Pos)	.95 x .04 = .038 (Fact B)	.10 x .96 = .096 (Fact C)
Negative Test result
Total	.04 (Fact A)	.96

All the remaining entries in the table can be determined by simple addition and subtraction. Adding across the top row, the total proportion of the population with Positive Test results is .038 + .096 = .134; the proportion with Negative Test results when the disease is present is .04 - .038 = .002; the proportion having a Negative Test result with no disease is .96 - .096 = .864; the total proportion having Negative test results is 1.00 - .134 = .866 and with that last item we can complete all the entries in the table.

	Disease Present (D)	No Disease (N)	Total
Positive Test result (Pos)	.95 x .04 = .038	.10 x .96 = .096	.134
Negative Test result	.04 - .038 = .02	.96 - .096 = .864	.866
Total	.04	.96	1.00

For our question of the probability of the Disease being present, given a Positive test result, we just need to focus on the top row of the table representing different ways to have a Positive test result.

.038 will test Positive when they actually have the disease

.096 will test Positive when they don’t have the disease (from false positive results)

In total, the proportion of the population that will have Positive results is .134 and of those .038 actually have the disease. Therefore, P(D|Pos) = .038/.134 = .284 so only 28.4% of those with a Positive test result will actually have the disease. This often strikes people as being too low. This is a common situation of imperfect tests and is known as the Paradox of the False Positive. From the table, we see that many more people had false positives than people who actually have the disease. Even though many medical tests have this False Positive error, the tests are still helpful. In our example, a person testing positive has an increase risk level compared to the general population (28.4% vs 4%) so the positive result would justify further testing and enhanced precautionary measures.

Returning to Bayes' Theorem, the inverse probability is given by:

P(D|Pos) = P(Pos|D) x P(D) / P(Pos)

The table given above provides the same answer as the Bayes' Theorem equation.

Additional Sources on Bayes:

CornellBlog: https://blogs.cornell.edu/info2040/2018/11/28/bayes-theorem-and-the-existence-of-god/

DanKopk article on Bayes and Price: https://qz.com/1315731/the-most-important-formula-in-data-science-was-first-used-to-prove-the-existence-of-god/

 Image details: Public Domain

• File:Michelangelo - Creation of Adam (cropped).jpg
• Created: circa 1511 date QS:P571,+1511-00-00T00:00:00Z/9,P1480,Q5727902

 

Twin Prime Sandwich

Building up on the lesson from a prior post, "Prime Numbers - a property rediscovered," it appears that all numbers between every pair of twin primes are multiples of 6. 

This isn't too surprising given that all primes are in the form 6n+/-1, but that fact alone doesn't guarantee that the number between a pair of twin primes is of the form 6n.

Here are the steps of the proof.

1. Let a, b, c, d, e represent 5 consecutive integers for which b and d are twin primes.

2. Consider the prime, b. Given the 6n+/-1 theorem, then either a is a multiple of 6 or c is a multiple of 6.

If a is a multiple of 6, then d cannot be prime because d=a+3 making d a multiple of 3. Therefore c must be a multiple of 6.

3. Likewise, consider d. Either d is prime because c is a multiple of 6 or e is a multiple of 6. However, if e is a multiple of 6 then b=e-3 so b would be a multiple of 3 and not prime. Therefore c must be a multiple of 6 if b and d are twin primes.

Prime Numbers - a property rediscovered

I stumbled across an interesting property of prime numbers last year. The numbers 2 and 3 are the only primes that cannot be expressed as a multiple of 6 plus or minus 1. For example, 7 is 6 + 1, 11 is 12 -1, 13 is 12+1 and so on. However, not all numbers of that are a multiple of 6 plus or minus 1 are prime. For example, 25 = 24 + 1 and 25 is not prime. Another example is 35 = 36 - 1 and 35 is not prime.

After a sleepless night, I put together a short proof of all primes greater than 3 being expressed as a multiple of 6 plus 1 or minus 1.

Steps of the proof:

1. Let S represent a multiple of 6. All numbers greater than 3 can be expressed as either S, S+/-1, S+/-2 or S+/-3. (Why not include S+/-4 and S+/-5? An S+4 number such as 10 is already covered by the next S-2. That is, 12-2=10. Likewise for S-4; for example the number 8 or 12-4, is already covered by the previous S, 6 or 8 = 6+2. Again, for the numbers expressed by S+/-5 are covered by next, or previous, S+/-1.)

2. Since all numbers greater than 3 can be expressed as either S, S+/-1, S+/-2 or S+/-3, then so can all prime numbers.

3. Eliminate reasons for S, S+/-2, S+/-3

    Any number that is S (a multiple of 6) is not prime.

    Any number that is S+/-2 is an even number greater than 2 and therefore is not prime.

    S, as a multiple of 6, is also a multiple of 3 so any number that is S+/-3 is also a multiple of 3 and therefore is not prime (note: in the case of 6-3=3, 3 is prime but we've stated this property is for primes greater than 3.)

4. From Step 2 we established all primes can expressed as either S, S+/-1, S+/-2 or S+/-3. From Step 3 we eliminated primes being expressed as S, S+/-2 and S+/-3, so therefore by elimination, all primes greater than 3 must be able to be expressed as S+/-1.

I wondered why I was never taught this before. Was I dozing through this lesson in math? Certainly someone has come across this property before me. I entered "primes as 6n +/-1" and was greeted by the following ego-deflating statement "this is the most re-discovered property of prime numbers." Nearly the same proof is presented by Chris Caldwell, University of Tennessee at Martin. 6n+/-1.

I didn't at all expect my approach to be unique, but it is always very satisfying to prove something on your own and that should always be encouraged.

Update 3-16-2021

One can build up on this property of prime numbers with the fact that the square of all prime numbers above 3 is 1 plus a multiple of 24. For example, 7 square is 49 which equals 1+2x24. Matt Parker proves this additional prime number fact two different ways in this Numberphile YouTube video.

Proofs of the Pythagorean Theorem

There are hundreds of proofs of the Pythagorean Theorem. I've been testing my own limited knowledge of geometry trying to master as many as I can. Many of the proofs use similar techniques (recombining various triangular, square and rectangular shapes in various ways to show the desired result), so sometimes it is difficult to say if a proof is truly unique.

Of the many reconstruction methods, a popular technique is arranging multiple copies of a right-angle tri angle into a square shape then calculating the area of the resultant square in two ways. First, the squaring the length of a side. Second, calculating the areas of the components that make up the square. Setting these two areas equal, produces the desired result, showing:

C² = A² + B²

An example of this approach is given by Bhaskara's First Proof: Bhaskara's First Proof

The most concise proof, in my opinion, is Bhaskara's second proof: 

Start with a right-angle triangle, ABC. Let the lengths of the sides be a, b, and c.

Draw the altitude line, from point C to the opposite side at point D. Line CD is perpendicular to side AB. 

The three triangles, ABC, CBD and CAD are all similar. Let the distance from A to D be x and then from B to D is c - x.

Since the ratio of sides of similar triangles are equal,

 

a/c = (c-x)/a                              b/c = x/b

Simplifying,  

a² = c² – cx                                 b² = cx

Adding these two equations together, results in (note, the term cx and -cx cancel),

 a² + b² = c²

The broken domino proof starts with a domino-shape tile that consists of two equal squares (each representing  c²). Three triangular pieces are broken from one of the squares and rearranged to form the squares representing a² + b². See proof 35 at Gary Zabel's page: Gary Zabel UMB

As a practical use of the broken tile proof, there are actual tiling patterns based on the Pythagorean Proof. Tiling with Pythagoras

A prior post outlines President James Garfield's proof of the Pythagorean Theorem. Garfield Proof

Wikipedia entry: Pythagorean Theorem