A conversation about mathematics inspired by a Spirograph set. Presented by Katie Steckles and Peter Rowlett.

]]>You’d be surprised how much math[s] you can learn by exploring some of the implications and ramifications of what may seem at first no more than a trivial brainteaser

Martin Gardner

Every so often, I get nerdsniped good and proper. In some cases, multiple times in a week. Here is Chris Smith doing just that:

Here’s Chris’s question in text form:

Can we find three different triangular numbers $a$, $b$ and $c$ so that every combination of the three gives a triangular number?

So \(a+b\), \(a+c\), \(b+c\) and \(a+b+c\) should all be triangular numbers.

In this case, it wasn’t the puzzle itself that did for me; I solved that fairly quickly. But in so doing, I dug up something deeper and more interesting, and – to the best of my knowledge – unsolved.

Welcome to the fascinating world of almost Pythagorean triples.

You, being an avid Aperiodical reader, might have come across Pythagorean triples before: three integers, \(a\), \(b\) and \(c\), such that \(a^2 + b^2 = c^2\). Examples of this have been known for about 4,000 years, and Euclid had a method for generating all of them. Been there, done that, and — like Hippasus — got the swimsuit.

In the 1980s, Orrin Frink wondered: what about triples that are off by one? That is, what if \(a^2 + b^2 = c^2 + 1\)? Quite naturally, he called these *almost Pythagorean triples*, or APTs.

These are not to be confused with *nearly Pythagorean triples*, in which \(a^2 + b^2 = c^2 – 1\). Different kettle of worms entirely.

Some examples of APTs include \(a=4, b=7, c=8\) and \(a=5, b=5, c=7\). To save space, I’m going to use brackets for my APTs from now on: instead of \(a=4, b=7, c =8\), I’ll simply write \((4,7,8)\).

Antalan and Tomenes also looked at APTs, and asked the key question I will answer in this article: “Is there an explicit formula to generate APTs?” The answer to this is “yes”, but we’ll need to do some work to get there.

First, some definitions: it’s useful to distinguish between *isosceles* APTs, such as \((5,5,7)\) and \((29,29,41)\), where \(a=b\) and *scalene* ones, where \(a \ne b\), as in \((4,7,8)\) and \((23,41,47)\).

The two shorter elements of an APT cannot both be even – if they were:

- \(a^2+b^2\) would be a multiple of 4
- \(c\) would be odd, so \(c^2+1\) would be two more than a multiple of 4

These two things can’t both be true! So at least one of \(a\) and \(b\) must be odd — and \( c \) is even if and only if one of the shorter elements is also even. Alternatively, if I suppose \( b \) is odd, which I will do presently, then \( c \) and \( a \) must have the same parity.

Without loss of generality, I can assume that \(b\) is odd. Define an *odd* APT as one where \(a\) and \(c\) also odd, and an *even* APT as one where \(a\) and \(c\) are both even.

So: \((4,7,8)\) is an even scalene APT, and \((5,5,7)\) is an odd isosceles APT. In fact, all isosceles APTs are odd.

Let’s apply some algebra to the key equation \(a^2 + b^2 = c^2 + 1\).

That can be rearranged to \(b^2 -1 = c^2- a^2\), and by difference of two squares to \((b-1)(b+1) = (c-a)(c+a)\).

I’ve asserted that \(b\) is odd and it follows that \(c\) and \(a\) are of the same parity, so all four of the factors here are even, and it makes sense to halve them: \( \left(\frac{b+1}{2}\right)\left( \frac{b-1}{2}\right) = \left(\frac{c+a}{2}\right)\left( \frac{c-a}{2} \right) \).

The algorithm I’m about to present treats each of these four factors as a product of two integers, so that the product on the left-hand side and the product on the right-hand side match. In particular, I’m going to generate four integers, \(s\), \( s’ \), \(t\) and \(t’\) such that:

- \(\frac{c+a}{2} = st\)
- \(\frac{c-a}{2} = s’t’\)
- \(\frac{b+1}{2} = ss’\)
- \(\frac{b-1}{2} = tt’\)

Note that \(ss’ – tt’ = 1\), which places a severe restriction on our values: \(\gcd(ss’, tt’)=1\). In fact, the ‘primed’ variables \(s’\) and \(t’\) can be deduced from any pair of coprime positive integers \(s\) and \(t\). I’ll explore how to generate the variables a little later (after we’ve talked about Bézout).

For example, if \(s = 10\) and \( t=3\), possible values for \(s’\) and \(t’\) are 1 and 3, respectively. Solving the simultaneous equations for \( a,b,c\) leads to the APT \( (27, 19, 33) \).

The goal is to begin with values of \( s \) and \(t \), find the corresponding values of \( s’ \) and \( t’ \), and use these to generate APTs. There are four major questions to answer:

- How can \(s’\) and \(t’\) be found?
- Does every pair of coprime \(s\) and \(t\) give an APT?
- Is every APT is generated by some \((s,t)\) pair?
- Is every APT is generated by exactly one \((s,t)\) pair?

If the answers to this are (respectively) “like so”, “yes”, “yes” and “yes”, then I’ll have an answer to Antalan and Tomenes’s challenge.

I’m going to give a **proposition**:

Given a pair of coprime integers, \(s\) and \(t\), there are infinitely many integer values of \(s’\) and \(t’\) such that \(ss’ – tt’=1\). However, there is only one of each such that \(0 \le t’ < s\) and \(0 \le s’ < t\), which is a lemma you may wish to prove.

Using the multiplicative inverse, let \(s’ = s^{-1} \pmod t\) and \(t’ = (-t)^{-1} \pmod s\).

Then \((st – s’t’, ss’ + tt’, st + s’t’)\) is an almost Pythagorean triple.

In the restriction \( 0 \le t’\), equality holds if and only if \( s = 1\).

A proof will follow later. You may provide your own if you can’t wait that long.

I thought you’d never ask. We need to take a quick detour into *modular arithmetic* — or rather, the world of remainders; modular arithmetic is fascinating and wonderful, but we’re only after one small aspect of it here.

A tl;dr, in case you just don’t recognise the term: if you’ve got two coprime integers, \( p \) and \( q \), then the multiplicative inverse of \( p \pmod q\) is the unique integer \(r \) such that \( 0 \le r < q \) and \( pr \) is one more than a multiple of \( q \).

Again, with the \( \le \) in the inequality, equality holds if and only if \( q = 1 \). We don’t usually do arithmetic modulo 1, but it helps to avoid an edge case with APTs.

For example, suppose \( q = 10 \) and \( p = 3 \). We’re looking for a number \( r \) that’s between 0 and 10 (exclusive), and with \( pr \) one more than a multiple of 10. The only option is \( 3 \times 7 = 21 \), so \( r = 7 \)

If you want to have a look at this in action, find (or write out) a times tables grid and find the entries that end in one. You will notice (I hope) that there is exactly one of these in each times table that doesn’t share a factor with 10.

Modulo 10, 7 is the *multiplicative inverse* of 3, and we write that as \(7 \equiv 3^{-1} \pmod{10}\). In general, \(r \equiv p^{-1} \pmod{q}\) whenever \(pr \) is one more than a multiple of \( q \), and \( 0 \le r < q \).

There’s no such restriction on \(p\): 43 has a perfectly good multiplicative inverse modulo 10 – it’s the same as the multiplicative inverse of 3.

So how do we find a multiplicative inverse if the numbers are too big to bother listing all the possibilities? We know an identity about that!

**Bézout’s identity**: Suppose \(\gcd(a,b)=d\). Then there exist integers \(a’\) and \(b’\) such that \(aa’ + bb’ = d\), and all multiples of \(d\) can be obtained this way.

In particular, if \(a\) and \(b\) are coprime, we can find \(a’\) and \(b’\) such that \(aa’ + bb’ = 1\) using the extended Euclidean algorithm, which I will explain only by example – suppose I want to find the multiplicative inverse of 3, modulo 10. I let \(a=10\) and \(b=3\):

- \(10 = (1 \times 10) + (0 \times 3)\)
- \(3 = (0\times 10) + (1 \times 3)\)
- Take away as many \(3\)s from \(10\) as possible: \(1 = (1\times 10 )- (3 \times 3)\)
- Take away as many \(1\)s from \(3\) as possible: \(0 = (-3 \times 10 )+ (10 \times 3)\), which makes sense, if you think about it.

I can add any multiple of \(0\) to the \(1\) to get another answer, so \(1 = 10(1-3k) + 3(10k – 3)\) for any integer \(k\). This gives \(a’ = 1-3k\) and \(b’ = 10k-3\). Since I want \( b’ \) to be the inverse of 3, modulo 10, I need \(0 \le b’ < 10\), so \(k=1\), \(a’=-2\) and \(b’=7\).

To double-check: \( (-2 \times 10) + (7 \times 3) = -20 + 21\), which is indeed one.

**TL;DR**, repeated for emphasis: Given that \(a\) and \(n\) are coprime, the multiplicative inverse \(a^{-1}\) of an integer \(a\), modulo \(n\) is the unique integer between \(0\) and \(n-1\) (inclusive) such that \(aa^{-1}\) is one more than a multiple of \(n\).

With regard to the algorithm, Euclid offers two-for-the-price-of-one here: if I work out \(s’ = s^{-1} \pmod t\) this way, the other coefficient (in absolute value) is the multiplicative inverse of \( -t \), modulo \(s \). Again, let’s check that with the example: I want \(t’ \) to be a number between \( 0 \) and \( 3 \), exclusive, such that \( -10 t’ \) is one more than a multiple of 3. Indeed, -20 is one more than -21, so it works!

Let’s pick two coprime integers at random to try this with: \( s= 14 \) and \( t = 9\).

**Step 1:** apply the Euclidean algorithm to find \( s’ \) and \( t’ \).

- \( 14 = 1s + 0t \)
- \( 9 = 0s +1t \)
- Take away as many \(9\)s as possible: \( 5 = 1s – 1t \)
- Take away as many \(5\)s as possible: \( 4 = -1s + 2t \)
- Take away as many \(4\)s as possible: \( 1 = 2s – 3t \)
- Take away as many \(1\)s as possible: \( 0 = -9s + 14t \), as one would hope.

In this case, the \( s \) coefficient is positive, so it doesn’t need to be adjusted: \( 14^{-1} \pmod 9 = 2\) and \((-9)^{-1} \pmod 14 = 3\), so \(s’ = 2 \) and \( t’ = 3 \).

Solving the simultaneous equations:

- \(\frac{c+a}{2} = st= 126\)
- \(\frac{c-a}{2} = s’t’ = 6\)
- \(\frac{b+1}{2} = ss’ = 28\)
- \(\frac{b-1}{2} = tt’ = 27\)

gives \( a = 120 \), \( b = 55 \) and \(c = 132\); checking, \( 120^2 + 55^2 = 17,425\), which is \( 132^2 + 1\). Therefore, \( s= 14\), \( t= 9 \) generates the APT \( (120, 55, 132 ) \).

These proofs have not been properly refereed, and I am not a proper number theorist. I would very much welcome constructive criticism about tightening them up.

**A quick note on an edge case**: as I mentioned previously if \( s = 1 \) the multiplicative inverse of \( t\) is not conventionally defined. However, the extended Euclidean algorithm *does* give a value, zero, for \( t’ \), which serves perfectly well for the algorithm. In fact, if \( s = 1 \) or \( t= 1\), the result is an isosceles APT of the form \( (a, 1, a) \) or \( (1, b, b) \), respectively.

In the proofs that follow, I adopt the convention that \( x^{-1}\pmod 1 = 0\) for all \( x \).

**Proposition**:

- If \(s\) and \(t\) are positive integers with \(\gcd(s,t)=1\)
- Let \(s’ = s^{-1}\pmod t\) and \(t’ = t^{-1}\pmod s\)
- If:
- \(a = st -s’t’\)
- \(b = ss’ + tt’\)
- \(c = st + s’t’\)

- Then \(a^2 + b^2 = c^2 + 1\).

**Proof**:

Note that for given coprime \(s\) and \(t\):

- \(ss’ = \lambda t + 1\) for some integer \(\lambda\)
- \(tt’ = \mu s – 1\) for some integer \(\mu\).

By inspection, \(\lambda = t’\) and \(\mu = s’\); this gives \(ss’ = tt’ +1\).

Now, let’s show that this gives an APT.

Consider \(c^2 – a^2 = (st + s’t’)^2 – (st – s’t’)^2\). By the difference of two squares, this works out to be \(4sts’t’\).

Similarly, \(b^2 – 1 = (ss’ + tt’)^2 – 1\). By the difference of two squares, this is \((ss’ + tt’ – 1)(ss’ + tt’ + 1)\).

However, \((ss’ + tt’ – 1) = 2tt’\) and \((ss’ + tt’ + 1) = 2ss’\), so \(b^2 -1 = 4sts’t’\).

So, \(c^2 – a^2 = b^2- 1\), which implies \(a^2 + b^2 = c^2 + 1\) as required. ∎

**Proposition**: For every triple of positive integers \((a,b,c)\) such that \(a^2 + b^2 = c^2 + 1\), where \(b\) is odd, there exist two positive integers \(s\) and \(t\) such that the algorithm outlined above produces the triple.

**Proof**:

Note that \(\frac{c-a}{2}\times\frac{c+a}{2} = \frac{b-1}{2}\times\frac{b+1}{2}\), so every factor on the left hand side must also appear on the right hand side.

Let:

- \(s = \gcd\left(\frac{b-1}{2},\frac{c+a}{2} \right)\)
- \(t = \gcd\left(\frac{b+1}{2},\frac{c+a}{2} \right)\)

Note that \(\gcd(\frac{b+1}{2},\frac{b-1}{2})=1\) , and every factor of \(\frac{c+a}{2}\) must be a factor of either \(\frac{b-1}{2}\) or \(\frac{b+1}{2}\), it must be the case that \(\frac{c+a}{2} = st\).

Let:

- \(p = \gcd(\frac{b-1}{2}, \frac{c-a}{2})\)
- \(q = \gcd(\frac{b+1}{2},\frac{c-a}{2})\)

By similar reasoning, \(pq = \frac{c-a}{2}\).

- Since \(\frac{c-a}{2} < \frac{c+a}{2}\), \(pq < st\).
- Since \(\frac{c-a}{2}\times\frac{c+a}{2} = pqst\), I can say \(\frac{b-1}{2}\times\frac{b+1}{2} = pqst\)
- So \(\frac{b-1}{2} = ps\) and \( \frac{b+1}{2}=qt\).

This means \(qt – ps = 1\) with \(pq < st\). The only solution to this in positive integers is \(p = s^{-1}\pmod t\) and \(q=t^{-1}\pmod s\).

Therefore, for any \((a,b,c)\) such that \(a^2 + b^2 = c^2 + 1\), there exists at least one \((s,t)\) pair for which the algorithm generates the APT. ∎

Suppose an APT \((a,b,c)\) is generated by applying the algorithm to coprime integers \((x,y)\) and to coprime integers \((X,Y)\).

This implies:

- \(\frac{c+a}{2} = xy = XY\)
- \(\frac{c-a}{2} = x’y’ = X’Y’\)
- \(\frac{b-1}{2} = xx’ = XX’\)
- \(\frac{b+1}{2} = yy’ = YY’\)

Note that \(yy’ – xx’ = 1\) [*].

Let \(p\) and \(q\) be such that \(X = \frac{p}{q}x\). This implies that \(q\) is a factor of \(x\).

Since \(XY = xy\), \(Y = \frac{q}{p}y\), so \(p\) is a factor of \(y\).

Since \(YY’ = yy’\), \(Y’ = \frac{p}{q}y’\), so \(q\) is a factor of \(y’\). However, since \(\gcd(x,y’)=1\) (by [*]), this implies \(q=1\).

Since \(X’Y’ = x’y’\), \(X’ = \frac{q}{p}x’\), so \(p\) is a factor of \(x’\). However, since \(\gcd(x’,y)=1 \), this implies \(p=1\).

Therefore, if \((x,y)\) and \((X,Y)\) generate the same APT, \(x=X\) and \(y=Y\). ∎

We’ve shown that the triple \((a,b,c)\) is only generated once. However, an odd scalene APT is also generated as \((b,a,c)\); odd scalene APTs are thus generated twice by this method.

I won’t work through the details here, but if an odd APT \((a,b,c)\) is generated by \((s,t)\), its counterpart \((b,a,c)\) is generated by \((S,T)\), where \(S = \gcd(\frac{1}{2}(s+t’)(t+s’), \frac{1}{2}(s-t’)(t+s’))\) and \(T= \gcd(\frac{1}{2}(s+t’)(t-s’), \frac{1}{2}(s-t’)(t-s’))\).

If we care about this, we can place restrictions on our inputs to insist that \(b>a\) if \(a\) is odd.

For \(b \ge a\), we require \(ss’ + tt’ \ge st – s’t’\).

I’m not aware of a simple restriction on \(s\) and \(t\) that satisfies this, although \((s+t’)(s’+t) \ge 2st\) is moderately pleasing as these things go.

Finding a nicer restriction here is left as an exercise for the interested reader.

Instead, we can simply reject any APTs where \(b < a\).

Since (1) every pair of positive coprime integers \((s,t)\) generates an APT, (2) every APT \((a,b,c)\) is generated by at least one seed, and (3) under the restriction every APT is generated by exactly one seed, there’s a bijection between (restricted) coprime integer pairs and almost Pythagorean triples.

As a result, by looping over the coprime positive integers and applying the given restriction, each possible APT is generated exactly once.

In this article, I’ve answered Antalan and Tomenes’s question: yes, there is a simple method to generate almost Pythagorean triples. It’s not clear to me that this is a new result (I haven’t been able to find any more on the topic, but I don’t know if that’s because (a) it’s not been done, (b) it’s been done and I’m looking in the wrong places, or (c) it’s been done and considered too trivial to be worth writing down.)

I haven’t answered Chris’s original question here, which is whether you can find a set of three triangular numbers \(\{x,y,z\}\) such that \(x+y\), \(x+z\), \(y+z\) and \(x+y+z\) are all triangular. The answer to that is also yes, but that’s for another post.

Neither have I explained the link between the puzzle and APTs. I’ll happily send a postcard to the first reader who figures it out!

Nerdsniping: the gift that keeps on giving.

This article refers to:

- Antalan, J.R.M. and Tomenes, M.D.:
**A Note on Generating Almost Pythagorean Triples**, International Journal Of Mathematics And Scientific Computing, Vol. 5, No. 2, 2015, pp. 100-102 - Frink, O.:
**Almost Pythagorean Triples**, Mathematics Magazine, Vol. 60, No. 4, October 1987, pp. 234-236

I’m grateful to Chris Smith, Dominika Vasilkova and Elizabeth A. Williams for their help with this article.

]]>A conversation about mathematics inspired by a balancing bird. Presented by Katie Steckles and Peter Rowlett, with special guest Alom Shaha.

]]>My son was born last September. While he doesn’t hate sleep as much as his sister did, he still needs a bit of help to drop off.

I’m not at all musically inclined, and I seem unable to remember more than a couple of lines from wordy songs (my version of “Papa’s gonna buy you a mockingbird” rapidly veers into nonsense as I try to think of a rhyme for the next line), so when the girl was little I hit upon the strategy of singing counting songs to lull her off to sleep.

At first, I went through the standards: “one man went to mow”, “ten green bottles on the wall”, and so on. The first problem I encountered is that when you start singing a counting-backwards song, you have to pick a number to start at.

The second problem is that they’re really, really boring, and it’s the baby I want to go to sleep, not me.

One day my wife the primary teacher taught me a song she uses at school:

Let’s count back in ones from 10,

Cock-a-doodle-doodle-doo,

10, 9, 8 and 7,

Cock-a-doodle-doodle-doo.

6, 5, 4 and 3,

Cock-a-doodle-doodle-doo,

2 and 1 and don’t forget the zero,

Cock-a-doodle-doodle-doo!

It’s another counting backwards song, but the great thing about it is that once you’re finished, you start again counting back in twos, then threes, and so on for as long as you need until the baby is asleep.

Let’s count back in \(N\)s from \(10N\),

Cock-a-doodle-doodle-doo,

\(10N\), \(9N\), \(8N\), \(7N\), …

My record for the girl is “let’s count back in twenty-twos”. I did say she doesn’t like going to sleep.

Eventually, I got bored of that too, so I started counting back in different sequences: squares, triangular numbers, fibonacci numbers, primes, binary, ternary.

Let’s count back in fibonacci numbers from 89,

Cock-a-doodle-doodle-doo,

89, 55, 34, 21,

Cock-a-doodle-doodle-doo.

13, 8, 5 and 3,

Cock-a-doodle-doodle-doo,

2 and 1 and don’t forget the other 1,

Cock-a-doodle-doodle-doo!

These didn’t quite work: if I take a while to work out the next number it disrupts the rhythm of the song. It’s also very easy to make an off-by-one error: while the normal song starts at \(10N\), you sing 11 numbers including the zero (which you mustn’t forget!)

So I wanted a counting-forwards song that had a bit of complexity to it, but not too much.

A few months ago, while pushing the boy around the block, I came up with a nice rule for an integer sequence:

For each \(N\), repeatedly subtract the largest power of 2 dividing \(N\) until you get to 0 (but don’t include the zero!)

It goes like this:

1

2

3, 2

4

5, 4

6, 4

7, 6, 4

8

9, 8

…

This works beautifully! It’s open-ended, so I can keep going if he takes a while to drop off, but the rule is just hard enough to keep my brain occupied without making me break the rhythm.

He’s normally asleep by the time I get to 64, but sometimes I keep going just for fun.

Last night he had a horrible time with teeth coming through, so he was ready for his first nap at 6am! I recorded myself singing the lullaby sequence as I pushed him around the block:

Of course, as a fan of the Online Encyclopedia of Integer Sequences, I checked if it was already in. It wasn’t! So I submitted it, and it’s now entry A343934.

]]>My aim is to collect examples of conventions in mathematical notation that lead to ambiguities, inconsistencies, or just make you feel yucky. This is largely a result of me wishing I had something to point to whenever I see $\sin^2$ or one of those viral “puzzles” relying on BODMAS.

I’ve set it up as a wiki, so you can edit it too. So far, it’s got a few dozen pages on some of the things that have troubled me over the years, along with some contributions from some of my maths pals, such as “There is no function application symbol”, “Parentheses are overused”, and “Juxtaposition means combine in the obvious way”.

My aim is to describe conventions, without prescribing a correct notation. Whenever I tweet a question about a notational convention, my aim is to find out the range of different opinions that people hold about it. I often get replies arguing authoritatively for a particular correct answer, usually followed up by an equally certain reply from someone else arguing for the opposite. Like all language, mathematical notation is just something we make up to help express our ideas, and opinions, abuses of notation, lapses in memory and convenience all work against consistency and clarity.

I’d like the site to collect all these difficult aspects of notation, so that they don’t trip up someone who thought they might have an easy day doing maths.

So, have a look, and if you can help to build it out, I’d be very happy!

]]>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.

]]>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.

]]>- Fields Medalist Peter Scholze has announced that his Liquid Tensor Experiment, a problem announced six months ago in a blog post that concerns condensed Abelian groups, has now reached the stage where proof assistance software has formally verified the hard part. Exciting!
*(via Jordan Ellenberg)*

- YouTuber and PhD physicist Derek Muller (Veritasium) has recently been involved in a physics-off with UCLA professor Alexander Kusenko, when they disagreed over the explanation behind a physical phenomenon, which escalated to a $10,000 bet over who was right. Long story short, Veritasium won the bet (as covered in this IFLScience news story) and will be using the money to fund a science communication contest. If you’ve got an under-a-minute maths/science video you can post on YouTube or TikTok, you could win a prize of up to $5,000. Props to Derek for encouraging more STEM communication and promoting new talent!

- It’s been formally announced that Neil Sloane is stepping down as president of the OEIS – Russ Cox will take over presidential duties, while Sloane steps down to Chairman of the Board so he can dedicate more time to his writing projects (which we’re assured ‘naturally involve sequences’). Cox has been involved with the OEIS for over 25 years and has been a major contributor to the backend software that makes the site run, so he’s a safe pair of hands to take the project on.

- The eleven 2021 LMS Prize winners were announced at the Society’s Meeting on 2nd July, and the prizes recognise contributions to mathematics in a variety of areas. (via @LondMathSoc)

On 23rd June the new Alan Turing £50 note was launched, featuring an image of Turing, a quote and various mathematical diagrams. Bletchley Park marked the occasion with a #Turing50Takeover, and the Bank of England has a whole page of info about the new polymer note on their website.

Meanwhile, in Turing-adjacent news, the National Museum of Computing has launched an online Virtual Enigma machine you can use to simulate the device behind the famous Enigma code, along with a video explaining the machine. This joins a host of other virtual historical computers they’ve built, including the Colossus that cracked the code, the Lorenz machine and even ERNIE the random number generator!

Another claimed proof of the Riemann Hypothesis, this time by Kumar Easwaran, emerged this month, and since like all big claims it would need thorough checking before acceptance by the mathematical community, there was some initial skepticism. (This didn’t stop the media from latching on to it as an exciting story though). Since claimed proofs of Riemann are like buses, many mathematicians don’t give them much attention, but Alex Kontorovich took the time to thoroughly debunk this one to save you the trouble.

If you want some actual Riemann Hypothesis news, here’s some: John Baez reports that Alain Connes and Caterina Consani have made some potential progress on part of the problem. In the words of Baez, “my interest is piqued”.

Thuses is a website for mathematicians to publish and discuss ideas of interest to the mathematical community. It’s described as “a perfect place to share new approaches, slick proofs, and surprising counterexamples. A place for ‘folklore results’ that are considered known but don’t actually exist in literature. A place for everything in math that just has to be shared.” *(via Piper H)*

The Royal Society has published a set of papers on modelling that shaped the early COVID-19 pandemic response in the UK as a special journal issue that’s free to access.

The BSHM (British Society for the History of Mathematics) has launched the Bibby Awards in the History of Mathematics, for “contributions to the popularization of the history of mathematics in education”. Named after (and funded by the legacy of) the late BSHM member Neil Bibby, up to four awards of £400 can be made each academic year, in return for which holders are expected to give two free talks in schools and produce four digital resources (videos, PDFs or interactives) for the BSHM website. *(via Sarah Hart)*

The Isaac Newton Institute in Cambridge is hosting The unity of mathematics: A conference in honour of Sir Michael Atiyah which will take place in September 2021 as a hybrid event with a mixture of in-person and virtual talks. The closing date for registration for physical participants is 8th August.

There’s just about still time to register for the People, Places, Practices History of Maths Conference (registration closes 9th July) taking place 12-15 July online (coordinated by the University of St. Andrews). With around 90 speakers contributing, the programme looks packed, and talks will be available to watch ahead, or at the specified time to be followed by a live Q&A.

Alexandre Borovik reports on his blog about Azat Miftakhov day, an event organised online by the Azat Miftakhov committee in solidarity with Azat Miftakhov – a graduate student from Moscow State University who was sentenced to six years in a medium-security penal colony and has already been arbitrarily detained by Russian state authorities for almost two and a half years. Fields medalist Cedric Villani made a speech at the event, and you can watch videos from the event on YouTube.

]]>A conversation about mathematics inspired by UUID 0412a969-5b27-4c28-9662-85ef2c201e0c. Presented by Katie Steckles and Peter Rowlett.

]]>A conversation about mathematics inspired by an auctioneer’s hammer. Presented by Katie Steckles and Peter Rowlett, with special guest Tim Harford.

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