Season 9 Episode 1
We've got some puzzles to solve in this episode, and a problem-solving strategy that might work for other puzzles too! Tamio brings insight from the Oxford Invariants society.

Further Reading
Game Night
The Jane Street puzzle that we were working on can be found here. Other puzzles are posted here. Some require a computer.
I also mentioned Project Euler which is an enormous collection of maths problems that require (1) mathematical insight (2) computational insight.
You might also like this episode of OOMC from last year where we discussed another multi-layered Jane Street puzzle; watch on YouTube.
Solitaire on a square board
I didn’t explain this very well live, so I’m going to have another go here!
There’s a classic game called Solitaire (or sometimes peg solitaire) that involves the sort of jumping move that we were investigating in this episode. Diagonal moves are not allowed on the square board.
For solitaire games on a square board, there’s an invariant that works like this; paint two out of every three diagonals on the board yellow. Now a move might start with two pegs on yellow squares, with one jumping over the other onto a blank square. Or maybe one of the original pieces was on a blank square and one was on a yellow square, with the move leaving us with a piece that’s on a yellow square.
Write $Y$ for the number of pieces on yellow squares. This number changes during the moves, but the odd-ness or even-ness (the "parity") of this number does not change, because in the first sort of move we lose two pieces on yellow squares, and in the second sort of move we lost a piece from a yellow square but gained a piece on a different yellow square.
I then showed you a way to paint squares blue instead, with the diagonal stripes offset by one. Crucially, I should have pointed out that yellow plus blue makes green, for the squares where the lenses overlap (the green squares count as yellow AND blue).
This is quite powerful. Let’s say you start with some complicated arrangement of pieces, and you want to end up with just one piece after all the jumping is done. Maybe you look at the puzzle with your yellow lens and see that 36 (an even number) of the pieces are on yellow squares. Since the parity is invariant, the final arrangement with one piece must have that piece NOT on a yellow square (zero is even, but one is odd!). If you then look at the puzzle with your blue lens and find that 24 (an even number) of the pieces are on blue squares, then you know that this hypothetical single final piece is not on a blue square either. Uh oh! All squares are either yellow or blue or both (green counts as both). So there is nowhere for this hypothetical final piece to be. So it must be impossible to end up with just one piece!
Isn’t that powerful? I didn’t need to tell you anything about where the pieces actually were or think about any of the possible sequences of jumps, and yet we could still prove that it is impossible to end up with a single piece.
Hexagonal grid
The snowflake puzzle uses a hexagonal grid instead of a square grid, but there’s a way to colour the spaces so that the same properties hold. The yellow spaces lie on the boundaries of tessellating hexagons, with each hexagon aligned with the grid. If you shift that by one unit for the blue spaces, and remember that yellow plus blue makes green, then you get this arrangement.

For our puzzle, we can count how many pieces are on yellow spaces and how many are on blue spaces.
In the live episode, I counted the pieces on green spaces separately, but Tamio corrected me, because the green spaces also count as yellow (and blue, as above, because by this point I’d forgotten that yellow plus blue makes green). We started with an even number of pieces on yellow spaces and an even number of pieces on blue spaces. Uh oh. As above, there’s no way for a single final piece to sit on an even number (zero?!) of yellow spaces and on an even number of blue spaces. So there's no way to be left with a single piece.
We didn't answer the following question in the live episode; is it possible to be left with just two pieces?
The answer might be yes (and maybe you could find a sequence of jumps that leaves you with two pieces). Or it might be no (and maybe you could find a proof that two pieces is impossible).
Lenses
Here’s one more problem that you can solve by looking at it through two different lenses.
Suppose $p>3$ is prime. Prove that $p^2-1$ is a multiple of $12$.
Hint: the lenses are something to do with 2 and something to do with 3.
Follow-up: Suppose $p>3$ is prime. Prove that $p^2-1$ is a multiple of $24$.
Invariants
The Art of Problem Solving website has another example of a problem that you can solve by spotting an appropriate invariant (something that doesn’t change during the process);
The numbers 1 to 10 are written on a blackboard. Evan is allowed to erase any three numbers $a$, $b$, and $c$ and replace them with $\sqrt{a^2+b^2+c^2}$. What is the greatest number that can appear on the board at any given point?
Hints: you might like to think about how this quantity is related to $a$ and $b$ and $c$. Separately, you should consider how many numbers are on the board; ten to start with... how many after Evan has done a few of these swaps?
The Invariants
If you'd like an idea of what the Oxford Invariants student society are up to, see their website here. Student societies are an important part of university life; they’re usually organised by students, for students, based on what they want to do. Oxford has many many other societies. There’s a partial list here but that's just the ones registered with the central university. There are also college societies and informal societies not registered on that list.
If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.