You want to know one of my favourite things about maths? It's the connections. Connections between things that don't, on the face of it, seem remotely connected. One just cropped up, and made me grin from ear to ear, so I thought I'd share it with you.
One of @standupmaths's favourite famous puzzles is the Magic Square Of Squares: can you place nine distinct square numbers in a three-by-three grid such that the diagonals, rows and columns all sum to the same number?
This is (to the best of my knowledge) an open problem: nobody has so far found an example, or a proof that no such magic square exists. (If one does, the numbers in it must be enormous).
I had a dabble with it, a few months ago; even though my level of number theory is almost certainly not enough to catch something nobody else has spotted, it's always good to give the muscles a bit of an exercise.
One of the rabbit-holes the puzzle led me down was the search for square numbers in arithmetic sequence -- for example, 1, 25 and 49 have a common difference of 24. (I believe there is a theorem that the longest run of square in arithmetic series is three, but that there are infinitely many such triples). As I remember, the search for a MSoS is equivalent to finding three triples of squares in arithmetic series, all with the same common difference - just in case you wanted to carry on the exploration from where I left off.
Some time later, this horror-show of a tweet landed in my timeline:
— ZedTeach (@ZedTeach) October 7, 2017
(In fairness to @zedteach, there is a proviso that Demi is just looking up to 100 - but that "always" really bugs me.)
Of course, my loyal band of followers were quick to point out counterexamples -- 120 (for which $119=7\times 17$ and $121 = 11^2$) was the first to spring to my mind, and 144 (one more than $11\times 13$, and one fewer than $29\times 5$) was a common answer.
Several good puzzles came out of this: @chrismaslanka wondered if there exists $k$ such that both $6k-1$ and $6k+1$ have at least three prime factors, which I'll leave you to explore.
In exploring it, @sxpmaths (Stuart Price) made the observation that $6n^2 + 5n - 1$ and $6n^2 + 5n + 1$ both factorise (as $(6n-1)(n+1)$ and $(3n+1)(2n+1)$, respectively), so making $n = 6m$ generates infinitely many multiples of six with neighbouring composites (although 120 and 144 are not caught by this.)
So I started investigating quadratics of the form $ax^2 + bx \pm 1$ to see whether they factorise. Quadratics of the form $Ax^2 + Bx + C$ factorise if $B^2 - 4AC$ is square, so we're looking for $a$ and $b$ such that $b^2 + 4a$ and $b^2 - 4a$ are both squares... oh look! Three squares in arithmetic sequence.
I haven't found the quadratic that corresponds to 120 yet - and part of me says "don't bother. You've found a Deeper Connection! Leave the trivial arithmetic for others."
Of course, other parts of my brain won't rest until I do find it... or one of you, loyal readers, tells me what it is.
* Thanks to @grey_matter for spotting that 119 is not, in fact, $7 \times 13$.