Listener Crossword No 4308 Sub-Prime More-Guess Relief By Ruslan

[BobHWW]

A fairly easy solve in contrast to the rather complicated preamble. I haven't tried to work out the title yet!

Farewell Ruslan - you will be missed.

[Postigod]

A very nicely constructed puzzle, so many thanks Ruslan. I agree this was on the simple side as Listener numericals go (yet still very fun). It could have been made more difficult (and still a unique solution, I believe) without the extra hint for 12A (as one has three possible pairs of simultaneous equations for u, v, only one of which has integer solutions).

[TheBear69]

The proof of the 1st sentence of the preamble is trivial.

Given a positive integer n, let k = n mod 6; n=6m+k for some m.

k=0 => n=6m is never prime
k=2 => n=6m+2=2(3m+1) is prime only if m=0
k=3 => n=6m+3=3(2m+1) is prime only if m=0
k=4 => n=6m+4=2(3m+2) is never prime

So other than 2 and 3 all primes are either 6m+1 or 6m+5 = 6(m+1)-1.
Of course the converse is NOT true; plenty of 6m+1's are not prime (e.g. 25).

[British_Olympics]

Even more simply, every prime number greater than 3 is odd (so every adjacent number (+1 or -1) has to be divisible by 2) and not divisible by 3 (so one adjacent number must be), therefore one adjacent number is divisible by (2 x 3) = 6.

[olichant]

Thank you, some very compact suggestions! Which have made me idly wonder if there's any milage for a competition to reduce famous mathematical proofs to a tweet. Other than the obvious "I have discovered a truly wonderful proof, which this tweet is too short to record". (Sorry, a bit off topic perhaps).