Ponder This Challenge - July 2026 - Return of the Superheroes
- Ponder This
This puzzle was suggested by Hugo Pfoertner - thanks Hugo!
In a now-famous 2004 article, Ben Green and Terence Tao proved that arbitrarily long arithmetic progressions exist in the primes. This holds true analogously for other sets of numbers if their density is sufficiently high, for example, numbers that are the sum of two squares. There are also enough numbers of the form x² + y² + x*y (with x,y being integers), called "Loesch numbers," that arbitrarily long arithmetic progressions can be found among them as well.
Example: The first Loesch numbers are 0, 1, 3, 4, 7, 9, 12, 13, 16, 19, 21, ..., and the numbers 1, 7, 13, 19 form an arithmetic progression with 4 terms and a difference of 6. This progression can be described by its initial value of 1 and the step size of 6.
Your goal: Find an arithmetic progression of 35 terms from the Loesch numbers such that the end of the progression becomes as small as possible, specifying the starting value and the step size.
A bonus "*" will be given for finding a progression with at least 42 terms, again specified by its starting value and step size (the end of the progression needs not be as small as possible).
An extra bonus "**" will be awarded to the participant(s) who find the progression with the most (>42) terms. The allocation of (**) to participants may change during the submission process.
The numerical solution for the case is
a = 219830911
d = 2709630
The best sequence, with , found by Hubert Puszklewicz and later independently by Daniel Bitin, is
a = 1456385415508399
d = 2639156742851970
A simpler solution achieving (actually, ) is
a = 43799686129
d = 8236888110
This is mostly a straight up search problem, relying on clever programming and theoretical tricks to speed up the search (and I do not know the tricks used by the top solvers; my own solution was far worse).
One example of a theoretical trick is that in our length 35 sequence, must be divisible by . Let's see why.
First of all, a known property of Löschian numbers is that if is a prime divisor of a Löschian number such that , then the power of in the prime factorization of must be even, and in particular whenever is a prime divisor of .
Now, suppose we have such that . Consider the linear modular equation . Since and are relatively prime, this equation has a unique solution, namely . Hence, looking at the sequence
We obtain a number divisible by - and so, by - at least once every steps. Take the first steps, and we obtain such that , meaning , and since we have which is impossible after only the first steps. Hence, if we consider a sequence of length 35, and we have such that , we must have . This criterion holds for .
A simpler argument shows that also : Consider . By going over all combinations of possible it is clear that , so in a progression of length 3 there are two elements equivalent modulo 3, and the same argument as before shows that . Hence must divide , reducing the search space size.
The ** holders were, in reverse order of appearance with their original, **-worthy, record: