Hi, Dryden, I think we know by now from previous replies that all the grid entries are primes of the form 4n+1. The mathematician Pierre de Fermat showed that any odd prime p can be expressed as the sum of two squares of positive integers if and only if p = 4n+1.
For the endgame, I checked whether any of the 20 9-digit numbers in the grid was divisible by 3 by calculating the digit sum of each of the numbers. If the digit sum is divisible by 3, then so is the number itself. In all of the symmetric pairs of 9-digit numbers, at least one of the pair was divisible by 3, and hence not a prime, except for the two numbers on the diagonals, which both had to be read downwards to satisfy the common feature.
Then came the slog. I took the smaller diagonal number and worked out the integer closest to the square root of half the number. I then took every 5-digit number from 10,000 to the integer I’d just calculated, and set up an Excel spreadsheet to calculate the square root of (the 9-digit number less the square of each 5-digit number) in my range, then inspected the results to look for the only answer that was itself an integer. Once I’d found the two 5-digit numbers in the grid, it was clear what the other pair of 5-digit numbers was, and I checked that the sum of their squares gave the 9-digit number on the other diagonal.
I’d been using a website for factorisation and checking whether numbers were primes, but the sum of squares website referred to in earlier replies would have saved a lot of tedious work on Excel. It will be interesting to see when the solution comes out if there was a more straightforward way of doing this puzzle than a lot of Excel calculations.