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One of the easier number puzzles and I have filled the grid quite quickly. However I must have made a mistake somewhere since I can only find one of the 9-digit numbers. Since this is symmetrically placed I think I know where to look for my mistake, but so far no luck. I have 9a as 85237 and 11a as 2309. If these are both correct then I will finish now. I don't have the patience to unpick all of my calculations.

I have just completed the grid and have the same for 9a and 11a as you.
Haven't found the 9 digit numbers yet. About to look.

catkin, those are correct. If you have one of the 9 digit numbers, you should be able to find one of the two pairs of positive integers. By inference that will show you where the other pair is, and as long as your error does not affect those cells, you might be able to derive the other 9 digit answer that way. Just a thought

I think I have the 9-digit numbers, on diagonals. It was difficult to be sure without first checking every row and column, but none of them shared the common feature.

So then I began looking for a 5-digit number in the mid- to upper-20,000s, hoping to add onto its square the relatively small square from its adjoining 4-digit number. But I've had no luck with this. Am I wrong to assume that the integer pairs are in the same row or column ?

I cannot find them either. Does one of the "diagonal numbers" begin 61798? At least I would be on the right rack.

I think your third digit should be 5. The others are correct.
Still looking for them myself.

You can eliminate all the 9-digit numbers formed by the rows and the columns because in every symmetric pair of numbers, at least one is divisible by 3 (including the pair formed by the middle row and column that form a cross). So the two 9-digit numbers we are looking for must be the two main diagonals.

I'm afraid I didn't enjoy this because it was largely a slog through the same set of calculations for each of the clues. A cleverly designed grid but not so much fun for solvers.

Beulah and Jazzlover42, I think that the 3rd digit must be 7.
Otherwise 10A would need to be 10585 and 9D would need to be 85. Both of which then would be divisible by 5.

I've been looking for a 5-digit number in the mid- to upper-20,000s, hoping to add onto its square the relatively small square from its adjoining 4-digit number. But I've had no luck with this. Am I wrong to assume that the integer pairs are in the same row or column ?

Hi wintonian. I just concluded they were the two diagonals too. But now we have to find two pairs of integers whose squares add up to each of these, according to the blurb. Yet more of a slog or is there a quicker route?

Yes, meursault, you are wrong. The whole puzzle has a pleasing symmetry. Use krauton’s link at #12