It is indeed quite complicated. But if you handwavily *estimate* the results of all that complexity -- the probabilities of divisibility by various things -- then the estimate you get is the one cousin_it gave earlier, because the Prime Number Theorem is what you get when you estimate the density of prime numbers by treating divisibility-by-a-prime as a random event. (Which for many purposes works very well.)

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There are 143 primes between 100 and 999. We can, therefore, make 2,924,207 3x3 different squares with 3 horizontal primes. 50,621 of them have all three vertical numbers prime. About 1.7%.

There are 1061 primes between 1000 and 9999. We can, therefore, make 1,267,247,769,841 4x4 different squares with 4 horizontal primes. 406,721,511 of them have all four vertical numbers prime. About 0.032%.

I strongly suspect that this goes to 0, quite rapidly.

How many Sudokus can you get with 9 digit primes horizontally and vertically?

Not a single one. Which is quite obvious when you consider that you can't have a 2, 4, 6, or 8 in the bottom row. But you have to, to have a Sudoku, by the definition.

It's a bit analogous situation here.

Here's another fun argument. The question boils down to "how common are primes?" And the answer is, very common. We can define a subset of positive integers as "small" if the sum of their reciprocals converges, and "large" otherwise. For example, the set of all positive integers is large (because the harmonic series diverges), and the complement of any small set is large. Well, it's possible to prove that the set of all primes is large, while the set of all numbers not containing some digit (say, 7) is small. So once you go far enough from zero, there are *way* more primes than there are numbers not containing 7. Now it doesn't sound as surprising that you can make squares out of primes, does it?

Say, that we have N-1 lines, with N-1 primes. Each N digits. What we now need is an N digit prime number to put it below.

Its most significant digit may be 1, 3, 7 or 9. Otherwise, the leftmost vertical number wouldn't be prime. If the sum of all N-1 other rightmost digits is X, then:

If X mod 3 = 0, then just 1 and 7 are possible, otherwise the leftmost vertical would be divisible by 3. If X mod 3 = 1, then 1, 3, 7 and 9 are possible. If X mod 3 = 2, then just 3 and 9 are possible, otherwise the leftmost vertical would be divisible by 3.

The probability is (1/3)*(((1+2+1)/5))=4/15 that the first digit fits. (4/15)^N, that all N digit fit.

Actually, we must consider the probability of divisibility by 11, which is roughly 1/11, which further reduces 4/15 per number to 40/165. And with 7 ... and so on.

For the divisibility with 3, we render out not only one permutation of N-1 primes but all of them. For the divisibilty with 11, some of them.

It's quite complicated.

Here's another fun argument. The question boils down to "how common are primes?" And the answer is, very common. We can define a subset of positive integers as "small" if the sum of their reciprocals converges, and "large" otherwise. For example, the set of all positive integers is large (because the harmonic series diverges), and the complement of any small set is large. Well, it's possible to prove that the set of all primes is large, while the set of all numbers not containing some digit (say, 7) is small. So once you go far enough from zero, there are *way* more primes than there are numbers not containing 7. Now it doesn't sound as surprising that you can make squares out of primes, does it?

Interesting line of inferring... I am quite aware how dense primes are, but that might not be enough.

I have counted all these 4x4 (decimal) **crossprimes**. There are 913,425,530 of them if leading zeros are allowed. But only 406,721,511 without leading zeros.

if leading zeros ARE allowed, then there are certainly arbitrary large crossprimes out there. But if leading zeros aren't allowed, I am not that sure. I have no proof, of course.

Was this meant to be a reply?

Well, I said something in line with "people may need some stuff to live and declaring that we should "put people before that stuff" is a silly way to present the situation". Maybe not as silly as it's a demagoguery.

But then I changed my mind and decided to not participate in a discussion at all. But somehow couldn't erase this now empty box.

I don't understand why it's silly.

I don't understand why you're comparing business to oxygen.

Lest you fall prey to the fundamental attribution error, I don't agree with the article, and I think a lot of it is applause lights. The core sentiment of humanity first isn't one I subscribe to either (I'm an individualist), but the philosophy behind the article is one I can under and appreciate. It's one I can pass an ideological Turing test for. You seem to be caricaturing the position in the article, and that isn't very epistemically healthy.

tl;dr

But I saw this:

Time to put humans before business.

Time to put humans before oxygen, too? Silly stuff.

I'm pretty sure arbitrarily large squares are possible. Here's an argument that assumes primes behave like random numbers, which is often okay to assume. By the prime number theorem, the chance that an N-digit number is prime is proportional to 1/N. So the chance that N^2 random digits arranged in a square will form 2N primes (N rows and N columns) is about 1/N^(2N). But the number of ways to select N^2 digits is 10^(N^2) which easily overwhelms that.

The bottom and the rightmost prime can both have only odd digits without 5. The probability for each prime to fit there is then only (2/5)^N times that. We can't see them as independent random numbers.

No, it gives plenty of non-symmetric solutions as well. Here's one:

8 1 8 9 9 1 4 3

6 5 4 0 4 9 0 3

6 0 2 7 2 9 5 1

8 2 7 8 7 4 1 3

3 9 2 9 3 8 6 1

2 4 7 8 9 7 6 7

2 2 0 6 6 3 3 7

9 9 7 3 7 9 3 3

Very well. What do you think, are there arbitrary large squares possible or not?

I think not. Even in binary notations NxN and above, probably don't exist for an N, large enough.

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A problem.