There are between 1 and N about perfect squares, which makes this set a quite sparse set, of density . Given that , it could be reasonable to think that , with some luck, cover all of or at least have positive density. In fact it’s not the case.

To estimate the size of , notice first that it contains all the primes congruent to 1 mod 4, which is a well-known theorem of Fermat, which is about one half of the primes, so that . This already a huge lot bigger than the squares alone.

In fact there are much more sums of two squares than these primes, because all their products are also sums of two squares. In fact, it contains all the products of such primes. A similar heuristics to the one leading to the prime number theorem suggests that the density of the numbers having no prime factor congruent to 3 mod 4 should be

, and the log of this product should be roughly

so we get a density . While this indeed is no proof, it should be believable. In general, a number n is a sum of two squares if and only if it is of the form where a is product of primes congruent to 1 mod 4 and b is a product of primes congruent to 3 mod 4. Thus, the total number of sums of two squares seem to be bounded above (using the union bound) by something like

where the denotes a sum restricted over number whose prime factors are all 3 mod 4. But this should be of the order of

and the sum is bounded, so certainly . In fact, Landau has proven in 1908 that the number of sums of two squares until capital N is asymptotic to

which fits quite well with what the heuristics indicate (the product over the restricted primes corresponds perfectly to the restricted sum).

**Repartition of the sums of squares in residue classes**

We see that the primes are analogous to the sums of two squares: indeed, the primes are the numbers which have no factor, and the sums of two squares are the numbers which have no factor congruent to 3 mod 4 (plus other ones of course, the products of such numbers by 9, 49, 121 etc. of which we saw that it simply multiplied the size of the set by a constant). The same heuristic provides a density which then turns out to be correct. So why not, as Dirichlet did for the primes, ask for the repartition of the sums of squares in residue classes?

A sum of two squares can only be congruent to 0,1 or 2 mod 4, because squares are 0 or 1 mod 4. Let’s show that these values are not equiprobable in fact. First notice that can be written as where is also a sum of two squares. So given the number of until which are sums of two squares, we get and . At least, the probabilities of being even or odd are the same. Another instance is that the sums of two squares are twice as dense in than in . Also notice that everybody in has so nobody there can be a sum of two squares. The final word is by Prachar. Putting the constant appearing above, he found

with

when ; thus the density is the same as the general density in APs, except when the modulus is divisible by 4 (which causes a multiplication by 2 of the density) and/or it has factors congruent to 3 mod 4.

**Additive quadruples in the squares**

Recall that by Cauchy-Schwarz, where is the number of additive quadruples, i.e. of quadruples of square with , which is also the number of quadruples with . Thus, we have at least additive quadruples.

We can try to do better. Indeed, rewrite the equation as

(we replaced by N here). It amounts to

We’ll count the positive solutions here, the other solutions are easily obtained from them. We partition such quadruples depending on the gcd . So . Then for some . So the number of add quadruples should be

By Cauchy-Schwarz, the inner sum ought to be

Now the second sum is not much bigger than

so the total number of additive quadruples should be

. Returning to the previous definition, we have the inequalities

Notice that in a random set of this density you would expect less solutions: of the order of magnitude of N. So if are squares, has increased probability to be a square.

Anyway the heuristics that for of density implies that that which is not the case here.

**Sums of two squares in **

We know that in , there are squares (including 0; here is a prime). This suffices to show, as a simple case of Cauchy-Davenport, that everybody is congruent mod p to a sum of two squares. This of course doesn’t extend to non-prime , as for instance not everybody is a sum of two squares modulo 4. Let us denote by the number of representations of a as sum of two squares mod d. Then we can see that for a prime congruent to 1 mod 4 (so that is a square mod p), , while for any other residue , we must have . On the other hand for , and hence for all other .

By the Chinese Remainder theorem, we know that . But it is far from obvious to estimate the . In fact it happens that

where is the only non-trivial character mod 4, which is also Legendre symbol mod 4.

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