5
$\begingroup$

Define the upper density of a set $A \subseteq \mathbf{N}$ to be

$$\bar{d}(A) = \limsup_{n \to \infty} \frac{|A \cap [1,n]|}{n}.$$

Let $A$ be the set of sums of two squares, i.e. $A = \{x^2 + y^2 : x,y \in \mathbf{Z}\}$.

I know that any prime congruent to 1 modulo 4 is the sum of two squares, as is the number $2$ itself. Also, for primes congruent to 3 modulo 4, I know that $p^{2n} = x^2+y^2$ for some $x,y$. Hence, since sums of two squares are closed under multiplication, any number, n, of the form:

$$n = 2^{\alpha} (\prod_{i \leq m_1} p_i^{\beta_i})^{2} \prod_{j \leq m_2} q_j^{\gamma_j},$$ Where $\alpha, \beta_i, \gamma_i, m_1, m_2$ are non-negative integers, $p_i$ is a prime congruent to 3 modulo 4, and $q_j$ is a prime congruent to 1 modulo 4.

I'm having difficulty passing to an argument about upper density about the set of all such $n$, though.

$\endgroup$
2
  • $\begingroup$ Modulo some prime-power/squarefreeness considerations, what you want to show is that almost all odd numbers have at least one prime $\equiv 3\pmod 4$ in their factorization. Heuristically, since the primes are equidistributed mod $4$, you're showing that the set of 'heads-only' sequences has density zero in the set of all coinflip sequences. $\endgroup$ – Steven Stadnicki Oct 21 '13 at 21:56
  • $\begingroup$ Also, your text does not make much sense, although the big formula is correct: some $n > 0$ is the sum of two squares if and only if the exponent for any prime divisor $p \equiv 3 \pmod 4$ is even. The true bound is this : the count up to large positive $x$ of sums of two squares is about $$ \frac{0.7642 \; x}{ \sqrt {\log x}} $$ $\endgroup$ – Will Jagy Oct 21 '13 at 22:04
5
$\begingroup$

Only primes $\equiv 3\pmod 4$ are "obstacles" against $n$ being the sum of two squares. For each such prime $p$ at least those numbers $n\equiv kp\pmod {p^2}$, $1\le k<p$, are not the sum of two squares. This alone would leave us with a density of $$\prod_{p\equiv 3\pmod 4}\left(1-\frac{p-1}{p^2}\right).$$ However, we neclected higher powers of $p$, i.e. we should additionally filter out $kp^3\pmod{p^4}$, $1\le k<p$, and so on, which leads to $$\prod_{p\equiv 3\pmod 4}\left(1-\frac{p-1}{p^2}-\frac{p-1}{p^4}-\frac{p-1}{p^6}-\ldots\right)=\prod_{p\equiv 3\pmod 4}\left(1-\frac{1}{p+1}\right).$$ The product does not converge (that is: the sequnce of partial products tends to zero) and therefore the density is zero. How can we see that thr product diverges? The reciprocal is $\prod_{p\equiv 3}\left(1+\frac 1p\right)>\sum_{p\equiv 3}\frac 1p$. As far as I know, at least $\sum_p\frac 1p$ diverges and by the "equidistribution" of primes modulo $4$, so should $\sum_{p\equiv 3}\frac 1p$.

$\endgroup$
2
  • $\begingroup$ This makes sense. Thank you for your answer. $\endgroup$ – mojambo Oct 21 '13 at 22:17
  • $\begingroup$ It is true that $\sum_{p \equiv 3}1/p=\infty$.It is a difficult result.It is possible to solve this Q without it. $\endgroup$ – DanielWainfleet Sep 30 '15 at 5:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.