2
$\begingroup$

Let $n$ be the product of $k$ distinct prime numbers of the form $4m+1$.

How can I prove that the number of solutions $n=a^2+b^2$ with integers $a,b$ satisfying $0<a<b$ is $2^{k-1}$ ?

I tried to use the idendity $$(a^2+b^2)(c^2+d^2)=(ad+bc)^2+(ac-bd)^2$$ and induction over the number of prime factors , but the problem is to show that the representations I get this way are actually distinct , so that the number of representations actually doubles with every new prime factor.

$\endgroup$
0
$\begingroup$

Let $$n=\prod_{j=1}^kp_j$$ Then $p_j$ has a unique representation $p_j=a_j^2+b_j^2=\lvert a_j+ib_j\rvert^2=\lvert q_j\rvert^2$ where $q_j=a_j+ib_j$, $a_j>b_j>0$. Let $r_j\in\{q_j,q_j^*\}$ and choose $0\le\ell\le3$ such that $$a+ib=i^{\ell}\prod_{j=1}^kr_j$$ and $a>|b|>0$. So there are $2^k$ representations. How can we tell them apart? Multiply by $q_j$. If $p_j\operatorname{|}\Re(a+ib)q_j$, then we know that $r_j=q_j^*$. If not, then $r_j=q_j$. If we now consider the representations with opposite signs of $b$ to be equivalent, then we are down to $2^{k-1}$ representations.

$\endgroup$
0
$\begingroup$

Let's show that the number of solutions $(a, b)$ with $0 \leq a, b$ equals $2^{k}$. Because $n$ is square-free, this implies that the number of solutions with $0 < a < b$ is $2^{k-1}$.

Call $R_2(n)$ the set of solutions $(a, b)$ with $0 \leq a, b$. Call $P(n)$ the set of ideals of $\mathbb Z[i]$ of norm $n$. The number of such ideals is equal to $2^k$, by unique factorization of ideals. We have a map $f : R_2(n) \to P(n)$ that sends $(a, b)$ to $(a+bi)$.

  • $f$ is surjective: Take an ideal of norm $n$, which is generated by some $\alpha \in \mathbb Z[i]$ of norm $n$, say $\alpha = a+bi$. If $a, b$ have different signs, multiply by $i$. If $a, b \leq 0$, multiply by $-1$. We may thus assume $a, b \geq 0$.
  • $f$ is injective: Suppose $a, b, c, d \geq 0$ and $a+bi = i^k (c+di)$ for some $k \in \{0, 1, 2, 3\}$. If $k = 0$, we are done. If $k = 1$, $a = -d = 0$ so that $n = b^2$ is a square, a contradiction. Similarly if $k = 2, 3$.
$\endgroup$

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.