- Use the formula for \(r _ { 2 } ( n )\) to prove the following:
If \(n = p ,\) where p is a prime of the form \(4 k + 1\) , then \(r _ { 2 } ( n ) = 8\) . This implies that n can be written in a unique way as \(n = n _ { 1 } ^ { 2 } + n _ { 2 } ^ { 2 }\) , except for the signs and reordering of \(n _ { 1 }\) and \(n _ { 2 }\) .
If \(n = q ^ { a }\) , where q is prime of the form \(4 k + 3\) and a is a positive integer, then \(r _ { 2 } ( n ) > 0\) if and only if a is even.
In general, n can be represented as the sum of two squares if and only if all the primes of the form \(4 k + 3\) that arise in the prime decomposition of n occur with even exponents.
Solution
Recall the two-squares formula \[ r_2(n)=4\bigl(d_1(n)-d_3(n)\bigr), \tag{1} \] where \(d_j(n)\) is the number of positive divisors of \(n\) congruent to \(j\pmod4\).
If \(p\equiv1\pmod4\) is prime, its divisors are \(1,p\), both congruent to \(1\pmod4\). Thus \[ r_2(p)=4(2-0)=8. \] For such an odd prime no representation can have a zero coordinate or two coordinates of equal absolute value. Therefore the eight sign-and-order variants come from one unordered pair of positive absolute values, proving uniqueness up to signs and reordering.
If \(q\equiv3\pmod4\) is prime, the divisors of \(q^a\) are \(1,q,\ldots,q^a\), and \(q^j\equiv(-1)^j\pmod4\). Hence \[ d_1(q^a)-d_3(q^a)= \begin{cases} 1,&a\text{ even},\\ 0,&a\text{ odd}. \end{cases} \] Thus \(r_2(q^a)>0\) exactly when \(a\) is even.
For the general case, let \(\chi\) be the nontrivial character modulo \(4\), extended by \(\chi(d)=0\) for even \(d\). Then \[ d_1(n)-d_3(n)=\sum_{d\mid n}\chi(d). \] The divisor sum of a multiplicative function is multiplicative. Write \[ n=2^e\prod_i p_i^{\alpha_i}\prod_j q_j^{\beta_j}, \] where \(p_i\equiv1\pmod4\) and \(q_j\equiv3\pmod4\). Then \[ \sum_{d\mid n}\chi(d) =\prod_i(\alpha_i+1) \prod_j\left(1-1+1-\cdots+(-1)^{\beta_j}\right). \] The second factor is \(1\) when \(\beta_j\) is even and \(0\) when \(\beta_j\) is odd. By (1), \(r_2(n)>0\) exactly when every prime \(q\equiv3\pmod4\) occurs with even exponent. This is precisely the two-squares criterion.