- Observe the following irregularities of the functions \(r _ { 2 } ( n )\) and \(r _ { 4 } ( n )\) as n becomes large:
\(r _ { 2 } ( n ) = 0\) for infinitely many n, while \(\limsup_{n\to\infty} r_2(n) = \infty\) ,
\(r _ { 4 } ( n ) = 2 4\) for infinitely many n while \(\limsup_{n\to\infty} r_4(n)/n = \infty\) .
[Hint: For (a) consider \(n = 5 ^ { k }\) ; for (b) consider alternatively \(n = 2 ^ { k }\) , and \(n = q ^ { k }\) using products of many distinct odd primes.]
Solution
By the two-squares and four-squares theorems, \[ r_2(n)=4\bigl(d_1(n)-d_3(n)\bigr), \qquad r_4(n)=8\sum_{\substack{d\mid n\\4\nmid d}}d, \tag{1} \] where \(d_j(n)\) counts divisors congruent to \(j\pmod4\).
For part (a), if \(n=3^{2k+1}\), the divisors are \(1,3,3^2,\ldots,3^{2k+1}\); exactly \(k+1\) are \(1\pmod4\) and \(k+1\) are \(3\pmod4\). Hence \(r_2(3^{2k+1})=0\), giving infinitely many zeros.
On the other hand, for \(n=5^k\), every divisor is \(1\pmod4\), so \[ r_2(5^k)=4(k+1)\longrightarrow\infty. \] Thus \(\limsup r_2(n)=\infty\).
For part (b), if \(n=2^k\) with \(k\ge1\), the only divisors not divisible by \(4\) are \(1\) and \(2\). Therefore \[ r_4(2^k)=8(1+2)=24, \] so the value \(24\) occurs infinitely often.
To make \(r_4(n)/n\) arbitrarily large, let \[ n_k=p_1p_2\cdots p_k \] be the product of the first \(k\) odd primes. Since \(n_k\) is odd and squarefree, (1) gives \[ \frac{r_4(n_k)}{n_k} =8\frac{\prod_{j=1}^k(1+p_j)}{\prod_{j=1}^kp_j} =8\prod_{j=1}^k\left(1+\frac1{p_j}\right). \] The product diverges: indeed \(\log(1+1/p)\ge 1/(2p)\) for \(p\ge2\), and Euler’s theorem \(\sum_p1/p=\infty\) implies \[ \sum_j\log\left(1+\frac1{p_j}\right)=\infty. \] Hence \[ \limsup_{n\to\infty}\frac{r_4(n)}n=\infty. \]