Exercise 4.3.11 (If $S(n) = \sum_{j \in [\![1, n]\!], \, j \wedge n = 1} j^2$, then $\sum_{j=1}^n j^2 = \sum_{d \mid n} d^2 S \left(\frac{n}{d}\right)$)

Proof. Let

denote the sum of the squares of the positive integers $j\le n$ and prime to $n$.

If $d$ is some divisor of $n$, we define the set

Since

we obtain

that is

Put $m=n∕d$. If $j\wedge n=d$, then there is a unique integer $i\in [[1,n∕d]]$ such that $j=\mathit{di},n=\mathit{dm}$ and $i\wedge m=i\wedge m∕2=1$. Conversely, if $i\wedge (n∕d)=1$, then $j=\mathit{di}$ satisfies $j\wedge n=d$. Therefore, the change of index $j=\mathit{di}$ gives

We have proved

We know that $\phi \{\begin{array}{ccc}\hfill \{d\in {\mathbb{N}}^{\text{∗}}:d\mid n\}\hfill & \hfill \text{→}\hfill & \{d\in {\mathbb{N}}^{\text{∗}}:d\mid n\}\hfill \\ \hfill d\hfill & \hfill \mapsto \hfill & n∕d.\hfill \end{array}$ is bijective (see Problem 6), therefore the change of index $\delta =n∕d$ gives

In conclusion,

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