Exercise 3.2.24* ( For every prime $p \equiv 1 \pmod 4$ , $\sum_{i=1}^{(p-1)/4} \left \lfloor \sqrt{ip} \right \rfloor = (p^2 - 1)/12$)

Proof. Let $p=4k+1$ be a prime number. Put

We first compute the last term $\lfloor \sqrt{\mathit{kp}}\rfloor =\lfloor \sqrt{\frac{p-1}{4}p}\rfloor$. Note that

Since these two last conditions are true, we obtain for any prime $p=4k+1$,

which proves that

Consider now the parabola $𝒫$ with equation $y^2=\mathit{px}$, and $𝒮=[[1,\frac{p-1}{4}]]\times [[1,\frac{p-1}{2}]]$ be the set of all ordered pairs of integers $(x,y)$ satisfying $1\le x\le \frac{p-1}{4},1\le y\le \frac{p-1}{2}$. The set $𝒮$ has ${(p-1)}^2∕8$ members. Separate this set into two mutually exclusive subsets ${𝒮}_1$ and ${𝒮}_2$ according as $y^2>\mathit{px}$ or $y^2<\mathit{px}$. Note that there is no pair $(x,y)\in 𝒮$ on the parabola $𝒫$, otherwise $y^2=\mathit{px}$, where $x,y$ are integers, thus $p\mid y^2$, $p\mid y$, so $p^2\mid y^2=\mathit{px}$, thus $p\mid x$. But $1\le x\le \frac{p-1}{4}$, so $p\nmid x$. This contradiction shows that

To summarize,

and $𝒮={𝒮}_1\cup {𝒮}_2$, where ${𝒮}_1\cap {𝒮}_2=\varnothing$, so that

The set ${𝒮}_2$ can be described as the set of all pairs $(x,y)\in {\mathbb{Z}}^2$ such that $1\le x\le k=(p-1)∕4$ and $1\le y<\sqrt{\mathit{px}}$. Since $(x,y)\notin 𝒫$, this last condition is equivalent to $1\le y\le \sqrt{\mathit{px}}$, so that $\mathrm{Card}\{y\in \mathbb{Z}\mid 1\le y<\sqrt{\mathit{px}}\}=\lfloor \sqrt{\mathit{px}}\rfloor$. Therefore

Similarly, the set ${𝒮}_1$ can be described as the set of all pairs $(x,y)\in {\mathbb{Z}}^2$ such that $1\le y\le \frac{p-1}{2}$ and $1\le x\le y^2∕p$. Thus

By equation (1),

(Formula verified with Sage for all primes $p\equiv 1(\mathit{mod}4)$ up to $1000$.)

This seems not to be a progress, because the value of this last sum is not obvious. But consider the Euclidean division of $j^2$ by $p$. For all $j\in [[1,\frac{p-1}{2}]]$,

Here $r_j=0$ if $p\mid j$, and is a quadratic residue in $[[1,p-1]]$ if $p\nmid j$. By the solution of Problem 3.1.14, every quadratic residue modulo $p$ is congruent to $1^2,2^2,\dots ,{((p-1)∕2)}^2$, thus $r_j$ defined in (2) take all possible values of quadratic residues in $[[1,p-1]]$ when $1\le y\le (p-1)∕2$. By Problem 3.1.15, the sum of all such residues is $p(p-1)∕4$, thus

By equation (3),

therefore, using $\underset{j=1}{\overset{n}{\sum <!--\; FUNCTION\; APPLICATION\; -->}}{j}^2=\frac{n(n+1)(2n+1)}{6}$ for $n=\frac{p-1}{2}$, we obtain

Finally, by equation (2)

We have proved, for all prime $p\equiv 1(\mathit{mod}4)$,

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