How many points $(x, y)$ with integer coordinates satisfy the inequality $x^2+y^2 \leq 25$? I had this questions from a previous exam that I couldn't answer, I am apologizing for any English mistakes or for any stupid questions, I tried to solve them and I searched the internet and I couldn't find answers or at least ones with explanations.  
1-if $$x^2 + y^2 \leq 25$$  How many INTEGER pairs of $x,y$ satisfy the inequality?
*I tried to think of combinatorics but I didn't know how it can help me, I had to use brute force in the end.  
Thanks for taking the time to read the question, if anyone has tips for my exam or know any challenging problems I would be really thankful if he/she could tell me about them!
Thanks!
 A: In the closed disc of radius $5$ there are
$$\bigl\lfloor\sqrt{25}\bigr\rfloor+\bigl\lfloor\sqrt{24}\bigr\rfloor+\bigl\lfloor\sqrt{21}\bigr\rfloor+\bigl\lfloor\sqrt{16}\bigr\rfloor+\bigl\lfloor\sqrt{9}\bigr\rfloor=20$$
lattice points satisfying $x\geq0$, $y>0$. The total number of lattice points in this disc therefore is $1+4\cdot20=81$.
A: This is an instance of Gauss' circle problem about lattice points in a circle.
If we set $r_2(n)=\left|\{(a,b)\in\mathbb{Z}\times\mathbb{Z}: a^2+b^2=n\}\right|$ and 
$$\chi(n)=\left\{\begin{array}{rcl}1&\text{if}& n\equiv 1\pmod{4}\\
-1&\text{if}& n\equiv 3\pmod{4}\\
0&\text{if}& n\equiv 0\pmod{2}\end{array}\right.$$
Lagrange's identity and the UFD property of $\mathbb{Z}[i]$ allow us to state that
$$ r_2(n) = 4\sum_{d\mid n}\chi(d) $$
so the number of lattice points inside a circle depends on the average value of $r_2(n)$.
In our case we have:
$$\begin{eqnarray*}\left|\left\{(a,b)\in\mathbb{Z}\times\mathbb{Z}:a^2+b^2\leq 25\right\}\right|&=&1+4\sum_{n=1}^{25}\sum_{d\mid n}\chi(d)\\&=&21+4\sum_{k=1}^{3}\left\lfloor\sqrt{25-k^2}\right\rfloor=\color{red}{81}.\end{eqnarray*} $$
It is reasonable to expect that the number of lattice points in $x^2+y^2\leq 25$ is pretty close to the enclosed area, namely $25\pi\approx 78.54$. Gauss' circle problem is indeed about estimating the difference between these objects.

A: There place were grid-lines of the graph cross "lattice points."
$x^2 + y^2 = 5$ cuts through 16 of the lattice points
$(\pm 5,0), (0,\pm 5), (\pm3,\pm4)$
If we construct an irregular octagon connecting these 8 points. It would have an area slightly less than the area of the circle.
The area of the circle is $25\pi$
Pick's theorem says that the number of lattice point on the interior (I) $+ \frac 12$ the number of lattice points on the perimeter(E) $- 1$ equals the area of any enclosed polygon.
$I+ 8 - 1 < 25\pi\\
I < 71$
Next, there a symmetry to the grid-line.  The center is stationary.  For every other point on the lattice, if we rotate 90 degree we will find another point on the lattice.
$I-1$ is divisible by $4$
$I=69$ is the largest number that meet the two constraints we have.
A: A brute force approach abusing symmetries. Notice that if $(x,y)$ is a solution then so is $(y,x)$ and if $(x,y)$ is a solution then so is $(\pm x,\pm y)$.
Solutions including zero are $(1,0),(2,0),(3,0),(4,0),(5,0)$ and each of the pairs has $4$ symmetries hence $5\cdot 4=20$ and plus $(0,0)$ so $21$ solutions with $0$.
Now solutions where $x=y\neq 0$ also have $4$ symmetries $(1,1),(2,2),(3,3)$ so $3\cdot 4=12$ solutions and the rest solutions have $8$ symmetries and thoose are $(4,3),(4,2),(4,1),(3,2),(3,1),(2,1)$ so $6\cdot 8=48$ so in total there are $48+12+21=81$ solutions.
A: This is not an answer but a comment or a hint (so please don't down-vote it).
Consider this figure:

A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

How many points $\ds{\pars{x, y}}$ with integer coordinates satisfy the inequality $\ds{x^{2} + y^{2} \leq 25}$ ?.



*

*The number of points $\ds{\pars{x, y}}$ with integer coordinates that satisfy
$\ds{x^{2} + y^{2} = n}$ for a given integer $\ds{n \geq 0}$ is given by
$$
\mc{N}_{n} \equiv \sum_{x = -\infty}^{\infty}
\sum_{y = -\infty}^{\infty}\bracks{q^{n}}q^{x^{2} + y^{2}} =
\bracks{q^{n}}
\pars{\sum_{x = -\infty}^{\infty}q^{x^{2}}}^{2} =
\bracks{q^{n}}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2}
$$

*Therefore, the answer $\ds{\mc{N}_{\leq\ 25}}$ of the question at the top is given by
$$
\mc{N}_{\leq\ 25} \equiv \sum_{n = 0}^{25}\mc{N}_{n} =
\sum_{n = 0}^{25}\bracks{q^{n}}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2}
$$

Then,
\begin{align}
\mc{N}_{\leq\ 25} & =
\bracks{q^{0}}\sum_{n = 0}^{25}\pars{1 \over q}^{n}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2} =
\bracks{q^{0}}{\pars{1/q}^{26} - 1 \over 1/q - 1}
\pars{1 + 2\sum_{x = 0}^{\infty}q^{x^{2}}}^{2}
\\[5mm] & =
\bracks{q^{25}}{1 - q^{26} \over 1 - q}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2} =
\bracks{q^{25}}\sum_{k = 0}^{25}q^{k}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2}
\\[5mm] & =
\sum_{k = 0}^{25}\bracks{q^{25 - k}}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2} =
\sum_{k = 0}^{25}\bracks{q^{k}}
\pars{1 + 2\sum_{x = 1}^{\infty}q^{x^{2}}}^{2}
\\[5mm] & =
\sum_{k = 0}^{25}\bracks{q^{k}}
\pars{1 + 2q + 2q^{4} + 2q^{9} + 2q^{16} + 2q^{25}}^{2} = \bbx{81}
\end{align}

