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!

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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}

