Exercise 0.3.7 ($a^2 + b^2
ot \equiv 3 \pmod 4$)

Question

Prove for any integers $a$ and $b$ that $a^2 + b^2$ never leaves a remainder of $3$ when divided by $4$ (use the previous exercise).

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} 
If $\overline{a}$ denotes the class of $a$ in $\Z/4\Z$, then by Exercise 6,
$$\overline{a}^2 \in \{\overline{0}, \overline{1}\}, \qquad \overline{b}^2 \in \{\overline{0}, \overline{1}\}.$$
Therefore
$$\overline{a}^2 + \overline{b}^2 \in \{\overline{0} + \overline{0}, \overline{0} + \overline{1}, \overline{0} + \overline{0}, \overline{1} + \overline{1}\} = \{\overline{0} , \overline{1} , \overline{2} \}.$$
So
$$\overline{a^2 + b^2} = \overline{a}^2 + \overline{b}^2 
e \overline{3},$$
thus, for any integers $a$ and $b$,
$$a^2 + b^2 
ot \equiv 3 \pmod 4.$$
In other words, $a^2 + b^2$ never leaves a remainder of $3$ when divided by $4$.
\end{proof}

Exercise 0.3.7 ($a^2 + b^2 \not \equiv 3 \pmod 4$)

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Exercise 0.3.7 ($a^2 + b^2 \not \equiv 3 \pmod 4$)

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