Exercise 3.6.11 (Sums of three squares)

Question

Show that if $n$ is a sum of three squares then $n 
ot \equiv 7 \pmod 8$. Show by example that there exist positive integers $m$ and $n$, both of which are sum of three squares, bu whose product $mn$ is not a sum of three squares.

Richard Ganaye · Accepted Answer

ewcommand{\Z}{\mathbb{Z}}

\begin{proof} For all integers $a, b ,c$, $a^2 \equiv 0, 1, 4 \pmod 8$, thus $a^2 + b^2 \equiv 0, 1,2,4,5 \pmod 8$, and $a^2 + b^2 + c^2 \equiv  0, 1, 2 , 3,4,5,6 \pmod 8$, so 
$$\forall (a,b,c) \in \Z^2,\ a^2 + b^2 + c^2 
ot \equiv 7 \pmod 8.$$

Note that $3 = 1^2 + 1^2 + 1^2$ is the sum of three squares, and so is $5 = 1^2 + 2^2 + 0^2$, but $15 = 3 \cdot 5 \equiv 7 \pmod 8$ is not the sum of three squares.
\end{proof}

Exercise 3.6.11 (Sums of three squares)

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Exercise 3.6.11 (Sums of three squares)

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