Let $S=${$a^2+b^2: a,b\in\mathbb{N}$}. Prove that $S$ is closed under multiplication.

Asked Oct 20 '16 at 12:03

Active Oct 20 '16 at 12:03

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Proof trying. Let $a_{1},a_{2},b_{1},b_{2}$ in $\mathbb{N}$. We will show that $\left( a_{1}^{2}+b_{1}^{2}\right) \left( a_{2}^{2}+b_{2}^{2}\right) \in S$.

So, $\left( a_{1}^{2}+b_{1}^{2}\right) \left( a_{2}^{2}+b_{2}^{2}\right)=a_{1}^{2}a_{2}^{2}+b_{1}^{2}b_{2}^{2}+a_{1}^2b_{2}^2+b_{1}^2a_{2}^2=\left( a_{1}a_{2}\right) ^{2}+\left( b_{1}b_{2}\right) ^{2}+\left( a_{1}b_{2}\right) ^{2}+\left( a_{2}b_{1}\right) ^{2}$. bla bla...

So, How can I show this in $S$?

asked Oct 20 '16 at 12:03

6

$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$ – Did Oct 20 '16 at 12:07
2

Let's not ignore the other solution: $(a^2+b^2)(c^2+d^2)=(ac \pm bd)^2+(ad \mp bc)^2$ – PM 2Ring Oct 20 '16 at 12:18
I have a question that how can we say $ad-bc>0$ for saying $ad-bc\in\mathbb{N}$? – Oct 21 '18 at 00:15
@did May you see my last comment? – Oct 21 '18 at 00:17
@PM2Ring May you see my last comment? – Oct 21 '18 at 00:17
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@KathySong We don't have to worry about that because we're working with squares. If $ad-bc<0$ then just use $bc-ad$. Or swap a&b and c&d. – PM 2Ring Oct 21 '18 at 00:38
@PM2Ring Yes, you are right. Thanks. – Oct 21 '18 at 00:45

Let $S=${$a^2+b^2: a,b\in\mathbb{N}$}. Prove that $S$ is closed under multiplication.

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