When is $(x+3)^2$ equal to $x^2 +9$?

Question

https://matheducators.stackexchange.com/a/1400/775

Someone commented that the equation in the above answer might sometimes be correct after I commented a correction (feel free to rewrite it appropriately in the question title, my phone does not have a superscript 2). In what situations is this the case?

In $\Bbb Z_6$ we have this equality. – Jun 22 '14 at 15:52 — , Jun 22 '14 at 15:52

score 6 · Accepted Answer · 2014-06-22T16:08:33.110

6

Expanding, $(x+3)^2=x^2+6x+9$. If this is equal to $x^2+9$, then $6x=0\implies x=0$ in the real numbers.

Alternatively, when "$6=0$", i.e., in $\mathbf{Z}_6,\mathbf{Z}_2$ and $\mathbf{Z}_3$, $6x=0\implies (x+3)^2=x^2+9\forall x$.

edited Jun 22 '14 at 16:08

answered Jun 22 '14 at 15:49

1

Also in $\mathbb{Z}{2}$ and $\mathbb{Z}{3}$. – coolpapa Jun 22 '14 at 16:05
Yes, I edited. Thanks. – coolpapa Jun 22 '14 at 16:08
I'm sorry, but can you please tell me what $\mathbb Z_6$ is? I'm new to this and did a quick google search but could not find its meaning! Please help me with this. Is it integers with base 6? – puru Jul 09 '14 at 03:09

When is $(x+3)^2$ equal to $x^2 +9$?

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