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If $a \mid(b+c)$ and $a \mid(b-c)$, where $a, b, c \in N$, then
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Verified Answer
The correct answer is:
$b^{2} \equiv c^{2}\left(\bmod a^{2}\right)$
$b \equiv c(\bmod a)$
So, $\frac{b+c}{a}$ and $\frac{b-c}{a}=\frac{(b+c)(b-c)}{a^{2}}$
$=\frac{b^{2}-c^{2}}{a^{2}} \text { or } \frac{a^{2}}{b^{2}-c^{2}}$
Here, $b^{2} \equiv c^{2}\left(\bmod a^{2}\right)$
So, $\frac{b+c}{a}$ and $\frac{b-c}{a}=\frac{(b+c)(b-c)}{a^{2}}$
$=\frac{b^{2}-c^{2}}{a^{2}} \text { or } \frac{a^{2}}{b^{2}-c^{2}}$
Here, $b^{2} \equiv c^{2}\left(\bmod a^{2}\right)$
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