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Question

Mathematics Question on Vector basics

If a(b+c)a | (b + c) and a(bc)a| (b-c) where a,b,cNa,b,c \in \, N then

A

c2a2(modb2)c^2 \equiv a^2(mod\, b^2)

B

a2b2(modc2)a^2 \equiv b^2(mod\, c^2)

C

a2+b2b2a^2+b^2-b^2

D

b2c2(moda2)b^2 \equiv c^2(mod \, a^2)

Answer

b2c2(moda2)b^2 \equiv c^2(mod \, a^2)

Explanation

Solution

Since a(b+c) a |(b+c) and a(bc)a|(b-c)
b+ca\Rightarrow \frac{b+c}{a} and bca \frac{b-c}{a}
b+cabca=b2c2a2\therefore \frac{b+c}{a} \cdot \frac{b-c}{a}=\frac{b^{2}-c^{2}}{a^{2}}
a2(b2c2)\Rightarrow a^{2} \mid\left(b^{2}-c^{2}\right)
b2c2(moda2)\Rightarrow b^{2} \equiv c^{2}\left(\bmod a^{2}\right)