A) \[{{b}^{2}}\equiv {{c}^{2}}(\bmod \,{{a}^{2}})\]
B) \[{{b}^{2}}\equiv {{c}^{2}}(\bmod \,{{a}^{2}})\]
C) \[{{a}^{2}}\equiv {{b}^{2}}(\bmod \,{{c}^{2}})\]
D) \[{{c}^{2}}\equiv {{a}^{2}}(\bmod \,{{b}^{2}})\]
Correct Answer: A
Solution :
Since a \[a|(b+c)\]and \[a|(b-c)\] \[\Rightarrow \] \[\frac{b+c}{a}\] and \[\frac{b-c}{a}\] \[\therefore \] \[\frac{b+c}{a}.\frac{b-c}{a}=\frac{{{b}^{2}}-{{c}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \] \[{{a}^{2}}|({{b}^{2}}-{{c}^{2}})\] \[\Rightarrow \] \[{{b}^{2}}={{c}^{2}}(\bmod \,{{a}^{2}})\]
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