Given that ${\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2$, if

${\log }_{\mathrm{b}+\mathrm{c}}\mathrm{a}+{\log }_{\mathrm{c}-\mathrm{b}}\mathrm{a}={\mathrm{klog}}_{\mathrm{b}+\mathrm{c}}{\mathrm{alog}}_{\mathrm{c}-\mathrm{b}}\mathrm{a}.$ 

Then find the positive integral value of k is

The given condition is equivalent to ${\mathrm{a}}^2\equiv (\mathrm{c}+\mathrm{b})(\mathrm{c}-\mathrm{b})$. We can take ${\log }_{\mathrm{c}+\mathrm{b}}$ and ${\log }_{\mathrm{c}-\mathrm{b}}$ of this equation to obtain two equations:

$2{\log }_{\mathrm{c}+\mathrm{b}}\mathrm{a}={\log }_{\mathrm{c}+\mathrm{b}}{\mathrm{a}}^2={\log }_{\mathrm{c}+\mathrm{b}}(\mathrm{c}+\mathrm{b})(\mathrm{c}-\mathrm{b})={\log }_{\mathrm{c}+\mathrm{b}}(\mathrm{c}-\mathrm{b})+1$

$2{\log }_{\mathrm{c}-\mathrm{b}}\mathrm{a}={\log }_{\mathrm{c}-\mathrm{b}}{\mathrm{a}}^2={\log }_{\mathrm{c}-\mathrm{b}}(\mathrm{c}+\mathrm{b})(\mathrm{c}-\mathrm{b})={\log }_{\mathrm{c}-\mathrm{b}}(\mathrm{c}+\mathrm{b})+1$

Subtracting 1 from each of these equations and multiplying gives

$\left(2{\log }_{\mathrm{c}+\mathrm{b}}\mathrm{a}-1\right)\left(2{\log }_{\mathrm{c}-\mathrm{b}}\mathrm{a}-1\right)={\log }_{\mathrm{c}+\mathrm{b}}(\mathrm{c}-\mathrm{b}){\log }_{\mathrm{c}-\mathrm{b}}(\mathrm{c}+\mathrm{b})=1$

Expanding this product, we obtain

$4{\log }_{\mathrm{c}+\mathrm{b}}{\mathrm{alog}}_{\mathrm{c}-\mathrm{b}}\mathrm{a}-2{\log }_{\mathrm{c}+\mathrm{b}}\mathrm{a}-2{\log }_{\mathrm{c}-\mathrm{b}}\mathrm{a}=0.$

Dividing by 2 and rearranging gives

$2{\log }_{\mathrm{c}+\mathrm{b}}{\mathrm{alog}}_{\mathrm{c}-\mathrm{b}}\mathrm{a}={\log }_{\mathrm{c}+\mathrm{b}}\mathrm{a}+{\log }_{\mathrm{c}-\mathrm{b}}\mathrm{a},$

as claimed.