<s>[INST] Prove: $\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$, where $a^2 + b^2 = c^2$ Could you help me proving this?

$$\log _{c+b}\left(a\right)+\log _{c-b}\left(a\right)=2\log _{c+b}\left(a\right)\cdot \log _{c-b}\left(a\right)$$
  where $c$ is the length of the hypotenuse of a right triangle, and $a$ and $b$ are the lengths of the other sides.

I tried that but not sure if it helps and what to do next:
$$\frac{\log\left(a\right)}{\log\left(c+b\right)}+\frac{\log\left(a\right)}{\log\left(c-b\right)}=\frac{2\log^2\left(a\right)}{\log\left(c+b\right)\log\left(c-b\right)}$$
Thank you in advance! 
 [/INST] We have $$a^2+b^2=c^2$$ so that $$(c+b)(c-b)=c^2-b^2=a^2.$$ Take the logarithm of both sides $$\log(c+b)+\log(c-b)=2\log(a).$$
Divide by $\log(c+b)$:
$$\frac{\log(c+b)}{\log(c+b)}+\frac{\log(c-b)}{\log(c+b)}=2\frac{\log(a)}{\log(c+b)}$$
Simplify, using the fact that $\frac{\log(x)}{\log(y)}=\log_y(x)$:
$$1+\log_{c+b}(c-b)=2\log_{c+b}(a).$$
Multiply by $\log_{c-b}(a)$:
$$\log_{c-b}(a)+\log_{c-b}(a)\log_{c+b}(c-b)=2\log_{c+b}(a)\log_{c-b}(a).$$
Finally,  $$\log_{c-b}(a)\log_{c+b}(c-b)=\frac{\log(a)}{\log(c-b)}\frac{\log(c-b)}{\log(c+b)}=\frac{\log(a)}{\log(c+b)}=\log_{c+b}(a).$$
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