Algebra Exercise Level 1

Exercise:

Let $p\in \mathbb{Z}$ such that $\frac{{{a}^{p}}+{{b}^{p}}}{{{c}^{p}}+{{d}^{p}}}=\frac{a+b}{c+d}$ where $a,b,c,d>0$

Show that , $p=-1$

Solution: Put $p=-k$ \(\Leftrightarrow \frac{{{a}^{p}}+{{b}^{p}}}{{{c}^{p}}+{{d}^{p}}}=\frac{{{a}^{-k}}+{{b}^{-k}}}{{{c}^{-k}}+{{d}^{-k}}}\times \frac{{{\left( abcd \right)}^{k}}}{{{\left( abcd \right)}^{k}}}=\frac{{{a}^{-k+k}}{{\left( bcd \right)}^{k}}+{{b}^{-k+k}}{{\left( acd \right)}^{k}}}{{{c}^{-k+k}}{{\left( adb \right)}^{k}}+{{d}^{-k+k}}{{\left( abc \right)}^{k}}}=\frac{{{\left( cd \right)}^{k}}\left( {{b}^{k}}+{{a}^{k}} \right)}{{{\left( ab \right)}^{k}}\left( {{d}^{k}}+{{c}^{k}} \right)}=\frac{{{a}^{k}}+{{b}^{k}}}{{{c}^{k}}+{{d}^{k}}}\)

Hence $\frac{{{a}^{-k}}+{{b}^{-k}}}{{{c}^{-k}}+{{d}^{-k}}}=\frac{{{a}^{k}}+{{b}^{k}}}{{{c}^{k}}+{{d}^{k}}}$

Now for $k=1$ we get $\frac{{{a}^{-1}}+{{b}^{-1}}}{{{c}^{-1}}+{{d}^{-1}}}=\frac{a+b}{c+d}$ thus $p=-1$