Gọi d là \(ƯCLN\left(a,b\right)\)
\(\Rightarrow\hept{\begin{cases}a⋮d\\b⋮d\end{cases}}\)
\(a=dm,b=dn\)với \(m,n\inℕ,\left(m,n\right)=1\)
\(\Rightarrow\hept{\begin{cases}a^2+b^2=d^2\left(m^2+n^2\right)\\ab=d^2mn\end{cases}}\)
\(a^2+b^2⋮ab\)
\(\Rightarrow d^2\left(m^2+n^2\right)⋮d^2mn\)
\(\Rightarrow m^2+n^2⋮mn\)
Do \(\left(m,n\right)=1\)
\(\Rightarrow\hept{\begin{cases}m^2+n^2⋮m\\m^2+n^2⋮n\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}n⋮m\\m⋮n\end{cases}}\)
\(\Rightarrow m=n\)
Mà \(\left(m,n\right)=1\)
\(\Rightarrow m=n=1\)
\(\Rightarrow P=\frac{a^2+b^2}{ab}=\frac{1+1}{1}=\frac{2}{1}=2\)
Vậy P=2