Let F_k(a, b)=(a+b)^k-a^k-b^k and let S=\ 1,2,3,4,5,6,7,8,9,10\ ....

$$\begin{aligned} \frac{\mathrm{F}_5(\mathrm{a}, \mathrm{~b})}{\mathrm{F}_3(\mathrm{a}, \mathrm{~b})} & =\frac{(\mathrm{a}+\mathrm{b})^5-\mathrm{a}^5-\mathrm{b}^5}{(\mathrm{a}+\mathrm{b})^3-\mathrm{a}^3-\mathrm{b}^3} \\ & =\frac{5}{3}\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}\right) \\ \text { and } \quad \mathrm{a} & =3 \mathrm{k}_1 \text { or } 3 \mathrm{k}_1+1 \text { or } 3 \mathrm{k}_1+2 \\ \mathrm{~b} & =3 \mathrm{k}_2 \text { or } 3 \mathrm{k}_2+1 \text { or } 3 \mathrm{k}_2+2 \end{aligned}$$ only when a and b give same remainder $\mathrm{a}^2+\mathrm{b}^2+\mathrm{ab}$ is divisible by 3 . \begin{array}{lll}\therefore & \begin{array}{ll}a=1, & b=1,4,7,10 \\ a=2, & b=2,5,8\end{array} & \rightarrow 4 \\ a=3, & b=3,6,9 & \rightarrow 3 \\ a=4, & b=4,7,10 & \rightarrow 3 \\ a=5, & b=5,8 & \rightarrow 2 \\ a=6, & b=6,9 & \rightarrow 2 \\ a=7, & b=7,10 & \rightarrow 2 \\ a=8, & b=8 & \rightarrow 1 \\ a=9, & b=9 & \rightarrow 1 \\ a=10, & b=10 & \rightarrow 1 \\ & \text { Total } & 22 \text { ordered pair }\end{array}$