Let a and b natural numbers such that 2 a-b, a-2 b and a+b are all...

2 \mathrm{a}-$\mathrm{b}=\mathrm{k}$_1^2$......(1)$a-2 b=k_2^2$......(2)$a+b=k_3{ }^2$......(3) Add (2) \& (3) we get$2 \mathrm{a}-$\mathrm{b}=\mathrm{k}$_2^2+\mathrm{k}_3^2 \Rightarrow \mathrm{k}_2^2+\mathrm{k}_3^2=\mathrm{k}_1^2 \quad$\left(\mathrm{k}_2 < \mathrm{k}_3\right)$ For least ' b ' difference of $\mathrm{k}_3{ }^2$ $\&$ $\mathrm{k}_2{ }^2$ is also least and must be multiple of 3 $\Rightarrow \mathrm{k}_2^2=\mathrm{a}-2 \mathrm{~b}=\mathrm{a}^2$ & $\mathrm{k}_3^2=\mathrm{a}+\mathrm{b}=12^2$ $\Rightarrow \mathrm{k}_3^2-\mathrm{k}_2^2=3 \mathrm{~b}=144-81=63 \Rightarrow \mathrm{~b}=21$ $\Rightarrow$ least $b$ is 21