If a, b, c, d, are distinct integers in AP such that $\mathrm{d}={\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\text{, then the sum }\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}$ equals

Let the common difference of the given AP be k

Given $d=a^2+b^2+c^2$

$\begin{array}{l}\Rightarrow a+3k=a^2+(a+k{)}^2+(a+2k{)}^2\\ \Rightarrow 5k^2+6ak-3k+3a^2-a=0\\ \Rightarrow 5k^2+3k(2a-1)+3a^2-a=0\\ \Rightarrow k\text{ is real; so }D\ge 0\\ \Rightarrow 9(2a-1{)}^2-4\times 5\left(3a^2-a\right)\ge 0\\ \Rightarrow 36a^2-36a+9-60a^2+20a\ge 0\\ \Rightarrow 24a^2+16a-9\le 0\Rightarrow a^2+\frac{2}{3}a-\frac{3}{8}\le 0\\ \Rightarrow {\left(a+\frac{1}{3}\right)}^2\le \frac{3}{8}+\frac{1}{9}\Rightarrow {\left(a+\frac{1}{3}\right)}^2\le \frac{35}{72}=\frac{70}{144}\\ \Rightarrow \frac{-\sqrt{70}}{12}\le \left(a+\frac{1}{3}\right)\le \frac{\sqrt{70}}{12}\\ \Rightarrow \frac{-4-\sqrt{70}}{12}\le a\le \frac{\sqrt{70}-4}{12}\end{array}$

$\Rightarrow a=-1,0,\text{ since }a\text{ is an integer, }$