KVPY 2020-21 SA Exam Previous Papers

$\because \mathit{p},\mathit{q},\mathit{r},\in \mathit{Q}$ and $\sqrt{\mathit{p}}+\sqrt{\mathit{q}}+\sqrt{\mathit{r}}\in \mathit{Q}$
$\Rightarrow {\left(\sqrt{\mathit{p}}+\sqrt{\mathit{q}}+\sqrt{\mathit{r}}\right)}^2\in \mathit{Q}\mathrm{}$ $\Rightarrow \sqrt{\mathit{p}\mathit{q}}+\sqrt{\mathit{q}\mathit{r}}+\sqrt{\mathit{r}\mathit{p}}\in \mathit{Q}$
Case I: Let exactly one of $\sqrt{\mathit{p}},\sqrt{\mathit{q}},\sqrt{\mathit{r}}$ is irrational
WLOG, $\sqrt{\mathit{p}},\notin \mathit{Q}$ but $\sqrt{\mathit{r}},\sqrt{\mathit{q}}\in \mathit{Q}$
From (i), $\sqrt{\mathit{p}}\left(\sqrt{\mathit{q}}+\sqrt{\mathit{r}}\right)\in \mathit{Q}$ (contradiction)
Case II: Let exactly two out of $\sqrt{\mathit{p}},\sqrt{\mathit{q}},\sqrt{\mathit{r}}$ are irrational.
WLOG, $\sqrt{\mathit{p}},\sqrt{\mathit{q}}\notin \mathit{Q}$ but $\sqrt{\mathit{r}}\in \mathit{Q}$
From(i), $\sqrt{\mathit{p}}\sqrt{\mathit{q}}+\sqrt{\mathit{r}}\sqrt{\mathit{p}}+\sqrt{\mathit{r}}\sqrt{\mathit{q}}+{\left(\sqrt{\mathit{r}}\right)}^2\in \mathit{Q}$
$\Rightarrow \left(\sqrt{\mathit{p}}+\sqrt{\mathit{r}}\right)\left(\sqrt{\mathit{q}}+\sqrt{\mathit{r}}\right)\in \mathit{Q}$
$\because$ both $\sqrt{\mathit{p}}+\sqrt{\mathit{r}}$ and $\sqrt{\mathit{q}}+\sqrt{\mathit{r}}$ are irrational, hence they must be conjugate of each other. (Contradiction)
Case III: Let all $\sqrt{\mathit{p}},\sqrt{\mathit{q}},\sqrt{\mathit{r}}$ are irrational.
Let $\sqrt{\mathit{p}}+\sqrt{\mathit{q}}+\sqrt{\mathit{r}}=\mathit{x}$ when $\mathit{x}\in {\mathit{Q}}^{+}$
From (i), $\sqrt{\mathit{p}}\left(\mathit{x}-\sqrt{\mathit{p}}\right)+\sqrt{\mathit{q}}\sqrt{\mathit{r}}\in \mathit{Q}$
$\Rightarrow \mathit{x}\sqrt{\mathit{p}}+\sqrt{\mathit{q}}\sqrt{\mathit{r}}\in \mathit{Q}$ (Contradiction)
Hence, all $\sqrt{\mathit{p}},\sqrt{\mathit{q}}$ and $\sqrt{\mathit{r}}$ must be rational.