Question Number 199184 by Tawa11 last updated on 29/Oct/23
Commented by AST last updated on 29/Oct/23
$${x}=\mathrm{9},{y}=\mathrm{4},{z}=\mathrm{1} \\ $$
Commented by Tawa11 last updated on 29/Oct/23
$$\mathrm{Workings}\:\mathrm{please}. \\ $$
Commented by AST last updated on 29/Oct/23
$${Let}\:{x},{y},{z}\in\mathbb{N} \\ $$$${If}\:{one}\:{of}\:{x},{y}\:{or}\:{z}\:{are}\:{not}\:{perfect}\:{squares},{then} \\ $$$${sum}\:{cannot}\:{be}\:{an}\:{integer} \\ $$$$\Rightarrow{z}\in\left\{\mathrm{1},\mathrm{4}\right\};{z}=\mathrm{1}\Rightarrow\sqrt{{x}}+{y}=\mathrm{7}\Rightarrow{y}\in\left\{\mathrm{1},\mathrm{4}\right\}\Rightarrow{y}=\mathrm{4} \\ $$$$\Rightarrow\sqrt{{x}}=\mathrm{3}\Rightarrow{x}=\mathrm{9} \\ $$$${z}=\mathrm{4}\Rightarrow\sqrt{{x}}+{y}=\mathrm{4}\Rightarrow{y}=\mathrm{1}\Rightarrow{x}=\mathrm{3}\:\left({contradicts}\:{iii}\right) \\ $$$${Hence}\:\left({x},{y},{z}\right)=\left(\mathrm{9},\mathrm{4},\mathrm{1}\right).\:{This}\:{is}\:{the}\:{only}\:{solution} \\ $$$${for}\:{the}\:{first}\:{line}. \\ $$