An algebra problem by Harshi Singh

$\large \begin{aligned} \begin{cases} x+y+\sqrt{z} = 670 \\ x+\sqrt{y}+z=940 \\ \sqrt{x} + y+z=544 \end{cases} \end{aligned}$

$\text{Each of the equations have one term involving square root on LHS and rest all including RHS are integers}$ $\text{The equations clearly imply that all of x,y,z are perfect squares .}$

$\begin{aligned} \text{Let } \begin{cases} x=a^2 \\ y=b^2 \text{ where a,b,c } \in \mathbb{N} \\ z=c^2\end{cases} \end{aligned}$

$\text{The modified equations are : }$

$\large \begin{aligned} \begin{cases} a^2+b^2+c = 670 \\ a^2+b+c^2=940 \\ a + b^2+c^2=544 \end{cases} \end{aligned}$

$\text{Now we subtract the equations to derive }$

$\begin{aligned} a(a-1) - c(c-1) = 670-544=126\end{aligned}$

$\begin{aligned} a(a-1) - b(b-1) = 940-544=396\end{aligned}$

$\begin{aligned} c(c-1) - b(b-1) = 940-670=270\end{aligned}$

$\text{We note that : } a(a-1) > c(c-1) > b(b-1)$

$\text{From } c(c-1)=b(b-1)+270 \text{ we conclude that } c(c-1) > 270$

$\begin{aligned} c(c-1)>270 \implies c^2-c-270>0\end{aligned}$

$\text{On solving we get }c\ge17 \text{ \& checking for first few values of c as smallest c is 17 we get}$

$\begin{aligned} \begin{cases} c=17\implies a(a-1) = 126+17.16=398 \text{ Not of the form a(a-1) } \implies \text{ rejected} \\ c=18\implies a(a-1) = 126+18.17=432 \text{ Not of the form a(a-1) } \implies \text{ rejected} \\ c=19\implies a(a-1) = 126+19.18=468 \text{ Not of the form a(a-1) } \implies \text{ rejected} \\ c=20\implies a(a-1) = 126 + 20.19 = 506 = \color{#624F41}{23.22} \implies \text{Selected} \end{cases}\end{aligned}$

$\text{Solving c=17 we get the other values as follows :}$

$c=20\implies z=c^2=400$

$a=23\implies x=a^2=529$

$b=11\implies y=b^2=121$

$\large \text{Finally we get } \boxed{x+y+z=1050}$