\[ \begin{array} { l l l l l l l l l } x & + & y & + & \sqrt{z} & = & 148 \\ x & + & \sqrt{y} & + & z & = & 82 \\ \sqrt{x} & + & y & + & z & = & 98 \\ \end{array} \]
Find the value of \( x + y + z \).
Note: \(x, y, z \) are positive integers.
Solution:
\(\begin{cases} x+y+\sqrt{z} = 148 & ...(1) \\ x+\sqrt{y} + z = 82 & ...(2) \\ \sqrt{x}+y+z = 98 & ...(3) \end{cases} \)
We note that \(\sqrt{x}\), \(\sqrt{y}\) and \(\sqrt{z}\) and positive integers. Let they be:
\(\sqrt{x} = a\), \(\sqrt{y} = b \) and \(\sqrt{z} = c\)
Then,
\(\begin{cases} a^2+b^2+c = 148 & ...(1) \\ a^2+b+c^2 = 82 & ...(2) \\ a+b^2+c^2 = 98 & ...(3) \end{cases}\)
\( \Rightarrow \begin{cases} Eq1-Eq2: & b^2-b - (c^2-c) = 66 & ...(4) \\ Eq3-Eq2: & b^2-b - (a^2-a) = 16 & ...(5) \\ Eq1-Eq3: & a^2-a - (c^2-c) = 50 & ...(6) \end{cases} \)
\( \Rightarrow \begin{cases} b(b-1) - c(c-1) = 66 & ...(4) \\ b(b-1) - a(a-1) = 16 & ...(5) \\ a(a-1) - c(c-1) = 50 & ...(6) \end{cases} \)
We note that \(c(c-1) < a(a-1) < b(b-1)\) and that:
\( \begin{cases} a(a-1) = c(c-1) + 50 & ...(6) \\ b(b-1) = c(c-1) + 66 & ...(4) \end{cases} \)
The smallest \(c\) is (2) and:
When \(c=2\Rightarrow \begin{cases} c(c-1) = 2(1) & = 2 & \\ a(a-1) = 2 + 50 & = 52 & \text { rejected -- not of a(a-1) form } \\ b(b-1) = 2 + 66 & = 68 & \text { rejected -- not of b(b-1) form} \end{cases} \)
When \(c=3\Rightarrow \begin{cases} c(c-1) = 3(2) & = 6 & \Rightarrow z = 9\\ a(a-1) = 6 + 50 & = 56 = 8(7) & \Rightarrow a = 8 \Rightarrow x = 64 \\ b(b-1) = 6 + 66 & = 72 = 9(8) & \Rightarrow b = 9 \Rightarrow y = 81 \end{cases} \)
Therefore, \(x+y+z = 64+81+9 = \boxed{154}\)
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