Difficult system of equations

$\begin{cases} x+y+\sqrt{z} = 148 & ...(1) \\ x+\sqrt{y} + z = 82 & ...(2) \\ \sqrt{x}+y+z = 98 & ...(3) \end{cases}$

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, $x+y+z = 64+81+9 = \boxed{154}$

It is easy to prove that x, y, z are perfect squares

Let x, y, z be a^2, b^2 and c^2 respectively

Then we have

a^2 + b^2 + c = 148 .............. (1)

a^2 + b + c^2 = 82 .................(2)

a + b^2 + c^2 = 98 .................(3)

Eq (3) - Eq (2)

a + b^2 - a^2 - b = 16

(b - a)(b + a - 1) = 16 = (1)(16) = (16)(1) = (2)(8) = (8)(2) = (4)(4) .......... (4)

We note that

(b - a), (b + a - 1) are not both even (and are not both odd) ............ (I)

We also note that

(b - a) <(b + a - 1)

Then

(b - a) = 1

(b + a - 1) = 16

Then

a = 8, b = 9, c = 3

Then

x = 64, y = 81, z = 9

Hence

x + y + z = 154

Note that $x,y,z$ are positive integers, and in particular integers. Thus $\sqrt{z} = 148-x-y$ is an integer, and so $z$ is a perfect square. Similarly from the other two equations we obtain that $x,y$ are also perfect squares.

Next, subtracting equation 1 with equation 3, we obtain $(x+\sqrt{z}) - (\sqrt{x}+z) = 50 > 0$ , or $(x-z) - (\sqrt{x}-\sqrt{z}) > 0$ , or $(\sqrt{x}-\sqrt{z})(\sqrt{x}+\sqrt{z}-1) > 0$ . Since $x,z \ge 1$ , we have $\sqrt{x}+\sqrt{z}-1 \ge \sqrt{1}+\sqrt{1}-1 = 1 > 0$ , so the other factor $\sqrt{x}-\sqrt{z}$ must be positive as well. This means $\sqrt{x} > \sqrt{z}$ , or $x > z$ . Similarly, we obtain $y > x$ from equations 3 and 2.

Also, since $x,y,z$ are positive integers, all terms in the equations are positive integers. So $x,z < 82$ and $y < 98$ . Remember that $x,y,z$ are perfect squares.

Suppose $x \le 49$ . Then $z \le 36$ because $z < x$ and $z$ is a perfect square. Also, $y \le 81$ because $y < 98$ and $y$ is a perfect square. But then $x+y+\sqrt{z} \le 49+81+6 < 148$ , contradiction with the first equation. Thus $x > 49$ . The only possibility is $x = 64$ and $y = 81$ (there are only two squares between $49$ and $98$ , exclusive, so the smaller is $x$ and the larger is $y$ ). This gives $z = 9$ , and thus the result $x+y+z = \boxed{154}$ .

All three unknowns must be perfect squares because there is only one square root in each equation. In an equation of the form [integer] + [integer] + [?] = [integer], the [?] must also be an integer. With that understanding, and the knowledge that all three addends must be positive, it was quite quick to evaluate possible solutions using guess-and-check. I played around with perfect squares less than 100 (they must all be less than 100 because two of the sums are less than 100) and found the answer after just a few tries.

The simplest method here would be the guess and check method (so long as you know that they all have to be square numbers)

Solve these as Diophantine equations using Wolfram Alpha and you will get X=64, Y=81 and Z=9, therefore, Answer is (X+Y+Z) = 154.

Everybody has provided a purely mathematical solution, so its better if I provide a computer science based solution:

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#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    int x=0,y=0,z=0,n;
    cin>>n;

    for(x=1;x<=n;x++)
        {
            for(y=1;y<=n;y++)
                {
                    for(z=1;z<=n;z++)
                        {
                            if(x + sqrt(y) + z == 82 && sqrt(x) + y + z == 98 && x + y + sqrt(z) == 148)
                                {
                                    cout<<x<<" "<<y<<" "<<z<<endl;
                                }
                        }
                }
        }
    return 0;
}

Entered $100$ .