Given: $$x+y+\sqrt{z}=148$$ $$x+\sqrt{y}+z=82$$ $$\sqrt{x}+y+z=98$$
Find the value of $x+y+z$ such that x, y, and z are positive integers.
The only significant step I did was to let $a^2=x, b^2=y, c^2=z.$ After that, I just played with the equations, adding and subtracting, letting $S=x+y+z$, finding equations for S. But I always come to a dead end. Can anyone show me the solution?
On
$$x + y + \sqrt{z} = 148$$ $$x + \sqrt{y} + z = 82$$ $$\sqrt{x} + y + z = 98$$ As $x, y , z$ are positive integers and each of the sum is positive integer, that means $\sqrt{x}$, $\sqrt{y}$, $\sqrt{z}$ are also integers.
So, that means $x, y, z$ are perfect squares less than $148$.
Thus, x, y, z can be one of the following {$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144$}.
If we see the 3 equation, it is clear that, the max possible value of $x = 64,$ $y = 81$ and $z = 16$.
Now, by hit n trail method, it can be determined within 2-3 guesses that the only possible solution will be as follows - $$ x = 64, y = 81, z = 9$$
So, the final answer is, $$x + y + z = 154$$
Following on from your step, we subtract the second equation from the third. We then have $$a-a^2 +b^2-b =16$$ or $$(b-a)(b+a-1)=16$$ No the possible combinations are: $\{1,16\}$, $\{2,8\}$, or $\{4,4\}$. A quick check tells you only the first will work. Thus $b-a=1$ and $b+a-1= 16$. From this we get $a=8$ and $b=9$. Subtracting the third equation from the first we have $$ a^2-a + c -c^2=50$$ from which we can solve for $c$ and we obtain $c(c-1)=6$ or $c=3$. Thus $x=64, y =81, z =9.$