## Find the smallest $x + y + z$ with integers $x \gt y \gt z \gt 0$ such that $x + y$, $x – y$, $x + z$, $x – z$, $y + z$, $y – z$ are all perfect squares.

### To solve this problem, we can use the method of systematically creating equations and checking for integer solutions. Let’s denote the required perfect squares as follows:

$x+y=a^2$,

$x-y=b^2$,

$x+z=c^2$,

$x-z=d^2$,

$y+z=e^2$,

$y-z=f^2$.

From these equations, we can create some additional equations. For example, adding the first and the second equation, we get:

$x+y+x-y=a^2+b^2 \Rightarrow 2x=a^2+b^2$,

From where we conclude that $x=\frac{a^2+b^2}{2}$.

Similarly, adding the third and the forth equations we get that $x=\frac{c^2+d^2}{2}$.

Also, adding the fifth and the six equations, $y=\frac{e^2+f^2}{2}$.

Equating the two values for $x$, we get $\frac{a^2+b^2}{2} = \frac{c^2+d^2}{2}$. Simplifying this, we can obtain that $a^2+b^2=c^2+d^2$.

The smallest possible solution with $a^2+b^2=c^2+d^2$ with the condition that $a>b>c>d$ is $5^2+1^2=4^2+3^2=26$.

From here, we get $x=13$, $y=13$, and $z=5$ by substituting these values to previously derived equations.

But the condition also states that $x>y>z>0$. So, let’s find a solution where $y<13$ and $x=y+z$. For that, let's use the equation $y=\frac{e^2+f^2}{2}$. A suitable solution is $y=5$, as we can get $5=3^2+1^2$. Next, substitute $y=5$ to the equation $x=y+z$, we get $x=5+z$. From one of our initial perfect square equations, $x+z=c^2$, we can subtract $z$ on both sides to get $x=c^2-z$. From there, substitute $x=5+z$, we get $5+z=c^2-z$. Rearrange to solve for $z$ and we get $z=\frac{c^2-5}{2}$. Consider $c=3$, so we get $z=2$, and from the equation $x=5+z$ we get $x=7$. So, the smallest $x + y + z$ is $7 + 5 + 2 = 14$ with $x=7$, $y=5$, and $z=2$. All the conditions are fulfilled with these values. Please let me know if you have any further questions.

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