Problem 1659

1. Given the equations:
$$x + y + z = 1$$
and
$$\arctan x + \arctan y + \arctan z = \frac{\pi}{4},$$
we start by using the identity for the sum of arctangents:
$$\arctan a + \arctan b + \arctan c = \arctan \left( \frac{a + b + c - abc}{1 - (ab + bc + ca)} \right).$$
Setting $a = x$, $b = y$, and $c = z$, we have:
$$\arctan x + \arctan y + \arctan z = \arctan \left( \frac{x + y + z - xyz}{1 - (xy + yz + zx)} \right).$$

2. Given that $\arctan x + \arctan y + \arctan z = \frac{\pi}{4}$, we equate:
$$\arctan \left( \frac{x + y + z - xyz}{1 - (xy + yz + zx)} \right) = \frac{\pi}{4}.$$
Since $\arctan \frac{1}{1} = \frac{\pi}{4}$, it follows that:
$$\frac{x + y + z - xyz}{1 - (xy + yz + zx)} = 1.$$

3. Substituting $x + y + z = 1$ into the equation, we get:
$$\frac{1 - xyz}{1 - (xy + yz + zx)} = 1.$$
Simplifying, we obtain:
$$1 - xyz = 1 - (xy + yz + zx).$$
Therefore:
$$xyz = xy + yz + zx.$$

4. We can rewrite the equation as:
$$xyz - xy - yz - zx = 0.$$
Factoring, we get:
$$(x - 1)(y - 1)(z - 1) = 0.$$
This implies that at least one of $x$, $y$, or $z$ is equal to 1. Without loss of generality, assume $x = 1$.

5. Substituting $x = 1$ into $x + y + z = 1$, we get:
$$1 + y + z = 1 \implies y + z = 0 \implies y = -z.$$

6. Now, we need to prove that $x^{2n+1} + y^{2n+1} + z^{2n+1} = 1$ for all positive integers $n$. Substituting $x = 1$ and $y = -z$, we have:
$$1^{2n+1} + y^{2n+1} + (-y)^{2n+1}.$$
Since $y = -z$, we get:
$$1 + y^{2n+1} + (-y)^{2n+1}.$$

7. Note that $(-y)^{2n+1} = -(y^{2n+1})$, so:
$$1 + y^{2n+1} - y^{2n+1} = 1.$$

8. Therefore, we have shown that:
$$x^{2n+1} + y^{2n+1} + z^{2n+1} = 1.$$

$\blacksquare$

The final answer is $\boxed{ x^{2n+1} + y^{2n+1} + z^{2n+1} = 1 }$.