I was reading Andrew Broker's paper on his breakthrough with the sum of three cubes problem. In it, I was confused by one result. Earlier in the paper, it is stated that:
$$ x^3 + y^3 + z^3 = k \mid k,x,y,z \in \mathbb{Z} \ni k \equiv 3 \textrm{ (mod 9)} \implies x \equiv y \equiv z \equiv 1 \textrm{ (mod 3)}$$
Which totally makes sense to me. However, later, at the very end, it states in the explicit case of ($k=3$) with $|x| \geq |y| \geq |z|$ without loss of generality, it follows that:
$$ z \equiv \frac{4(x+y)^2}{3} + 3((x+y)^2-1) \textrm{ (mod 162)}$$
However, I do not see how $z \equiv 1 \textrm{ (mod 3)}$ is ever possible then. For $\frac{4(x+y)^2}{3}$ to be integral, 3 must divide $(x+y)$, which implies that $\frac{4(x+y)^2}{3} = 3k$ for some integer $k$. Meanwhile, $3((x+y)^2-1)$ is trivially a multiple of 3. Since this is under a modulus which is a multiple of three, wouldn't this imply that $z$ must also be a multiple of three, contradicting the first result? They did a computer search for solutions, so I assume I'm missing something, but I don't see what.
There are two small, previously found solutions, (1,1,1 & -5,4,4), but for neither part does the second claim hold true. I would presume it is only for larger solutions, or perhaps solutions where $x > y > z$, but at this point I don't know exactly what's going on.
So, in summary, why did they search for primes if $z$ cannot be 1 mod 3? Did I misread something or what?