a. $3$ is poor, as $6$ can't be represented as $3x^2-y^2-z^2$. Proof: mod 3 argument. b. $2$ is rich, as $2x^2-(x+1)^2-(x-2)^2=2x-5$ so every odd number is attainable, $2x^2-(x+1)^2-(x-3)^2=4x-10$ so every number of the form $4k+2$ more than $2$ is achievable. For $2$ write $2=2*81-144-16$. To get the multiples of $4$ just multiply the numbers in the triples of odd numbers and those of the form $4k+2$ by $2$ enough times to reach the desired number.

Fun questions: Does there exists infinitely many poor integers? Does there exists infinitely many rich integers?

For $a$ part, the answer is yes. $6$ can never be represented as $(9k+3)x^2-y^2-z^2$ hence $9k+3$ is poor for all $k$.

The answer is also yes for $b$ part. Consider a primitive pythagorean triple $a,b,c$ such that $a^2+b^2=c^2$. The number $c^2+1$ is rich. First note that $(c^2+1)x^2-c^2x^2-(x-1)^2=2x-1$. This makes all odd numbers except $1$. To make 1, we have $1=c^2+1-a^2-b^2$. Now note that $(c^2+1)x^2-(cx-1)^2-(x+c-2)^2=4x+c^2-4c+5$. Using the fact that $c$ is odd, we can see that all numbers of the form $4k+2$ can also be represented. We get all multiples of 4 by simply mulyiplying the elements of the triples of numbers by 2 enough times to get a multiple of $4$. Hence, $c^2+1$ is rich for $a^2+b^2=c^2$ where it is a primitive python triple.

This could be an interesting question, continuing the idea: The set $A$ contains all rich integers, the set $B$ contains all poor integers. Determine all the elements of set $A$ and all the elements of set $B$.

Look this problem: https://artofproblemsolving.com/community/q1h1473536p8562327 and the article there for a generalization: https://www.emis.de/journals/JIS/VOL18/Lam/lam2.pdf

We can show $4$ is poor by considering the equation $4x^2-y^2-z^2=1$ under $\pmod 4$. We claim $10$ is rich. For $n=1$, we can take $(x,y,z)=(3,5,8)$. For any $x>1$, by letting $y=3x$ and $z=x-1$, we have $10x^2-y^2-z^2=2x-1$ so a suitable triplet exists for all odd $n$. Setting $y=3x-1$ and $z=x+1$ yields $10x^2-y^2-z^2=4x-2$, so a suitable triplet exists for all $n\equiv 2\pmod 4$. And for $x>3$, setting $y=3x$ and $z=x-2$ yields $10x^2-y^2-z^2=4x-4$ so a suitable triplet exists for all $n\equiv 0\pmod 4$. Thus there exists the required triplet $(x,y,z)$ for any $n$, hence $10$ is rich.

Claim:M=5 is rich integer Proof: For odd n,take (X,Y,Z)=(a+1,2a+2,a) then 5x²-y²-z²=2a+1 For even n ,take (X,Y,Z)=(a+6,2a+13,a+3) then 5x²-y²-z²=2a+2.

If $4\mid m$ then m is poor. If $p\mid m$ where p is prime and $p\equiv 3 \pmod 4$ then m is poor. In all other cases m is rich.