$$1) \text{ Let there be a lot M included in } \Bbb C \text{ with the following properties} \\ i\in M \\ \text {if } z\in M \text { then } 1+z^2\in M \\ \text {if } 1+z \in M \text { then } z \in M$$ Prove that $17\in M$ and $1-2i\in M$

  • $\begingroup$ What have you tried ? For instance, can you get $0 \in M$ from the second hypothesis ? Then $-1 \in M$, since $0 = 1+(-1)$, and $2 = 1 + (-1)^2 \in M$. You can try to show that $5 \in M$ to get $4 \in M$ and finally $17 = 1 + 4^2 \in M$. $\endgroup$
    – Watson
    Dec 26 '15 at 13:47

It is easy to show that every integer $n \in M$.

$i \in M$, therefore, $1+i^2=0 \in M$. Now applying the same rule we can get arbitrary large integer $m \in M$.

E.g. $0\rightarrow 1\rightarrow 2\rightarrow 5\rightarrow 26 \rightarrow 677 \rightarrow \dots$

Now applying the second rule we see that all integers less than $m$ are also in $M$.

Now lets prove for $1-2i$.

$i \in M$, therefore, $i-1 \in M$, therefore, $1+(i-1)^2 = 1-2i \in M$.


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