Let us consider the following set $A:=\{(r,t)| r \in \mathbb N \cup \{0\}, t \in \mathbb Z, t \leq r-5\}$

My question is the follow : Does there exists any pair $(r,t)$ belonging to $A$ such that it satisfies the inequality : $r^2-2r-2 \leq rt+1$?

It can be easily seen that $(1,-4)$ is one such pair. I have tried upto $r=10$, for which there is no such $t$. But I somehow feel that for large enough $r$ we will have many such pairs $(r,t)$ satisfying the inequality. Is there a way to explicitly deduce such pairs?

Any help from anyone is appreciated


1 Answer 1


If $(r,t) \in A$ then $$ r^2-2r-2 \le r(r-5)+1=r^2-5r+1 \implies 3r \le 3 $$ hence $r \in \{0,1\}$. If $r=0$ then $t\le -5$ and $-2\le 1$. If $r=1$ then $t\le -4$ and $-4 \le t$. Hence $$ A=\{(1,-4)\} \cup \{(0,t): t\le -5\}. $$

  • $\begingroup$ Thank you very much for the answer. $\endgroup$
    – HARRY
    Mar 29, 2022 at 10:39

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