pohoatza wrote: Find all nonzero subsets $A$ of the set $\left\{2,3,4,5,\cdots\right\}$ such that $\forall n\in A$, we have that $n^{2}+4$ and $\left\lfloor{\sqrt{n}\right\rfloor}+1$ are both in $A$. $n\in A$ $\Rightarrow$ $m=n^{2}+4\ \in A$ $\Rightarrow$ $\left\lfloor{\sqrt{m}\right\rfloor}+1\ \in A$. But, for $n>1$, $\left\lfloor{\sqrt{n^{2}+4}\right\rfloor}+1=n+1$. So, $n\in A$ $\Rightarrow$ $n+1\in A$ $\Rightarrow$ $A=\{a,a+1,a+2, ...\}$ and we must have $\left\lfloor{\sqrt{a}\right\rfloor}+1\geq a\geq 2$, which means $a=2$ Then the only nonempty subset is $\left\{2,3,4,5,\cdots\right\}$ itself. -- Patrick

Quote: and we must have $ \left\lfloor{\sqrt {a}\right\rfloor} + 1\geq a\geq 2$, which means $ a = 2$ how?

$ a \ge 2 \implies \sqrt{a} \ge 1 \implies \lfloor\sqrt{a}\rfloor \ge 1 \implies \lfloor\sqrt{a}\rfloor +1 \ge 2$ So, $ 2 \ge a \ge 2 \implies a = 2$

Math4tots wrote: $ a \ge 2 \implies \sqrt {a} \ge 1 \implies \lfloor\sqrt {a}\rfloor \ge 1 \implies \lfloor\sqrt {a}\rfloor + 1 \ge 2$ So, $ 2 \ge a \ge 2 \implies a = 2$ It's wrong. For example: $ x>=a>=2$ and $ x=100,$ so $ x>=2 =>a=2???$ We have $ a<=\left\lfloor{\sqrt{a}\right\rfloor}+1<=\sqrt{a}+1 => a<=\sqrt{a}+1=>a>=2$

earldbest wrote: Quote: and we must have $ \left\lfloor{\sqrt {a}\right\rfloor} + 1\geq a\geq 2$, which means $ a = 2$ how? $ A = \{a,a + 1,a + 2, ...\}$ for some integer $ a\ge 2$ $ a\in A$ $ \implies$ $ [\sqrt a] + 1\in A$ $ \implies$ $ [\sqrt a] + 1\ge a$ $ \sqrt a\ge[\sqrt a]$ so $ \sqrt a + 1\ge a$ and so $ a - \sqrt a - 1\le 0$ and so $ \sqrt a\le \frac {1 + \sqrt 5}2$ $ \implies$ $ a\le \frac {3 + \sqrt 5}2$ $ \implies$ $ 2.618\ge a\ge 2$ So $ a = 2$ Is there something you dont understand ?

pco wrote: earldbest wrote: Quote: and we must have $ \left\lfloor{\sqrt {a}\right\rfloor} + 1\geq a\geq 2$, which means $ a = 2$ how? $ A = \{a,a + 1,a + 2, ...\}$ for some integer $ a\ge 2$ $ a\in A$ $ \implies$ $ [\sqrt a] + 1\in A$ $ \implies$ $ [\sqrt a] + 1\ge a$ $ \sqrt a\ge[\sqrt a]$ so $ \sqrt a + 1\ge a$ and so $ a - \sqrt a - 1\le 0$ and so $ \sqrt a\le \frac {1 + \sqrt 5}2$ $ \implies$ $ a\le \frac {3 + \sqrt 5}2$ $ \implies$ $ 2.618\ge a\ge 2$ So $ a = 2$ Is there something you dont understand ? oh, I see. Thanks!