Ahiles wrote: Let $ a,b$ be real nonzero numbers, such that number $ [an + b]$ is an even integer for every $ n \in \mathbb{N}$. Prove that $ a$ is an even integer. A very good problem . To solve this problem ,we need a lemma . Lemma If $ a$ is a irrational number then $ \{na\}$ is dense in $ (0,1)$ Proof of this lemma quite hard ,it can be found in a number theory book . Now consider problem . $ x_n = an + b$ Suppose $ [a_n]\equiv 0 (\mod 2 ),\forall n\in N$ then $ [x_{n + 1}]\geq [x_n] + 2$ We only need to consider $ a\in [0,2)$ First step we will prove that a is a rational number. Suppose a is a irrational number then $ x_{n + 1} - x_n = a$ Exist an integer n_0 such that $ \{an_0 + b\} < 2 - a$ So $ [x_{n_0 + 1}] = [an_0 + b + a] = [x_{n_0}] + [\{an_0 + b\} + a] < [x_{n_0}] + 2$ It gives contradiction . So a must be a rational number. We also only to consider $ b\in [0,2)$ Let $ a = \frac {p}{q},\gcd (p,q) = 1$ and $ q > 1$ ${ [x_n] = [\frac {np}{q}] + [\frac {np}{q}} + b]$ Take $ n = kq$ then easy to check that $ p\equiv 0 (\mod 2)$ Case 1 $ b\in [0,1)$ Exist an integer n such that $ np\equiv 1 (\mod q )$ Because $ p\equiv 1 (\mod 2)$ so $ [\frac {np}{q}]\equiv 1 (\mod 2)$ $ \Rightarrow [\frac {1}{q} + b] = 1,3$ Because $ 0 < \frac {1}{p} + b < 2$ so $ [\frac {1}{p} + b] = 1$ So $ 1 > b > 1 - \frac {1}{p}$ It follows that $ [b + \frac {np}{q}] = 1$ $ \Rightarrow [\frac {np}{q}]\equiv 0 (\mod 2 ),\forall n\in N$ It gives $ q = 1$ Case 2 :$ 1 < b < 2$ Consider $ b_1 = b - 1$ and $ y_n = x_n - 1$ then $ [y_n]\equiv 1 (\mod 2)$ Prove as case 1 ,it gives $ q = 1$ So $ a = p = 2k$ and $ [b]\equiv 0(\mod 2)$ Problem claim .

TTsphn wrote: Lemma If $ a$ is a irrational number then $ \{na\}$ is dense in $ (0,1)$ Proof of this lemma quite hard [...] This is not hard! See http://www.mathlinks.ro/Forum/viewtopic.php?t=27475 (scroll down for grobbers proof, post #16). The hard one is http://www.mathlinks.ro/viewtopic.php?t=172978 .

The lemma $ \{na\}$ is dense in $ (0,1)$ with $ a$ is a irrational is true.

But not that it is _hard_.

I tihnk I have reached a solution.Hope it is OK. Let a=2k+r (k integer and with r non negative and r smaller than 2). Then if we substitue in the original question n=1 we have that [2k+r+b]=2m (m is an integer) [r+b]=2(m-k) =2t ....by doing the same thing we obtain that [nr+b]=2P P is an integer. Suppose that b=2h+q q q can be in [0,1) or in [1,2) if it is in [0,1) we say that it "adds 0 to the equation" . Then we have [nr]=2P' then if n=1 we have [r]=0 so there will be an n such that [nr]=1 if r>0 so we have that r=0 so a=2k as we wish to prove. Then if q is in [1,2) we say that "it adds 1 to the equation" so [nr+1]=2P then repating the procedure we have [r]=1 then we can se that there will be an n such that [nr+1]=2L+1 ( you can start doing things like these: the numbersin [1,1.5) whem we set n=2 will contradict the condition of the problems, if you continue doing that it is easily checked that they can't be an answer to the problem) So we conclude that a=2k Excellent problem! I hope you understand the idea and I wish peple can improve the parts where are not completly proven Daniel[/code]

TTsphn wrote: Ahiles wrote: ${ [x_n] = [\frac {np}{q}] + [\frac {np}{q}} + b]$ Take $ n = kq$ then easy to check that $ p\equiv 0 (\mod 2)$ Case 1 $ b\in [0,1)$ Exist an integer n such that $ np\equiv 1 (\mod q )$ Because $ p\equiv 1 (\mod 2)$ so $ [\frac {np}{q}]\equiv 1 (\mod 2)$ $ \Rightarrow [\frac {1}{q} + b] = 1,3$ Because $ 0 < \frac {1}{p} + b < 2$ so $ [\frac {1}{p} + b] = 1$ So $ 1 > b > 1 - \frac {1}{p}$ It follows that $ [b + \frac {np}{q}] = 1$ $ \Rightarrow [\frac {np}{q}]\equiv 0 (\mod 2 ),\forall n\in N$ It gives $ q = 1$ why ? show me please