How prove $a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$ Contains infinitely even numbers. 
let sequence $$a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$$
where $[x]$  is  the largest integer not greater than $x$
show that  $\{a_{n}\}$  Contains infinitely   even numbers.
also I guess contains infinitely odd numbers.

before I have ask this How prove this sequence $S_{n}=[2^n\cdot \sqrt{2}],n\in N$ contains infinitely many composite numbers
I found :
 $$a_{2}=[2\sqrt{2}]+[2\sqrt{5}]=2+4=6$$
$$a_{3}=[3\sqrt{2}]+[3\sqrt{5}]=4+6=10$$
$$a_{5}=[5\sqrt{2}]+[5\sqrt{5}]=7+11=18$$
$$a_{7}=[7\sqrt{2}]+[7\sqrt{5}]=9+15=24$$
$$a_{8}=[8\sqrt{2}]+[8\sqrt{5}]=11+17=28$$
$$a_{9}=[9\sqrt{2}]+[9\sqrt{5}]=12+20=32$$
$$a_{10}=[10\sqrt{2}]+[10\sqrt{5}]=14+22=36$$
$$a_{12}=[12\sqrt{2}]+[12\sqrt{5}]=16+26=42$$
$$a_{14}=[14\sqrt{2}]+[14\sqrt{5}]=19+31=50$$
$$a_{15}=[15\sqrt{2}]+[15\sqrt{5}]=21+33=54$$
and so on
 this problem is my found it. It seem this is interesting problem,and How prove it?
Thank you 

When you increase $n$ by one, $\lfloor \sqrt 2 n \rfloor$ increases by either $1$ (with frequency $2-\sqrt 2$ ) or $2$ (with frequency $\sqrt 2 - 1$), while $\lfloor\sqrt 5n \rfloor$ does by either $2$ (with frequency $3-\sqrt 5$) or $3$ (with frequency $\sqrt 5 - 2$).
If your sequence only contains finitely many odd (or even) terms, then it has to increase by $4$ at every step past a certain point.
This means that whenever $\lfloor\sqrt2 n\rfloor$ increases by $2$ (resp. $1$), $[\sqrt 5 n\rfloor$ increases by $2$ (resp. $3$).
However, the frequency at which those increments occur are all different. Therefore this is impossible, and so you must obtain infinitely many even / odd terms .
You can make this proof more precise and show that there can't be more than $4$ consecutive terms of the same parity.
