ehsan2004 wrote: Let $a,b,c\in Z$ and $a$ be the even number and $b$ be the odd number. Show that for every integer $n$ there exist one positive integer $x$ such that $2^n\mid ax^2+bx+c$ Consider the values of the polynomial $f\left(x\right)=ax^2+bx+c$ for $x\in\left\{0;\;1;\;...;\;2^n-1\right\}$. These are $2^n$ values, and all of them have different remainders $\mod 2^n$; in fact, if, for two different numbers x and y from the set $\left\{0;\;1;\;...;\;2^n-1\right\}$, the values f(x) and f(y) would have the same remainder $\mod 2^n$, then f(x) - f(y) would be divisible by $2^n$, but since $f\left(x\right)-f\left(y\right)=\left(ax^2+bx+c\right)-\left(ay^2+by+c\right)=\left(x-y\right)\left(a\left(x+y\right)+b\right)$, and since the right factor a (x + y) + b is odd (since a is even and b is odd), this would mean that the number x - y would be divisible by $2^n$, what is impossible (since x and y are different integers between 0 and $2^n-1$). So the values of the polynomial $f\left(x\right)=ax^2+bx+c$ have $2^n$ different remainders $\mod 2^n$. But there are only $2^n$ possible remainders $\mod 2^n$, and thus all of these remainders are represented. Particularly, there must be some x such that f(x) has the remainder 0 $\mod 2^n$. But this means that $2^n\mid f\left(x\right)$; in other words, $2^n\mid ax^2+bx+c$. The problem is solved. darij