<s>[INST] proving there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$ by induction $n$ is a given positive integer.
how to prove  by using mathematical induction that there exists an integer $ k$ such that $ak^2+bk+c\equiv 0 \pmod{2^n}$, where $b$ is odd and at least one of $a, c$ is even
 [/INST] First check directly that the result holds for $n=1$. 
We do the induction step.  Let  $x$ be an integer such that
$$ax^2+bx+c\equiv 0\pmod{2^{n-1}}.$$
We would like to show that there is an integer $y$ such that $ay^2+by+c\equiv 0\pmod{2^n}$.  
If $2^n$ already divides $ax^2+bx+c$, we are finished.  So we need only worry about the case where $ax^2+bx+c\equiv 2^{n-1}\pmod{2^n}$.
Try for a $y$ of the shape $x+t$. So we want
$$a(x+t)^2+b(x+t)+c\equiv 0\pmod{2^n}.$$
But $ax^2+bx+c\equiv 2^{n-1}\pmod{p}$. So we want the difference 
$$[a(x+t)^2+b(x+t)+c]-[ax^2+bx+c]$$
to be congruent to $2^{n-1}$ modulo $2^n$. That is, we want to choose $t$ so that
$$a(2xt+t^2)+bt\equiv 2^{n-1}\pmod{2^n}.$$
Note that $t=2^{n-1}$ works, since $b$ is odd and $2t$ and $t^2$ are each divisible by $2^n$. 
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