Bump bump bump

As I know, in contest was using the version then $P(x)=x(x+1)/2$. Here is a solution for problem in #1 from Kvant no 5. I will write only case with $deg P \geq 2$, because if $P(x)=ax+b$ it is not hard. The answer is yes. Suppose a contrewise. Then there exist some number $m_1$ such a all numbers $m>m_1$ are representable in such form. Lemma: let $R(x)$ be a nonconstant polynom. Then for all $d>0$ there exist $x_0$ such a in any segment with length $d$ on the right side of $x_0$ there are at most $x$ such a $R(x)$ is power of $2$. Proof of lemma: if leading coefficient of $P$ is negative, then it is obvious. In other case we can choose $x_0$ such a on inteval $[x_0;+\infty)$ $R(x)$ is increases and $R(x+d)/R(x)<2$. This works. Return to problem. It is known that coefficients of $P$ are rational. Then for some natural number $N$ we have $Q(x)=N \cdot P(x) \in Z[x]$. Let $Q(x+t)-Q(x)=tQ'(x)+t^2S(x)+...$ be a polynom of $t$. Then $deq Q' \geq 1$. Let for some natural $m_0$ (integer) number $Q'(m_0) \not = 0$. Let $r$ be number such a $2^r \not | Q'(m_0)$. Then for all $m=m_0+t \cdot 2^r$ also $2^r \not |Q'(m)$. There exist $x_0>0$ such a for all $x \in [x_0;+\infty)$ we have $P(x+1)-P(x)>2^r$ and $P(x)>P(y)$ for all $0 \leq y < x$. Now we will look at $m_0<m \equiv_{2^r} m_0$ such a $m>m_0,x_0$ and $P(m)$ is representable in form $P(k)-2^n$. Then $k>m$. Let's estimate the difference $d=k-m$. We have $P(m+d)-P(m)=2^n$. Then $Q(m+d)-Q(m)=N \cdot 2^n$ and $dQ'(m)+d^2S(m)+...=2^n \cdot N$. So, $d|2^n \cdot N$. (*) In other hand, let's prove that $2^r \not | d$. Suppose a contrewise. Let $d=2^ts$ with $t \geq r$ and $2 \not | s$. In equality (*) $2^{t+r}|dQ'(m)$ and $2^{t+r}|2^{2t}|d^2|d^3|...$. It is enough to prove that $2^{t+r}|2^n \cdot N$. Really, $2^n=P(m+d)-P(m)=\left ( P(m+d)-P(m+d-1) \right ) + \left ( P(m+d-1)-P(m+d-2) \right ) + ... + \left ( P(m+1)-P(m) \right ) > d \cdot 2^r \geq 2^t \cdot 2^r$. So, $n \geq t+r$. So, $d$ is divisor of $2^n \cdot N$ which is not divisible by $2^r$. Then $d<d_0=2^r \cdot N$. From this we get that for all $m_0<m \equiv_{2^r} m_0$ at least one of $d_0$ polynomials $P(x+1)-P(x),P(x+2)-P(x),...,P(x+d_0)-P(x)$ equals when $x=m$ value, which equals to power of $2$. Segment $A=[m;m+d_0 \cdot 2^r]$ has $d_0+1$ values in form $m_0+t \cdot 2^r$. By this reason at least one of these polynomials $P_1(x)$ on segment $A$ has at least two points $a,b$ such a $P_1(a), P_1(b)$ are both powers of $2$. But if $m$ is suffieciently large, then there are contradicts the lemma. So, the problem is solved.