Can $a_1+2a_2, a_2+2a_3, a_3+2a_4,…, a_n+2a_1$ all be powers of $2$ simultaneously?

Can the numbers (all positive integers) $$a_1+2a_2, a_2+2a_3, a_3+2a_4,..., a_n+2a_1$$ all be powers of $$2$$ simultaneously?

I tried to subtract the second from the first, and the third from the second, but i didnt approach anything. I tried to show it is divisible by some other number than $$2$$ as well, but am unsuccessful. Help please !

Well, it seems that this question is interesting, if the original question is solved, a generalised question-

Is there any natural number $$k$$ such that $$a_1+ka_2, a_2+ka_3,...a_n+ka_1$$ are all powers of $$2$$ ? If there is, what are the possible forms of $$k$$?

Consider the 2-valuation function $$v_2:\Bbb N\to \Bbb N$$, where $$v_2(x)$$ is the exponent of largest power of $$2$$ that divides $$x$$. For instance, $$v_2(24) = 3$$, since $$2^3\mid 24$$, but $$2^4\nmid 24$$.

Note that if $$v_2(x)\neq v_2(y)$$, then $$v_2(x+y)=\min(v_2(x), v_2(y))\leq v_2(x)$$.

We have $$a_1+2a_2 = 2^k$$ for some natural $$k$$. Because $$a_1<2^k$$, and $$2^k$$ is the smallest number for which $$v_2$$ gives the value $$k$$, we have $$v_2(2^k) >v_2(a_1)$$. However, by the note above, we must therefore have $$v_2(a_1) = v_2(2a_2)$$. Which is to say, $$v_2(a_2) = v_2(a_1) - 1$$.

Similarily, $$v_2(a_3) = v_2(a_2) - 1$$. And so on. Until we get to the final sum, $$a_n + 2a_1$$, which shows that $$v_2(a_1) = v_2(a_n) - 1$$. Putting all these together, we find that $$v_2(a_1) = v_2(a_1) - n$$, which is impossible.

• This is the most splendid and compact one. Kudos!! Any idea on how to attact the more general part? – Peter joshua Oct 9 at 11:57
• @Peterjoshua For even $k$, the same argument works (although the $-1$ that appears is actually $v_2(k)$, so that will have to change to be correct). Other than that, no, I don't have any idea. – Arthur Oct 9 at 11:58
• Thanks anyway. Btw have you ever been at the IMO? – Peter joshua Oct 9 at 12:02
• @Peterjoshua Once, over a decade ago. I performed below median, so no medal for me. – Arthur Oct 9 at 12:04
• Can ypu tell your name please? – Peter joshua Oct 9 at 12:07

If they're all positive integers, then each power of $$2$$ is at least $$4$$, and hence is divisible by $$4$$. This implies that $$4 \mid (a_i + 2a_{i+1}) \implies a_i \equiv 2a_{i+1} \mod 4 \implies a_i \equiv0 \text{ or }2 \mod 4.$$ This in turn implies $$a_i$$ is even. The same can be said for $$a_n$$ as well, due to the last relation. Thus, $$a_i$$ is even for all $$i$$.

But then, this leaves us open to infinite regression. Replace $$b_i = \frac{a_i}{2}$$, and the new $$b_i$$ numbers are now another smaller solution.

• Try The generalised??? Thanks btw ;) – Peter joshua Oct 9 at 11:48
• @Peterjoshua Not sure. My argument works for any even $k$. Certainly $a_i = 1$ for all $i$ works when $k = 3$, or indeed when $k + 1$ is a power of $2$. I'm not sure about other $k$s. – Theo Bendit Oct 9 at 11:49

Interesting question. The answer is no.

We first observe that solving the system $$a_1 + 2a_2 = m_1,a_2 + 2a_3 = m_2,\dots,a_{n-1} + 2a_n = m_{n-1},a_n + 2a_1 = m_n$$ is same as solving for the solution here:

\begin{align*} \left( \begin{matrix} 1 & 2 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 2 \\ 2 & 0 & 0 & \cdots & 0 & 1 \\ \end{matrix} \right) \left( \begin{matrix} a_1 \\ a_2 \\ \vdots \\ a_{n-1} \\ a_n \\ \end{matrix} \right) = \left( \begin{matrix} m_1 \\ m_2 \\ \vdots \\ m_{n-1} \\ m_n \\ \end{matrix} \right) \end{align*} One can prove that the matrix on the left is always invertible. This tells us that if a solution exists, it must be unique.

We now tackle the main problem, i.e. we want to find integer solutions for $$a_1,\dots,a_n$$ in the following equation: \begin{align*} \left( \begin{matrix} 1 & 2 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 2 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 & 2 \\ 2 & 0 & 0 & \cdots & 0 & 1 \\ \end{matrix} \right) \left( \begin{matrix} a_1 \\ a_2 \\ \vdots \\ a_{n-1} \\ a_n \\ \end{matrix} \right) = \left( \begin{matrix} 2^{k_1} \\ 2^{k_2} \\ \vdots \\ 2^{k_{n-1}} \\ 2^{k_n} \\ \end{matrix} \right) \end{align*} where $$k_1,\dots,k_n \in \mathbb{N}$$. If $$k_1 = \cdots = k_n = 0$$, then observe that $$a_1 = \cdots = a_n = \frac{1}{3}$$ is a solution. Since the solution is unique, there exists no other solutions. We now assume that $$k_i > 0$$ for some $$i$$, and assume WLOG that $$k_1 > 0$$. We can manipulate the equation to get the following: \begin{align*} a_1 + 2a_2 &= 2^{k_1} \\ a_1 - 4a_3 &= 2^{k_1} - 2^{k_2 + 1} \\ a_1 - 8a_4 &= 2^{k_1} - 2^{k_2 + 1} + 2^{k_3 + 2} \\ &\vdots \\ a_1 + (-1)^n2^{n-1}a_n &= \sum_{i=1}^{n-1}(-1)^{i+1}2^{k_{i}+i-1} \\ a_1 + (-1)^{n+1}2^na_1 &= \sum_{i=1}^{n}(-1)^{i+1}2^{k_{i}+i-1} \end{align*} We thus have: \begin{align*} (1 + (-1)^{n+1}2^n)a_1 &= \sum_{i=1}^{n}(-1)^{i+1}2^{k_{i}+i-1} \end{align*} This identity can be proved by induction. Now since $$k_1 > 0$$, and the remaining powers of $$2$$ have positive powers, we can deduce that the number on the right is even. Meanwhile, since $$n \geq 2$$, we have the coefficient of $$a_1$$ on the left is odd. This means that $$a_1$$ is definitely not an integer, so no integer solutions to this problem exist.