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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 $k$?

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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.

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  • $\begingroup$ This is the most splendid and compact one. Kudos!! Any idea on how to attact the more general part? $\endgroup$ – Peter joshua Oct 9 at 11:57
  • $\begingroup$ @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. $\endgroup$ – Arthur Oct 9 at 11:58
  • $\begingroup$ Thanks anyway. Btw have you ever been at the IMO? $\endgroup$ – Peter joshua Oct 9 at 12:02
  • $\begingroup$ @Peterjoshua Once, over a decade ago. I performed below median, so no medal for me. $\endgroup$ – Arthur Oct 9 at 12:04
  • $\begingroup$ Can ypu tell your name please? $\endgroup$ – Peter joshua Oct 9 at 12:07
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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.

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  • $\begingroup$ Try The generalised??? Thanks btw ;) $\endgroup$ – Peter joshua Oct 9 at 11:48
  • $\begingroup$ @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. $\endgroup$ – Theo Bendit Oct 9 at 11:49
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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.

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