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In the set of numbers generated by $2^a - 2^j$ (where both variables are integers and $a >= j$), it appears that the numbers which are divisible by 3 appear when $a - j$ is even. The sequence starts with $2^0 - 2^0$, $2^2 - 2^0$, $2^3 - 2^1$, $2^4 - 2^2$... However, I don't know why this is the case. Additionally, it would be helpful to have a formula or procedure to find these numbers, instead of just brute force.

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    Hint: rewrite as $2^j\times (2^{a-j}-1)$. So, it comes down to asking "for which $n$ is $2^n-1$ divisible by $3$". – lulu Apr 12 '23 at 13:29
  • Hint: compare the congruence classes of $a,2a$ for $a\in [0,1,2]$... – abiessu Apr 12 '23 at 13:30
  • I recommend you calculate $2-1, 2^2-1, 2^3-1, 2^4-1$ and $2^5-1$ modulo 3 the pattern should start to become clear and you form a conjecture from there. – CyclotomicField Apr 12 '23 at 13:45
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    If $(a,n)=1$ then $\bmod n!:\ a^j \equiv a^k \iff a^{j-k}\equiv1 \iff j \equiv k \pmod{!{\rm ord}_n a}$ by the linked Order Theorem. Here $,\bmod 3!:\ 2\equiv -1,$ has order $\color{#c00}2$ so $,2^j\equiv 2^k\iff \color{#c00}2\mid j-k\ \ $ – Bill Dubuque Apr 12 '23 at 19:01

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$2^k$ is never divisible by $3$, so $2^k \equiv \pm 1 \bmod 3$. Thus $2^a-2^j$ is divisible by $3$ only if $2^a \equiv 2^j \bmod 3$.

Note that $2^1 \equiv -1 \bmod 3$, and $2^2 \equiv 1 \bmod 3$. From this we can see that $2^{2n}=(2^2)^n \equiv 1 \bmod 3$ and $2^{2n+1}=2\cdot (2^{2n}) \equiv -1 \bmod 3$

Thus, differences of powers of $2$ are divisible when their exponents have the same parity, i.e. $a\equiv j \bmod 2$

Keith Backman
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