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Suppose $a$ and $b$ are odd positive integers. Must $v_2(a^n+b^n)$ reach a maximum value as $n$ varies ?

Clearly if we change $2$ for any other prime the problem becomes way easier, just use lifting the exponent lemma to determine $(p-1)^{p^n}+1^{p^n}$ is a multiple of $p^{n+1}$.

How do we solve it for $2$? I am looking for alternative solutions as I feel my solution might be overly complicated.

My solution:

It is easy to show using the lifting the exponent lemma that the order of $7\bmod 2^n$ is $2^{n-2}$ for $n\geq 3$, since the subgroup generated by $7$ $\bmod 2^n$ only contains elements congruent to $1$ and $7\bmod 8$ we conclude that $-1$ is inside this subgroup. We conclude that $v_2(7^n+1^n)$ is unbounded.

edit: turns out the solution was wrong, but I think the following works (thanks to Thomas Andrews).

Suppose $2^n | a^k+b^k$, suppose $b=ca\bmod 2^n$ with $n\geq 2$, we conclude $a^k \equiv -(c^k)a^k\implies c^k\equiv -1 \bmod 2^n$. Notice that $-1$ is not a perfect square $\bmod 2^n$ as odd squares are $q\bmod 4$. If $k$ is odd then the map $x\rightarrow x^k\bmod 2^n$ in the multiplicative group is injective, because the multiplicative group is isomorphic to $\mathbb Z_{2^{n-2}}\times \mathbb Z_2$. Since $-1^k\equiv -1\bmod 2^n$ it follows that if $x^k \equiv -1\bmod 2^n$ then $x\equiv -1\bmod 2^n$. We conclude that $a\equiv -b\bmod 2^n$.

So if $2^n| a^k+b^k$ then $2^n|a+b$ for $n\geq 2$. So $v_2(a^k+b^k)$ reaches a maximum-

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    yes, that is what I mean. – 2017-01-22
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    Also, did you mean $(p-1)^{p^n}+1^{p^n}$ is a multiple of $p^n$? – 2017-01-22
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    oh yeah.${}{}{}$ – 2017-01-22
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    Also, is the order of $7$ modulo $8$ really $4$? If you replace $5$ Indeed, $7^2\equiv 1\pmod{16}$, so $7$ is order $2$ modulo $2^4$. – 2017-01-22
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    good point, it should be $2^{n-2}$, but this is strong enough to conclude, $7$ isn't a primitive root, but it covers half of the odd residues. – 2017-01-22
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    Actually, it is $2^{n-3}$. – 2017-01-22
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    You will now need an actual argument for why $-1$ is in the group generated by $7$... – 2017-01-22
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    Indeed, it isn't true that $7^n\equiv -1\pmod{16}$ for some $n$, so it can't be true for $\pmod {2^k}$ for $k\geq 4$. – 2017-01-22
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    dang, it works for $5$, I wonder why it won't work for $7$. – 2017-01-22
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    oh of course, it because $4$ doesn't divide $7-1$... – 2017-01-22
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    $-1$ is not a perfect power of anything other than $-1$ modulo $2^n$ . – 2017-01-22
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    @ThomasAndrews Oh, good point. – 2017-01-22
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    So then, we can do as follows right? suppose $b\equiv ca \bmod 2^n$, suppose $a^k \equiv -b^k \bmod 2^n$, then $a^k\equiv -c^k a^k \implies c^k \equiv -1 \bmod 2^n \implies c\equiv -1 \bmod 2^n \implies a\equiv-b \bmod 2^n$, which is clearly impossible for all $n$. – 2017-01-22
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    see new answer http://math.stackexchange.com/questions/2108240/prove-or-disprove-that-exists-a-b-c-in-mathbbz-forall-n-in-mathbbz/2108860#2108860 which appears to settle this – 2017-01-22
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    @WillJagy oh, that's simpler than my group-thoeretic aproach, although I think both work. in fact I was going to use this to answer that question, but I couldn't find it later. Thanks. – 2017-01-22
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    If you have large amounts of questions/answers to remember, including things that might be deleted, there is a "favorites" option on one's activity. If there are too many, it helps to make an ordinary text file on home computer. I have a file I call Topics.txt , for a few dozen topics that i have answered multiple times, i have the url addresses of several answers that I thought were good, not necessarily mine. When I want, I can paste in several of these into a new answer, very little additional effort. – 2017-01-22
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    @WillJagy that sounds like good advice. Thank you, I will do that. – 2017-01-22

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