Prove or disprove that $\exists a,b,c\in\mathbb{Z}_+\ \forall n\in\mathbb{Z}_+\ \exists k\in\mathbb{Z}_+\colon\ a^k+b^k+c^k = 0(\mathrm{mod}\ 2^n)$

Question

Prove or disprove that there exist a triplet of positive integers $(a,b,c)$ with $\mathrm{gcd}(a,b,c)=1$ such that for any positive integer $n$ there exist such positive integer $k$ that $$ a^k+b^k+c^k = 0(\mathrm{mod}\ 2^n) $$

First thing that I've noticed is that one and only one of the numbers $a,b,c$ is even, otherwise either all of them are even witch contradicts the condition $\mathrm{gcd}(a,b,c)=1$ or even amount of them are even which would mean that $a^k+b^k+c^k = 1(\mathrm{mod}\ 2)$ for any $k$ and so $a^k+b^k+c^k \neq 0(\mathrm{mod}\ 2^n)$. So without loss of generality we can assume that $c$ is even and $a,b$ are odd.

Now since $c$ is even we have $c^k = 0 (\mathrm{mod}\ 2^n)$ for any large enough $k$ (definitely for all $k>n$). This means that if for any positive integer $n$ there exist such positive integer $k$ that $$ a^k+b^k = 0(\mathrm{mod}\ 2^n) $$ then the problem is solved.

Other than that I have nothing, so I would really appreciate any help. Thank you very much in advance!

Answer 1

Your statement can be disproved. Note that if $k$ is even then $a^k=b^k=1(4)$ and hence $a^k+b^k=2(4)$ which means that $a^k+b^k\neq 0(2^n)$ for all $n>1$. Now let $k$ be odd. Then we have $$ a^k+b^k = (a+b)(a^{k-1}+a^{k-2}b+\ldots+ab^{k-2}+b^{k-1}) $$ The second multiple here is the sum of odd amount of odd numbers which means it's odd. Thus $a^k+b^k$ is divisible by $2^t$ as long as $a+b$ is divisible by $2^t$. So for large enough $n$ the number $a^k+b^k$ is not divisible by $2^n$ for any $k$.

Thus we can conclude that

for any triplet $(a,b,c)$ with $\mathrm{gcd}(a,b,c)=1$ there is such $n$ that for all $k$ $$ a^k+b^k+c^k \neq 0(\mathrm{mod}\ 2^n) $$