3
$\begingroup$

So as part of a former question I already proved that perfect cubes $\pmod 7$ can only have the remainders $0, 1$ and $6$ - Where the only cubes with remainder $6$ are cubes of the numbers of form $(7k + 3)$ or $(7k+6)$.

Do I now show that $(7k+3)^4 $ and $(7k+6)^4$ are $4 \pmod 7$ and $1\pmod 7$ respectively so $b^4$ can't fit this condition?

Sorry, I have read through a lot of similar questions but I'm still just not really getting it.

Also, why was modulo $7$ chosen? Modulo $13$ was also suggested (which leaves remainders of $1, 5, 8$ and $12$) but I'm wondering if it matters.

$\endgroup$
3
  • $\begingroup$ Not that it really matters, but $5^3\equiv 6\pmod 7$ also. $\endgroup$
    – user289958
    Commented Apr 3, 2016 at 10:19
  • 1
    $\begingroup$ @user188201 Thanks so much for picking that up! $\endgroup$
    – Inazuma
    Commented Apr 3, 2016 at 10:24
  • $\begingroup$ There are no solutions mod $13$. In fact, it seems that $13$ is the only prime that works. $\endgroup$
    – lhf
    Commented Jun 22, 2021 at 21:47

2 Answers 2

3
$\begingroup$

Modulo 7 may be helpful because only comparatively few remainders are cubes. For example, modulo 5 any reaminder is potentially a cube: $0^1\equiv 0$, $1^3\equiv 1$, $3^3\equiv 2$, $2^3\equiv 3$, $4^3\equiv 4\pmod 5$. The reason behind this again is that $7\equiv1\pmod 3$ and $5\not\equiv 1\pmod 3$. (Why is this a reason? The $p-1$ non-zero remainders modulo $p$ form a cyclic group of order $p-1$, and in such a cyclic group it is always possible to "take $n$th roots" of every element for any $n$ except when $n$ is a divisor of $p-1$.)

So as we are here dealing with both third and fourth powers, working modulo $13$ may be even more promising as $13\equiv 1$ both $\pmod 3$ and $\pmod 4$.

Now, possible remainders of cubes modulo 13 are: $0,1,5,8,12$. And possible remainder of fourth powers modulo 13 are: $0,1,3,9$. If we add $6$ to the latter we arrive at $6,7,9,2$ and observe that none of the values occurs among the list of possible cubes - so 13 was a lucky choice for our endeavor.

$\endgroup$
1
$\begingroup$

I thought I would leave this as a hint. You can take $b = 7k, 7k+1,7k+2,..., 7k+6$, and go through each case with your observation there.

$\endgroup$
2
  • $\begingroup$ I am just confused why they would suggest to find the remainders of perfect cubes if we needed to find the remainders of perfect fourth numbers? Or do we need to use both? $\endgroup$
    – Inazuma
    Commented Apr 3, 2016 at 10:08
  • $\begingroup$ Use both, I think. That's what I did it to my self before posting this. $\endgroup$
    – DeepSea
    Commented Apr 3, 2016 at 10:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .