Greatest $n$ such that $2^n \mid 63^{2018}-1$

Question

Find the greatest $n$ such that $2^n\mid 63^{2018}-1$

I got to this question and don't know how to solve it, I'm not familiar with questions in this type. I feel that we should get how many factors of $2$ we have in the $RHS$, so it would make it easier. I thought too in:

$63^{2018}-1=\underbrace{(63^{1009}-1)}_x \underbrace{(63^{1009}+1)}_y \implies n=\nu_2(xy)=\nu_2(x)+\nu_2(y)$, where $\nu_p(a)$ is the p-adic valuation. But $x$ and $y$ are not looking good.

I would be glad for help.

On a whim, taking $63=64-1$, I see $(64-1)^{2018}-1=(2^{12}-2^7+1)^{1009}-1$, but I haven't gotten much further than that. — abiessu, Jun 18 '22 at 03:39
Perhaps a further step would be $(k2^7+1)^{1009}-1=\dots+k^22^{14}\binom{1009}2+1009k2^7$, but do you know if $2^8$ divides the original? — abiessu, Jun 18 '22 at 03:47
Follows immediately by LTE as in the linked dupes. This is the master theorem for problems like this so one should learn it asap. — Bill Dubuque, Jun 18 '22 at 09:11

John Omielan · Accepted Answer · 2022-06-18T04:50:36.690

4

As stated in abiessu's comment, note that $63 = 64 - 1$. Thus, using what you did with modulo $64$ gives that

$$x = 63^{1009}-1 \equiv (-1)^{1009} - 1 \equiv -2 \equiv 62 \pmod{64} \tag{1}\label{eq1A}$$

Thus, $\nu_2(x) = 1$. Next, we have

$$y = 63^{1009} + 1 = (63 + 1)(\color{blue}{63^{1008} - 63^{1007} + \ldots - 63 + 1}) \tag{2}\label{eq2A}$$

Note each term in $\color{blue}{\text{blue}}$ is odd, and that there are $1009$ (which is an odd #) of them, so their sum would also be odd. This means that $\nu_2(y) = \nu_2(63 + 1) = 6$. Altogether, we therefore get

$$n = \nu_2(x) + \nu_2(y) = 1 + 6 = 7 \tag{3}\label{eq3A}$$

Update: As indicated in Jyrki's comment, the Lifting-the-exponent lemma could have been used instead to get that $\nu_2(x) = \nu_2(63-1) = 1$ and $\nu_2(y) = \nu_2(63+1) = 6$. Actually, I could've used that lemma with the original expression to more directly get $\nu_2(63^{2018} - 1) = \nu_2(63-1) + \nu_2(63+1) + \nu_2(2018) - 1 = 1 + 6 + 1 - 1 = 7$.

edited Jun 18 '22 at 04:50

answered Jun 18 '22 at 03:53

John Omielan

47,976

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Wow, $(2)$ is clever. By the way, does this factorization have a "name"? Thank you for the help! – Ian Barquette Jun 18 '22 at 04:01
@IanBarquette You're welcome for the help. Regarding there being a specific name for the factorization, i.e., where for odd positive integer $m$ we have $x^m + y^m = (x + y)(x^{m-1} - x^{m-2}y \ldots - xy^{m-2} + y^{m-1})$, I don't know of any. – John Omielan Jun 18 '22 at 04:05
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@IanBarquette I think that lifting the exponent lemma is the most general umbrella idea. At least one of them. +1, of course. – Jyrki Lahtonen Jun 18 '22 at 04:24
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@JyrkiLahtonen Thanks for mentioning that. I forgot about that lemma when I wrote the answer, but I could have used it instead to get both $\nu_2(x)$ and $\nu_2(y)$. I've added an update to my answer about this. – John Omielan Jun 18 '22 at 04:40
Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 18 '22 at 09:11
@BillDubuque I'm aware of that site policy announcement. Note I suspected this was a duplicate and searched for an appropriate question to use to vote to close as a duplicate, but I couldn't find anything, with this likely being because I was checking just for the specific problem of $2^n\mid 63^{2018}-1$. – John Omielan Jun 18 '22 at 09:15
@John Searching via SE for lifting exponent yields 200 matches, including the linked dupes (and some of your answers using LTE). – Bill Dubuque Jun 18 '22 at 09:19

score 0 · Answer 2 · answered Jun 18 '22 at 04:23

Alternative approach:

$\underline{\text{Lemma 1}}$
$4 \mid 63^n - 1 \iff 4 \mid 63^{n+2} - 1.$
$4 \mid 63^n + 1 \iff 4 \mid 63^{n+2} + 1.$
Proof
For both assertions, it is sufficient to show that
$~\displaystyle 4 \mid \left(63^2 - 1\right) = 4 \times 992.$

$\underline{\text{Lemma 2}}$
$64 \mid 63^n + 1 \iff 64 \mid 63^{n+4} + 1.$
$128 \mid 63^n + 1 \iff 128 \mid 63^{n+4} + 1.$
Proof
For both assertions, it is sufficient to show that
$~\displaystyle 128 \mid \left(63^4 - 1\right) = 128 \times 123070.$

Let $A = 63^{1009} - 1, ~B = 63^{1009} + 1 \implies A \times B = 63^{2018} - 1.$

$2 \mid A.$
By Lemma 1, since $4 ~\nmid (63^1 - 1), ~4 ~\nmid ~(63^{1009} -1).$

$64 \mid (63^1 + 1)$ and $128 \nmid (63^1 + 1).$
Therefore, by Lemma 2, $~64 \mid B~$ and $~128 \nmid B.$

Therefore,

the largest exponent $~\alpha~$ such that $~2^\alpha \mid A~$ is $~\alpha = 1.$
the largest exponent $~\beta~$ such that $~2^\beta \mid B~$ is $~\beta = 6.$

Therefore,

$$2^7 | (A \times B) = 63^{2018} - 1, ~~\text{and}~~ 2^8 \nmid (A \times B).$$

Greatest $n$ such that $2^n \mid 63^{2018}-1$

2 Answers2