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I stumbled upon a solution to the so-called "The Math Problem of the Month" for October 2018. https://www.aus.edu/sites/default/files/PastProblems_October18.pdf

Now, I have to admit I am very confused by the last few steps.

It says there:

"If $n$ is a multiple of $3$, then $2^{7n}-1$ has the factor $2^{3}-1$"

How do you get the factor out of there? If you set $n=3$ for example,you get $2^{7*3}-1$, but you can't factor out $2^{3}-1$, so what is the method behind that argument?

"When $n$ is not a multiple of $3$, it is easy to see that $7$ does not divide $2^{7n}-1$"

Why? Because you can't extract the factor which you couldn't do in the first place?

"A similar argument shows that

There are $2*953$ values for $m= 1, 2$"

First, shouldn't it be $m=1,3$?

Second, shouldn't it be $2*476$ or am I completely missing something here?

I appreciate any help.

Physics Guy
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2 Answers2

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$2^7\equiv2\pmod7$

$2^{7n}=(2^7)^n\equiv2^n\pmod7$

Now $2^1\equiv2,2^2\equiv4,2^3\equiv1\pmod7$

lab bhattacharjee
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  • And how can you show the opposite,namely if $n$ is not a multiple of three, then $2^{7n}$ is not congruent to 1 mod 7 ? – Physics Guy Dec 18 '18 at 18:37
  • As shown in this answer, $2^{7(3a+b)} \equiv 2^{3a+b} \equiv (2^3)^a\cdot2^b \equiv 1^a\cdot2^b \equiv 2^b \pmod 7,$ so if $b=1$ or $2$ then $2^{7(3a+b)}$ has remainder $2^1$ or $2^4$ when divided by $7.$ – David K Dec 18 '18 at 18:54
  • @PhysicsGuy, Generalization : https://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Dec 19 '18 at 01:59
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In general, $$ x^r - 1 = (x^{r-1} + x^{r-2} + \cdots + x^2 + x + 1)(x - 1). $$

So $ x^r - 1$ has a factor $x - 1.$

Now let $n$ be divisible by $3,$ so $n = 3k,$ and set $x = 2^3$ and $r = 7k.$ Then $$ 2^{7n} - 1 = (2^3)^{7k} - 1 = x^r - 1, $$ which has a factor $x - 1 = 2^3 - 1.$

Is it "easy" to see that if $n$ is not divisible by $3$ then $2^{7n} - 1$ is not divisible by $7$? Yes, if you follow the modulo-$7$ arithmetic from this other answer to its logical conclusion. I don't think it is very well explained in the posted solution at https://www.aus.edu/sites/default/files/PastProblems_October18.pdf.

I agree the "$m = 1$ or $2$" part should be $m = 1,3.$ It is not so clear how the argument for either of these cases is "similar" to the argument already given.

David K
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