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$N=2^{1224}-1,A=2^{153}+2^{77}+1,B=2^{408}-2^{204}+1$

Show that $A$ and $B$ both divides $N$ .

This problem is of binomial theorem,

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    "Show that A and C..." well where is "C"? – 2017-02-11

2 Answers 2

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I can't figure out a way to phrase it as a slick invocation of the binomial theorem, but here is a relatively short argument that what you seek is true:

We have $$ \begin{align} A(2^{153}-2^{77}+1) &= (2^{153}+1+2^{77})(2^{153}+1-2^{77}) \\ &= (2^{153}+1)^2 - (2^{77})^2 \\ &= 2^{306} + 2\cdot 2^{153} + 1 - 2^{154} \\ &= 2^{306}+1 \end{align} $$ (and if you ever need to factor another number of the form $2^{4n+2}+1$, this is one way).

Now, since $1224=4\cdot 306$, let's set $P=2^{306}$, and then $$ P+1 \text{ divides } P^4-1 $$ is a simple matter of polynomial division.

Similarly $1224=6\cdot 204$, so set $Q=2^{204}$, and then $$ Q^2-Q+1 \text{ divides } Q^6-1 $$ is also just polynomial division.

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    To show that B divides N was easy but to show A divides N was hard . I want to ask you how did you knew that that approach will work , I mean that I could have never thought of the approach that you used( to show that A divides N ) , so my question is what made you think that you should try out that approach. – 2017-02-14
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    @Raunii: I started reducing modulo $A$ by hand, and after 3 or 4 steps I noticed that I was left with $-2^{918}-1$. Thus $2^{1224-918}+1$ must be a multiple of $A$. – 2017-02-15
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For $B|N$:

$2^{204}=x$. Then $N=x^6-1, B=x^2-x+1$. We can see $B|N$ because $N=(x^3-1)(x^3 + 1)= (x^3 -1)(x+1)(x^2-x+1)$