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I understood everything up until "BUT". My question is how are the factors on the right side for the bolded line of equation under the word "BUT" obtained? I tried to get these factors using property (5), but I was unsuccessful, can someone please show me how to get those factors? Thanks in advance.

This is example 2 from "Problems of Number Theory in Mathematical Competitions" by You Hong-Bing on page 2.

Some properties the example will be using are:

(1) If $b\mid c$, and $c\mid a$, then $b\mid a$, that is, divisibility is transitive.

(5) If $n$ is a positive odd number, then

$x^n-y^n= (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$

The problem, example 2 is as follows:

Let $\mathbf{m > n}$ and $\mathbf{n\ge 0}$, show that $\mathbf{(2^{2^n}+1)\mid(2^{2^m}-1)}$

The following is the proof from the book:

Take $x=2^{2^{n+1}}$,$\,\,y=1$ in factorization (5), and substitute $n$ by $2^{m-n-1}$

WE GET:

$2^{2^m}-1 = (2^{2^{n+1}}-1) \left[(2^{2^{n+1}})^{2^{m-n-1}-1} +\ldots + 2^{2^{n+1}}+1 \right]$

THUS,

$(2^{2^{n+1}}-1) \mid (2^{2^m}-1)$

BUT,

$\mathbf{(2^{2^{n+1}}-1) = (2^{2^n} - 1)(2^{2^n} + 1)}$
This is the equation I'm having trouble with so everything else after this didn't make much sense to me

HENCE,

$(2^{2^n} + 1)\mid(2^{2^{n+1}} - 1)$

FURTHER, BY PROPERTY (1) WE HAVE:

$(2^{2^n} +1)\mid(2^{2^m} - 1)$

REMARK: sometimes it is difficult to prove $b\mid a$ directly when dealing with divisibility problems. Therefore, we can attempt to choose an "intermediate number" $c$ and prove $b\mid c$ and $c\mid a$ first, then use property (1) of divisibility to deduce the conclusion.

Hunter Batley
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schoolwithschool
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2 Answers2

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Note that $2^{2^{n+1}}=2^{(2^{n+1})}=\left(2^{2^n}\right)^2$
So $2^{2^{n+1}}-1=\left(2^{2^n}\right)^2-1$
and you can factor it as the difference of squares into

$\left(2^{2^n}-1\right)\left(2^{2^n}+1\right)$

Ross Millikan
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More generally, $x^{2^m}-1 =(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1) $. For this, put $x = 2$.

Proof.

For $m=1$ this is $x^{2^1}-1 =(x-1)\prod_{n=0}^{0}(x^{2^n}+1) $ or $x^2-1 =(x-1)(x+1) $.

If it is true for $m$ then

$\begin{array}\\ (x-1)\prod_{n=0}^{(m+1)-1}(x^{2^n}+1) &=(x-1)\prod_{n=0}^{m}(x^{2^n}+1)\\ &=(x-1)(x^{2^m}+1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\ &=(x^{2^m}+1)(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\ &=(x^{2^m}+1)(x^{2^m}-1) \qquad\text{(induction hypothesis)}\\ &=x^{2^{m+1}}-1 \qquad (x^a+1)(x^a-1)=x^{2a}-1 \text{ and } 2(2^m) = 2^{m+1}\\ \end{array} $

marty cohen
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