how to prove the identity that $2^{ab}-1=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+.....+2^a+1)$

Question

I am really confused. The identity is given but I really want to know why$?$

Show that if $2^n -1$ is a prime , then $n$ is a prime . [Hint : Use the identity]

$$2^{ab} -1 = (2^a-1)\cdot(2^{a(b-1)} + 2^{a(b-2)} + \cdots + 2^a +1 )$$

Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. — José Carlos Santos, Dec 07 '19 at 08:31

score 4 · Answer 1 · answered Dec 07 '19 at 08:32

4

You know that $x^b-1=(x-1)(x^{b-1}+x^{b-2}+\cdots+1)$, right?! Now, replace $x$ with $2^a$.

answered Dec 07 '19 at 08:32

José Carlos Santos

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score 2 · Answer 2 · answered Dec 07 '19 at 09:32

A different way for understanding this identity is to use binary notation.

Let us write $2^{ab}-1$ with binary digits, taking for illustration the particular case $a=3$ and $b=6$ (it works in the same way in the general case) :

$$\underbrace{111111111111111111}_{ab \ \text{digits}}=\underbrace{111}_{a \ \text{digits}} \times \underbrace{1001001001001001}_{2^{15}+2^{12}+2^{9}+2^{6}+2^{3}+1}$$

score 2 · Accepted Answer · edited Jun 12 '20 at 10:38

Consider the following G.P :

$$1,2^a , 2^{2a} ,\cdots ,2^{a(b-2)} , 2^{a(b-1)} $$

We use the Formula to find it's sum :

$$S = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}$$ Or

$${\color{#d05}{1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)} = \dfrac{1\cdot(2^{ab}-1)}{2^a-1}}}$$

Rearranging this gives us the deisred result :

how to prove the identity that $2^{ab}-1=(2^a-1)(2^{a(b-1)}+2^{a(b-2)}+.....+2^a+1)$

$$2^{ab} -1 = (2^a-1)\cdot(2^{a(b-1)} + 2^{a(b-2)} + \cdots + 2^a +1 )$$

3 Answers3

$${\color{#20f}{({2^a-1})(1+2^a + 2^{2a}+\cdots +2^{a(b-2)} + 2^{a(b-1)}) = {(2^{ab}-1)}}}$$