Prove this?

Number Theory Level 2

If $k = 2^p - 1$ , where $p >1$ , and $N = k^2 - 1$ , is $2^{p + 1}$ a factor of $N$ ?

In your solution, provide a complete proof

No Yes

2 solutions

Chew-Seong Cheong
Jun 5, 2019

$\begin{aligned} N & = k^2 - 1 \\ & = (k+1)(k-1) & \small \color{#3D99F6} \text{As }k = 2^p - 1 \\ & = 2^p\color{#3D99F6}(2^p - 2) & \small \color{#3D99F6} \text{Note that }p > 1 \\ & = 2^p \cdot \color{#3D99F6}2(2^{p-1} - 1) \\ & = {\color{#D61F06} 2^{p+1}}(2^{p-1} - 1) \end{aligned}$

Yes , $\color{#D61F06}2^{p+1}$ is a factor of $N$ .

Can you elaborate on the last step after stating that p > 1? Like how did you come to the fact that $2^p (2^p - 2) = 2^{p+1} (2^{p-1} - 1)$

Syed Hamza Khalid - 2 years ago

Take a 2 common from the bracket

Mr. India - 2 years ago

U mean factorize $2^p - 2$ ? How to do that

Syed Hamza Khalid - 2 years ago

I have added a line. Hope that it helps.

Chew-Seong Cheong - 2 years ago

Sir, can you please post a solution for this: https://brilliant.org/problems/confusing-question-no-way-out/ ; posted by @Syed Hamza Khalid

I am unable to understand it. Tbh, your solutions are the best; I always understand it thx to your amazing talent and use of colours

Jake Tricole - 2 years ago

Oh yes sure thanks a lot

Syed Hamza Khalid - 2 years ago

Vin Benzin
Jun 5, 2019

Brilliant - but there's a small typo; you should have $2^{p-1}-1$ at the end.

Chris Lewis - 2 years ago

Prove this?

2 solutions

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