Divisible by?

Number Theory Level 3

For $P \in N$ , $3^{4P} - 2^{4P}$ is always divisible by.......

Both 5 & 13 None of these 13 15 5

1 solution

Zach Abueg
Aug 11, 2017

$\displaystyle \begin{aligned} 3^{4P} - 2^{4P} & = 81^P - 16^P & \small \color{#3D99F6} a - b \mid a^n - b^n \ \forall \ n \in \mathbb{N} \\ \implies 65 & \mid 81^P - 16^P \\ \implies 5 & \mid 81^P - 16^P \\ \implies 13 & \mid 81^P - 16^P \end{aligned}$

What is it that you have written in blue? I mean, what is its mathematical name? so that i could search and read more about it...

Ojasee Duble - 3 years, 10 months ago

Interestingly, it doesn't have a name. You can read about several proofs of it here . Basically, for any positive integers $a, b, n$ , $a^n - b^n$ is divisible by $a - b$ .

Zach Abueg - 3 years, 10 months ago

Ohh Ok... Thanks a lot :)

Ojasee Duble - 3 years, 10 months ago

@Ojasee Duble – Of course! :)

Zach Abueg - 3 years, 10 months ago

Divisible by?

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