It's for all $n$

If $a+b=c$ where $c\in\mathbb Z^+;b\in\mathbb Z^+$ and $a\in\mathbb Z^+$ then for all $n\in\mathbb Z^+$ either $a^n+b^n$ or $a^n-b^n$ (not both at the same time) will be divisible by $c$

Proof $a+b=c..........[1]$

Case 1 ( $n$ is odd) $b\equiv b \pmod c$ $b^n\equiv b^n \pmod c..........[A_1]$ $a\equiv a \pmod c ..........[A_2]$ From $[1]$ and $[A_2]$ : $a\equiv -b \pmod c$ $a^n\equiv (-b)^n \pmod c$ $a^n\equiv -b^n \pmod c ..........[A_3]$ Adding $[A_1],[A_3]$ $a^n+b^n\equiv 0 \pmod c$ $\therefore c|(a^n+b^n)$

Case 2 ( $n$ is even) $b\equiv b \pmod c$ $b^n\equiv b^n \pmod c ..........[B_1]$ $a\equiv a \pmod c..........[B_2]$ From $[1]$ and $[B_2]$ : $a\equiv -b \pmod c$ $a^n\equiv (-b)^n \pmod c$ $a^n\equiv b^n \pmod c ..........[B_3]$ Subtracting $[B_1],[B_3]$ $a^n-b^n\equiv 0 \pmod c$ $\therefore c|(a^n-b^n)$

Bonus :

Prove or disprove the converse of the above theorem

#NumberTheory

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Comments

Nice!

Yajat Shamji - 11 months, 2 weeks ago

In more general for KC i.e a+b=kc

A Former Brilliant Member - 11 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

It's for all nnn

Comments

It's for all $n$