# 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

Note by Zakir Husain
8 months ago

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

- 7 months, 4 weeks ago

Nice!

- 8 months ago