It's for all nn

If a+b=ca+b=c where cZ+;bZ+c\in\mathbb Z^+;b\in\mathbb Z^+ and aZ+a\in\mathbb Z^+ then for all nZ+n\in\mathbb Z^+ either an+bna^n+b^n or anbna^n-b^n (not both at the same time) will be divisible by cc

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

Case 1 (nn is odd) bb(modc)b\equiv b \pmod c bnbn(modc)..........[A1]b^n\equiv b^n \pmod c..........[A_1] aa(modc)..........[A2]a\equiv a \pmod c ..........[A_2] From [1][1] and [A2][A_2] : ab(modc)a\equiv -b \pmod c an(b)n(modc)a^n\equiv (-b)^n \pmod c anbn(modc)..........[A3]a^n\equiv -b^n \pmod c ..........[A_3] Adding [A1],[A3][A_1],[A_3] an+bn0(modc)a^n+b^n\equiv 0 \pmod c c(an+bn)\therefore c|(a^n+b^n)

Case 2 (nn is even) bb(modc)b\equiv b \pmod c bnbn(modc)..........[B1]b^n\equiv b^n \pmod c ..........[B_1] aa(modc)..........[B2]a\equiv a \pmod c..........[B_2] From [1][1] and [B2][B_2] : ab(modc)a\equiv -b \pmod c an(b)n(modc)a^n\equiv (-b)^n \pmod c anbn(modc)..........[B3]a^n\equiv b^n \pmod c ..........[B_3] Subtracting [B1],[B3][B_1],[B_3] anbn0(modc)a^n-b^n\equiv 0 \pmod c c(anbn)\therefore c|(a^n-b^n)

Bonus :

  • Prove or disprove the converse of the above theorem

Note by Zakir Husain
3 months, 2 weeks ago

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Nice!

Yajat Shamji - 3 months, 2 weeks ago

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

Kriti Kamal - 3 months, 2 weeks ago

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