Question

Prove that:
\[\Large3^{1000k+n}-2^{1000k+n}\equiv3^n-2^n\mbox{(mod 1000)}\]

Answer 1

can you show \(3^{1000}\equiv1\mod 1000\)

Answer 2

yes

Answer 3

but \(2^{1000}\equiv376\mbox{(mod 1000)}\)

Answer 4

Disproved:
\[3^1-2^1\equiv1\mod1000\]\[3^{1001}-2^{1001}\equiv251\mod1000\]