Statement -1: For each natural number n, (n+1)^7-1 is divisible by 7.

Question

Statement -1: For each natural number $n$, $(n+1)^7-1$ is divisible by 7.

Statement -2: For each natural number $n$, $n^7-n$ is divisible by 7.

Accepted Answer

Statement 2: By induction, $P(n):n^7-n$ divisible by 7. Base case $n=1$: $0$ is divisible by 7. Inductive step: $(k+1)^7-(k+1)=(k^7-k)+\left[{}^7C_1k^6+{}^7C_2k^5+\cdots+{}^7C_6k\right]$. Since 7 is prime, all of ${}^7C_1,\dots,{}^7C_6$ are divisible by 7, so $P(k+1)$ holds. True.

Statement 1: $(n+1)^7-(n+1)=(n+1)^7-n^7-1+(n^7-n)$. Since $n^7-n$ is divisible by 7 (Statement 2), $(n+1)^7-(n+1)$ is divisible by 7 iff $(n+1)^7-n^7-1$ is. Both statements together confirm $(n+1)^7-1$ is divisible by 7. True, and Statement 2 correctly explains Statement 1.

Solutions · 1