Find all integers n such as $7^{n+1}-(n+1)7^n-1 = 0[4]$

Question

we know that $ 7^2=1[4]$ But I don't know how to continue from here! I tried the fact that

$7=7= -1 [4]$

$7^{2k} = 1 [4]$

$7^{2k+1} = -1 [4]$

The order is 2,

I get when considering $n=2k$ :

$-1-(n+1)×1-1 = -n-3 [4]$

The $ n $ keeps popping up !! I cant get rid of it, I need something $4k$ instead of $-n-3$

Plz help out !

if $n=2k$ then $n=0$ or $n=2$ so you just showed that even integers can't be a solution — MIO, Apr 07 '21 at 22:07
Have you tried computing $7^{n+1}-(n+1)7^n-1 \bmod 4$ for $n=1$ to $10$? — J. W. Tanner, Apr 07 '21 at 22:08

score 1 · Answer 1 · answered Apr 07 '21 at 22:26

then $7^{n+1}-(n+1)7^n-1\equiv 1-1-1=-1\equiv1\bmod2$,

so $7^{n+1}-(n+1)7^n-1\equiv 1$ or $3\bmod4$.

then $7^{n+1}-(n+1)7^n-1\equiv(-1)^{n+1}-2(-1)^n-1\equiv1+2-1=2\bmod4. $

then $7^{n+1}-(n+1)7^n-1\equiv(-1)^{n+1}-0-1\equiv 1-0-1=0\bmod4$.

1 Answers1