solve this congruence please.

Question

$7^{n+1} -(n+1)\times 7^{n} -1 ≡ 0 $ (mod 4)

with the variable n as an exponent you can't use a modulo 4 table, which is why it bothers me a bit.i tried messing around with it and i got that this equation is verified when 7^(n+1) ≡ 3^(n+1) (modulo 4) which then tempts me to say that this is verified for every integer.but it doesn't seem right.i could use some help.

Answer 1

We know that $7\equiv (-1)(\mod 4)$, hence we have $(-1)^{n+1}-(n+1)\times (-1)^n-1\equiv 0 (\mod 4)$. If $n$, is even then we have:

$(-1)-(n+1)-1\equiv 0\rightarrow (-n-3)\equiv 0(\mod 4)$, but this is impossible, because it implies $4|(n+3)$, but $n$ is an even number and hence $n+3$ is odd and not dividable by an even number.

If $n$ is an odd number then:

$1+(n+1)-1\equiv n+1\rightarrow (n+1)\equiv 0(\mod 4)\rightarrow 4|(n+1)$. As $n$ is odd, it is possible that $4|(n+1)$ and it is when $n=4k-1$

Answer 2

Hint: show that

$7^n = 1 \mod 4$ if $n = 0 \mod 2$

$7^n = 3 \mod 4$ if $n = 1 \mod 2$