Prove by the method of Congruences that 31 \mid $2^{5n}-1

Question

I'm being asked to prove, by method of congruences, that $$ 31\mid2^{5n} - 1$$

Now I'm trying to this via mathematical induction (is this the correct way to go about this?). And the method of congruences is $$a \equiv b \mod m $$ in case $m | b - a$.

First, the base case. For $n = 1$, the statement simplifies to:

$$31\mid2^{5(1)} − 1$$ $$31\mid32 − 1$$

$$1 ≡ 32 \mod\ 31$$

which is true.

Next, we will prove the statement must also be true for $n = k + 1$:

$$31\mid2^{5(n + 1)} − 1$$

which, when rewriten, is

$$1 \equiv 2^{5n + 5} \mod 31)$$ Now this is the part where I think my induction skills aren't quite there yet to see patterns or something I should swap. I asked another question here recently that used induction; should I be trying to play with the exponents above the 2 to end up with something exchangeable in the earlier statements? Or is induction the wrong way to go altogether?

Answer 1

Using Property $\#(10)$ of this, $$2^5\equiv1\pmod{31}\implies (2^5)^n\equiv1^n$$

Answer 2

Continuing your proof:

As $32 \equiv 1$ and as $a \equiv b$ implies $ac \equiv bc$ for any $a,b,c \in {\mathbb Z}$, from the induction assumption, $3^{5n} \equiv 1$, one have that $$3^{5n+5} \equiv 3^{5n}.3^5 \equiv 3^{5n}.32 \equiv 3^{5n}.1 \equiv 1. $$