Answer:
See below
Step-by-step explanation:
We shall prove that for all
. This tells us that 3 divides 4^n+5 with a remainder of zero.
If we let
, then we have
, and evidently,
.
Assume that
is divisible by
for
. Then, by this assumption,
.
Now, let
. Then:

Since
, we may conclude, by the axiom of induction, that the property holds for all
.