<s>[INST] Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that 
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$, show that $3^{16}-33$ and $3^{15}+5$ are divisible by 4.
My proof:
$$p^{2n}-2nq-1=(1+q)^{2n}-2nq-1$$
$$=[1+2nq+\frac{(2n)(2n-1)}{2!} q^2+\frac{2n(2n-1)(2n-2)}{3!}q^3+...]-2nq-1$$
$$=n(2n-1)q^2+\frac{2}{3} n(2n-1)(n-1)q^3+...$$
$$=q^2[n(2n-1)+\frac{2}{3} n(2n-1)(n-1)q+...]$$
Hence the expansion has a common factor $q^2$
Taking $n=8$ and $p=3$, and given that $p=q+1, q=2$, by substitution,
$$3^{16} -33=4[120+1120+...]$$
By factoring a 3:
$$3(3^{15}-11)=4[120+1120+...]$$
Dividing both sides by 3 and adding 15 to both sides:
$$3^{15} +5=4[\frac{1}{3}(120+1120+...)+4]$$
Then it is proven that it is also divisible by 4.
The only problem I have with the proof is that how do I know that each term in the brackets $(120+1120+...)$ are divisible by 3?
 [/INST] By the Euclid's lemma: $3(3^{15}-11)=4\cdot [120+1120+...]$ implies that $3$ must divide either $4$ or the sum. It does not divide $4$, hence the conclusion.
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