Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem

Question

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?

Answer 1

It's better if you use $$ (1+q)^{2n}=1+2nq+\sum_{k=2}^{2n}\binom{2n}{k}q^k $$ so $$ p^{2n}-2nq-1=q^2\sum_{k=2}^{2n}\binom{2n}{k}q^{k-2} $$ is divisible by $q^2$.

With $q=2$ and $n=8$ you have $p=3$ and $p^{2n}-2nq-1=3^{16}-33$.

For the second case, consider $$ 3^{15}+5=3(3^{14}-29)+3\cdot29+5 $$ and set $q=2$, $n=7$, noting that $3\cdot 29+5=92$, which is divisible by $4$.

Answer 2

Use the Euclid lemma in last implication: $$\begin{eqnarray} 3(3^{15}-11)=4[120+1120+...]&\implies &3\mid 4[120+1120+...] \\ &\implies & 3\mid 120+1120+... \end{eqnarray}$$

Answer 3

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.