Vtyjkyuk

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:
$$p^2n-2nq-1=(1+q)^2n-2nq-1$$

$$=[1+2nq+frac(2n)(2n-1)2! q^2+frac2n(2n-1)(2n-2)3!q^3+...]-2nq-1$$

$$=n(2n-1)q^2+frac23 n(2n-1)(n-1)q^3+...$$

$$=q^2[n(2n-1)+frac23 n(2n-1)(n-1)q+...]$$

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:
$$3^15 +5=4[frac13(120+1120+...)+4]$$

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?

It's better if you use
$$
(1+q)^2n=1+2nq+sum_k=2^2nbinom2nkq^k
$$
so
$$
p^2n-2nq-1=q^2sum_k=2^2nbinom2nkq^k-2
$$
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
$$
3^15+5=3(3^14-29)+3cdot29+5
$$
and set $q=2$, $n=7$, noting that $3cdot 29+5=92$, which is divisible by $4$.

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

By the Euclid's lemma: $3(3^15-11)=4cdot [120+1120+...]$ implies that $3$ must divide either $4$ or the sum. It does not divide $4$, hence the conclusion.