Have you considered and tried to come up with counter-examples for (B), (C), (D) and (E)? Also, regarding $8n^2(2n^2 - 5) = 9m$, a counter-example to that is $n = 2$.

@JohnOmilelan Cannot prove without counterexamples then?

You can prove it without using explicit counter-examples, but I believe either using counter-examples, or showing it by using modular arithmetic, in your case may be easiest.

@JohnOmielan please see my edit, and suggest how to prove using modular arithmetic.

As you requested, I posted an answer using modular arithmetic.

Hint: your $\rm\color{#c00}{quartic}$ polynomial $\,f(n)\,$ can have at most $\color{#c00}4$ roots mod $p$ since $\,\Bbb Z_p\,$ is a field, so there are values of $n$ that are not roots for any prime $p>3$, i.e. values of $n$ not divisible by $p.\ \ $

@BillDubuque Request answer, as unable to make with the hint. Thanks anyway.

Do you know that $\Bbb Z_p$ is a field, and polys over a field can have no more roots than their degree? Btw, accepting a first (quick) answer is almost never a good idea since it highly discourages later answers (and a first answer is only rarely the best).

@BillDubuque But, then aren't you stating that take for each $p=3,5,7$. And simply ignoring $9$.

^^^ typo: above I meant to write "values of $n$ with $f(n)$ not divisible by $p$". Yes, you still need to consider $3$ and $9$ but that's very easy.

@BillDubuque Hope you meant: Hint: your quartic polynomial $f(n) $ can have at most $4$ roots $\mod {p}$ since $\mathbb{Z_p}$ is a field, so there are values of $n$ (with $f(n)$) that are not roots for any prime $p>3$, i.e. values of $f(n)$ not divisible by $p$.

@BillDubuque but isn't $9$ not a prime, so need be left from your approach?

Yes I meant "with $f(n)$" in both places, and yes, as I said, that doesn't handle $3$ and $9$ but they are easy. Using that idea allows us to easily generalize widely, whereas ad-hoc arguments may not generalize.

@BillDubuque please provide answer, as am unable to pick.

You will likely get many further answers if you unaccept (many users skip questions that already have accepted answers). And Sat night/Sun morn is a slow time, so be patient. Generally it's best to wait a day or two before accepting.

@BillDubuque am too tiny to correctly interpret your last remark/comment.

You can unaccept the current answer by clicking on the checkmark (you can accept it again later if you like). That will motivate many more users to view your question, so you may receive a larger variety of answers (advantageous to both you and other students). After a couple of days the activity will slow down and that is a much better time to decide acceptance.

Are you supposed to choose all of the choices that are correct, or do you know in advance that only one is correct (i.e. precisely what does "must be" mean here)? Please give some knowledge-level context, e.g. do you know modular arithmetic (congruences), and any ring or field theory? (we may be able to infer that from prior activity, but we should not have to waste time doing so).

@BillDubuque it is not given if more than one option can be chosen. I know modular theory at basic level, and am good at groups, but know basic idea of rings, and fields. But, you can use results and can cover fast.

You should put the context into the question, not in the comments, so readers can quickly see the context,

@BillDubuque Can only understand that $\mathbb{Z_p}$ is a field, for prime $p$, as only then (under the second operation of multiplication) all elements (in $\mathbb{Z_p}$) have inverse. But fail to understand: "there are values of $n$ (with $f(n)$) that are not roots for any prime $p>3$, i.e. values of $f(n)$ not divisible by $p$."

@BillDubuque hope my level is enough to understand, so must be missing some detail, so facing issues.