Understanding of Tao's proof example which uses vacuous implications So, in his analysis book in appendix for logic he gives a proof that

if $n$ is a an integer, then $n(n+1)$ is an even integer.($Theorem A.2.4.$)
Since $n$ is an integer, $n$ is even or odd. If $n$ is even, then... . If $n$ is odd, then... . Thus in either case $n(n+1)$ is even, and we are done.
Note that this proof relies on two implications: "If $n$ is even, then $n(n+1)$ is even" and "If $n$ is odd, then $n(n+1)$ is even". Since $n$ cannot be both odd and even, at least one of these implications has a false hypothesis and is therefore vacuous.

After that he shows a corollary of this theorem with some big $N$.

Let $n = (253+142)*123 - (423+198)^{342} + 538 - 213$. Then $n(n+1)$ is an even integer.
In this particular case, one can work out exactly which parity $n$ is - even or odd - and then use only one of the two implications in the above Theorem, discarding the vacuous one.

What does he mean by "discarding the vacuous one"?(And why would i want to do this?)
After that he says:

This may seem like it is more efficient, but it is a false economy, because one then has to determine what parity $n$ is,...

but why, because of that "discard", do i have to determine parity then?
 A: At the end of that section, Tao goes on to summarize the point of this particular exercise:

So, somewhat paradoxically, the inclusion of vacuous, false, or
  otherwise “useless” statements in an argument can actually save you
  effort in the long run! [...] All I’m saying here is that you need not be
  unduly concerned that some hypotheses in your argument might not be
  correct, as long as your argument is still structured to give the
  correct conclusion regardless of whether those hypotheses were true or
  false.

There are two implications:


*

*If $n$ is even, then $n(n+1)$ is even

*If $n$ is odd, then $n(n+1)$ is even


We could look at $n = (253+142)*123 - (423+198)^{342} + 538 - 213$ and say that $n(n+1)$ is an even integer because $n$ is odd (after expending some effort determining the parity directly), and know this thanks to implication 2.
We "discarded" (i.e. cast aside, didn't mention, ignored, etc) the first implication because it didn't apply. We found that $n$ was odd, so why bother bringing up details about unrelated scenarios involving even $n$?
In actuality, it would have saved us some effort if we had included both implications, despite the fact that one of them would be vacuous and not apply directly to this particular value of $n$. Since both implications cover both parity cases and yield the same outcome, we didn't need to actually check the parity directly to know that $n(n+1)$ is even.
A: Imagine you are travelling along a road. You come to a junction with two choices.
On the left is a sign saying:

Choose me if $n$ is even

On the right, a sign says:

Choose me if $n$ is odd

Whichever route you choose, the other route becomes vacuous.
