Proof related to pigeon hole principle to be done with induction 
since the question is about a positive integer m, it's obvious that the use of mathematical induction needed, but to prove the fact for n = k+1 we have to use the pigeon hole principle, i am so confused in using these both at once, some one help me in solving this proof 
 A: Hint:
The base case is easy, so let's look at the inductive step.  Assuming $P(m)$ is true, let's consider any set $S$ of $2m+3$ distinct integers from among $[-2m-1, 2m+1]$. If less than $3$ of those are from among $R = \{\pm(2m+1), \pm2m\}$, the induction hypothesis takes care of it. So that leaves cases where $|S \cap R| \in \{3, 4\}$. Here use Pigeonhole for each distinct case separately. 
There are two distinct cases to consider, accounting for symmetry in signs.  Either $\{2m+1, -2m-1\} \subset S$ or $\{2m+1, 2m, -2m\} \subset S$ but $-2m-1 \not \in S$.
Case 1: $\pm (2m+1) \in S$
If $0 \in S$ it is trivial.  Else take as holes the pairs $(1, 2m), (2, 2m-1), \dots (m, m+1)$, their negatives and you will have $2m$ holes and $2m+1$ pigeons to roost.  
Case 2: $\{2m+1, 2m, -2m\} \subset S$ but $-2m-1 \not \in S$
Again if $0, -1 \in S$, it is trivial.  Else we take pairs $(1, 2m-1), (2, 2m-2), \dots(m-1, m+1)$ and the pairs $(-2, -2m+1), (-3, -2m+3), \dots (-m, -m-1)$ to get $2m-2$ pairs for $2m-1$ pigeons (excluding $m$ if its there).
