Proof by cases: If n is a positive integer and there is a perfect square in $\{k \in \mathbb{Z} | n \leq k \leq 2n\}$ If n is a positive integer and there is a perfect square in $\{k \in \mathbb{Z} | n \leq k \leq 2n\}$, then there is a perfect square in $\{k \in \mathbb{Z} | n + 1 \leq k \leq 2n + 2\}$.
There seems to be 3 cases here.
I have found the first and second case, 
Case 1: $n$ is not a perfect square.
Case 2: If $n$ is the only perfect square.
I seem to be unable to find a 3rd case here, but I do not think I can complete this proof with just these 2 cases. 
--
I have an example proof here, which makes several mistakes, and I must identify some of the mistakes. 
-- example --
Proof: Case 1: If $n$ is not a perfect square, then we must be done.
Case 2: If $n$ is the only perfect square, then we can say $n=a^2$
The next perfect square is $(a+1)^2 = a^2+2a+1$, which is definitely in $\{k \in \mathbb{Z} | n + 1 \leq k \leq 2n + 2\}$.
In either case, we have shown that if $n$ s a positive integer and there is a perfect square in $\{k \in \mathbb{Z} | n \leq k \leq 2n\}$, then there is a perfect square in $\{k \in \mathbb{Z} | n + 1 \leq k \leq 2n + 2\}$.
-- end of example --
This proof is incomplete, and I need to edit it
There are 2 non-trivial claims in this example that are correct, but not explained. I need to identify them, and provide the argument that should have been written.
I think the 2 non trivial claims are:
1: If n is not a perfect square, then we must be done. 
2: $a^2+2a+1$ is definitely in $\{k \in \mathbb{Z} | n + 1 \leq k \leq 2n + 2\}$.
I'm not fully sure if these are non-trivial, and I don't know how to fully explain them if they are.
Any feedback or help would be appreciated here. Thank you.
 A: We are told there is a perfect square between n and 2n inclusive.  This means two possible cases:
Case 0: There is a perfect square between n + 1 and 2n inclusive.  In this case there is nothing to do.
Case 1: Not case 0.  Then $n$ must be the perfect square. And n + 1 through 2n are not. 
There doesn't seem to be a third case.
Let $n = k^2$  Then $n = k^2 < (k+1)^2 = k^2 + 2k + 1 = n + 2k + 1$.
Case 1a: n = k = 1, $(k+1)^2 = 4 = 2n + 2$
Case 2a: k = 2, n = 4. $n + 2k + 1 = 2n + 1$
Case 2c: k > 2, $n + 2k + 1 < 2n + 1$ (which is a subset of case 0; and we didn't really need to use that we were assuming the strict case that there were no squares in n + 1 through 2n). 
=================
"I think the 2 non trivial claims are:"
"1: If n is not a perfect square, then we must be done."
This is trivial!  if n is not a perfect square we were told there is a perfect square between n and 2n so it must be between n+1 and 2n which is a subset of the numbers between n+1 and 2n + 2.
"2: $a^2+2a+1$ is definitely in {k∈Z|n+1≤k≤2n+2}."
This isn't trivial but it is easy for a>2.  (then $a^2 + 2a + 1 = n + 2a + 1 < n + a^2 + 1 = 2n + 1).
And we can do a = 1,2 on a case by case basis.
