Proof: If n is a perfect square, $\,n+2\,$ is NOT a perfect square "Prove that if n is a perfect square, $\,n+2\,$ is NOT a perfect square." I'm having trouble picking a method to prove this. Would contraposition be a good option (or even work for that matter)? If not, how about contradiction?
 A: Hint $\rm\ f(n) = n^2\:$ is increasing on $\Bbb N$ so the difference between two different values is at least the difference between two consecutive values, which is $\rm\:(n\!+\!1)^2 - n^2\:\! =\, 2n\!+\!1 > 2\:$ for $\rm\:n > 0.$
Remark $\ $ This has a natural presentation by  telescopy, e.g.
$$\rm f(4)\!-\!f(1)\ =\ f(4)\!-\!f(3)\ +\ f(3)\!-\!f(2)\ +\  f(2)\!-\!f(1) $$
and since each RHS difference is $> 2,\,$ so too is the LHS.
A: Hint: Every perfect square is either $0$ or $1$ modulo $4$.
A: You don't even need algebra to do this – think squared paper. The only thing to specify about a square is its side length, so if you want to show a bigger square is a perfect square, the only thing you can do is increase the side length. But if you increase the side length by 1, you need to add at least 1 square to each side, and 1 on the corner, so you need to add at least three squares.
This is aside from the case where $n = 0$. But $2$ isn't a perfect square, so that's fine.
A: I don't know that you should pick a proof strategy before you have played with the hypotheses a little bit.  In my experience, thinking about the hypotheses tends to suggest natural proof strategies.
In this case, I would think like this: perfect squares are the partial sums of the series of positive odd numbers, so you can't get to the next perfect square by adding $2$.  This suggests contradiction.
Alternatively -- and this would suggest proof by contraposition -- if you have arranged $n$ tiles in a square, and you remove two, can you rearrange those into a smaller square?
A: More generally, for every positive integer $m$,
there are only a finite number of positive integers $n$
such that $n^2+m$ is a perfect square.
Proof: Suppose $n^2+m = (n+k)^2$.
Then $m = 2nk + k^2 > 2nk \ge 2n$,
so $n < m/2$.
If $m = 2$, then $n < 1$, so there are no solutions.
To find all solutions for a particular large $m$,
congruences can greatly reduce the number of cases that have to be tested.
This readily generalizes to any function that increases at least linearly:
If $f(n+1)\ge f(n)+cn$ for some positive $c$,
then, for any positive $m$, 
there are only a finite number of $n$
such that $f(n)+m = f(k)$ for some $k > n$.
Proof: $f(n+k) \ge f(n)+cnk$ (you can do better, but this is enough),
so if $f(n+k)-f(n)=m$,
$m \ge cnk \ge cn$
or $n \le m/c$.
Generalizing further
(my goal is a theorem so general it has no particular application),
the result holds
if $f(n+1)-f(n) \ge g(n)$
where $g$ is strictly increasing
and unbounded
(an example of a slowly growing $g$ is $\log$).
Proof: $f(n+k)-f(n) \ge g(n)+g(n+1)+ ... +g(n+k-1) > kg(n)$
so, if $f(n+k)-f(n) = m$,
$m > kg(n)$. 
The assumptions about $g$
imply that. for any particular $m$,
 there are only a finite number of $n$
satisfying this.
A: Suppose $n=m^2$ and $n+2=k^2$. Clearly $k>m$, so $k\ge m+1$. But then 
$$n+2\ge (m+1)^2=m^2+2m+1\ge m^2+3=n+3$$
a contradiction.
A: $$n=a^2\,\,,\,\,n+2=b^2\Longrightarrow 2=(n+2)-n=b^2-a^2=(b-a)(b+a)$$
Now check that this is impossible (and $\,b>0\,$ )
A: If $n$ is odd and perfect square, then it is of the form $n=8k+1$. But $n+2=8k+1+2$ is also an odd number, so it must be of the form $8k'+1$ which is not. 
If $n$ is even, It is of the form $n=4k^2$ while $n+2=4k^2+2$ is even but not of the form $4k'^2$. Therefore $n+2$ is not a perfect square.
In each case such $n$ does not exist.
A: Any integer $\ n$, upon division by $\ 4$ leaves one of the remainders $\ 0,1,2$ and $\ 3$.  
Therefore $\ n$ will be one of the forms $\ 4k, 4k+1, 4k+2$ and $\ 4k+3$, $\ k\in\mathbb{Z}$
Since $\ n$ is a perfect square, $\ n$ cannot be in the form $\ 4k+2$ and $\ 4k+3$ 
$ \bbox[tan,5px, border:2px solid red] {{\text{To prove it, we assume}  \sqrt{n}=p. \\ 
\text{Then  p may be even or odd.} \\ 
\text{Let} \; p=2q\;\text{(for even) or}\;  p=2q+1\;(\text{for odd})\\   
\text{Now} \;p^2=(2q)^2=4q^2=4k,\; k=q^2\in\mathbb{Z} \\ 
\text{or}\,  p^2= (2q+1)^2=(4q^2+4q+1)=4(q^2+q)+1=4k+1,\; k\in\mathbb{Z}\\  
\text{So any perfect square p is of the form 4k or 4k+1}, k\in\mathbb{Z}}}$ 
Thus $\ n$ will be one of the forms $\ 4k $ or $\ 4k+1$ 
If $\ n=4k$, then $\ n+2=4k+2$ which is not a perfect square $\ \mathbf{(just\; mentioned \;above)}$
If $\ n=4k+1$ then $\ n+2=4k+3$ which is not a perfect square $\ \mathbf{(same \;reason)  }$
Therefore in whatever form a perfect square $\ n$ may be, $\ n+2$ is not perfect square. 
