"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?
Proof: If n is a perfect square, $\,n+2\,$ is NOT a perfect square
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1I was an Assistant Professor there. And we were in Swain. Some fond memories. – 2012-09-23
9 Answers
Hint: Every perfect square is either $0$ or $1$ modulo $4$.
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0@LouisRhys: Consider $(2k)^2$ and $(2k + 1)^2$. – 2013-01-01
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.
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0@celtschk Yes, of course. This generalizes to large differences than $2$ as well, if you check finitely many cases. – 2012-09-23
$n=a^2\,\,,\,\,n+2=b^2\Longrightarrow 2=(n+2)-n=b^2-a^2=(b-a)(b+a)$
Now check that this is impossible (and $\,b>0\,$ )
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0Just to avoid the option $\,(-2)(-1)\,$ , and everything is positive, that's all. – 2012-09-23
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.
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0This is certainly the best answer. Good one. – 2017-08-31
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.
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?
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.
Here's the key fact: the sum of consecutive odds make all perfect squares. For example, $1=1^2$ $1+3=2^2$ $1+3+5=3^3$ $1+3+5+7=4^2$ and so on. To prove this, you can either draw a picture or use this algebraic/inductive method. Let $n^2=1+3+5+\cdots +(2n-1),$ then $(n+1)^2 = n^2+2n+1=1+3+5+7+\cdots (2n-1)+(2n+1).$ Therefore, if $n=k^2=1+3+5+\cdots (2k+1)$ then $n+2=k^2+2=1+3+5+7+\cdots (2k-1)+(2k+3)...$ note that it's not $2k+1...$. This clearly won't be a perfect square.
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.