Proof: If n is a perfect square, $\,n+2\,$ is NOT a perfect square

Question

"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?

Answer 1

Hint: Every perfect square is either $0$ or $1$ modulo $4$.

Answer 2

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.

Answer 3

$$n=a^2\,\,,\,\,n+2=b^2\Longrightarrow 2=(n+2)-n=b^2-a^2=(b-a)(b+a)$$

Now check that this is impossible (and $\,b>0\,$ )

Answer 4

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.

Answer 5

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?

Answer 6

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.

Answer 7

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.