I have been told that this proof is incorrect, however, I'm having a hard time seeing the issue.
Prove that for any integer $n > 0$ which is a perfect square, $n+4$ is not a perfect square:
Proof:
$n > 0$
Since $n$ is a perfect square, $a^2=n$
The subsequent perfect square can be written as ${b^2=\left(a+1\right)}^2$
The difference between consecutive squares is then ${|b}^2-a^2|=|2a+1|$
$\forall a\in\mathbb{Z}(|2a+1|\neq 4)$
$\square$
Edit: Thanks everyone, appreciate it 🙂
Best Answer
What's wrong is no-one said it was the next perfect square.
So if $n = a^2$ and $n+4 = b^2$ we have no reason to assume $b = a+1$.
But we can assume $b \ge a+1$ and get a similar problem.
$n=a^2$ and $n+4 = b^2 \ge (a+1)^2 = a^2 + 2a + 1$
So $4 \ge 2a+1$ so $a \le 1.5$. Well nothing wrong with that. So $a = 1, 0$ and $a=0\implies n =a^2 = 0$ that's out (but it is a case where $n =0^2$ and $n+4 = 2^2$) and $a=1\implies n=1^2 = 1$ and $n+4 = 5$ not a perfect square.
So that can fix the proof.
Or we can do:
You have $n=a^2$ and let's suppose $n+4 = b^2$ then
$b^2 - a^2 = 4$ and
$(b-a)(b+a) =4$ and... now we are talking. Assume $a,b$ are positive and $b > a$ we get $(b-a)(b+a) = (1,4), (2,2)$.
The first means $2b = 5; 2a = 3$ and $a = 1.5$ and $b = 2.5$ (and indeed $1.5^2 = 2.25$ and $2.5^2 = 6.25 = 2.25 + 4$. But $a, b$ are not integers.
The second means $a=0$ and $b =2$. So $n = 0^2 = 0 \not > 0$. So that is out.