(solved)Proof by contradiction: Let $n > 1$ be an integer. if $n$ is a perfect square, then $n+3$ cannot be a perfect square.

Question

My Work so far:

Contradiction: Assume that both $n$ and $n + 3$ are perfect squares.

$n = a^2,\ n+3 = b^2$

Now,

$3 = (n+3) - n = b^2 - a^2 = (b+a)*(b-a)$

so we assume that $b+a = 3$ and $b-a=1$ which means $a = 3-b $ so $b+b-1=3 \Rightarrow 2b=4 \Rightarrow b = 2$ and $a = 1$, so $n = 1$ which is not possible, thus the contradiction is false which proves the original statement to be true. Thanks guys!

I don't get where you get $b^23$... Can you try rewriting it in a clearer way? — user88595, Feb 20 '14 at 16:34
Plus you wrote $n+3 = ... = (n+3)-n$. Typo or incorrect reasoning? — user88595, Feb 20 '14 at 16:35
I don't think there's anything wrong, a priori, with $b$ being an even number. But with that information, you can figure out what $a$ and $n$ must be... — Dustan Levenstein, Feb 20 '14 at 16:39

score 0 · Accepted Answer · answered Feb 20 '14 at 16:47

Note that when you conclude that $b=2$ you can go on and establish that $a=1$, which means that $n=1$. That isn't exactly a contradiction, because you haven't assumed anything at the start to contradict. (and there are other cases for $b+a$ and $b-a$ which have to be eliminated a little more carefully)

For a contradiction you might try to argue that if $m^2-n^2=3$ then $m\ge n+1$ so that $(n+1)^2-n^2 \le 3$ and see where that leads you assuming that $n\gt 1$.

(solved)Proof by contradiction: Let $n > 1$ be an integer. if $n$ is a perfect square, then $n+3$ cannot be a perfect square.

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