prove that if n is a perfect square, then n+2 is not a perfect square.
I approached this way:
hypothesis: n+2=k^2
conclusion: n=k;
But i tried both direct and indirect proof, but neither works. Or my approach is not right? pls help
prove that if n is a perfect square, then n+2 is not a perfect square.
I approached this way:
hypothesis: n+2=k^2
conclusion: n=k;
But i tried both direct and indirect proof, but neither works. Or my approach is not right? pls help
Hello, zpwnchen!
Prove that if $\displaystyle n$ is a perfect square, then $\displaystyle n+2$ is not a perfect square.
$\displaystyle n$ is a perfect square: .$\displaystyle n \:=\:a^2\,\text{ for some positive integer }n.$
Suppose $\displaystyle n+2$ is a perfect square: .$\displaystyle n+2 \:=\:b^2\,\text{ for some positive integer }b.$
We have: .$\displaystyle \begin{array}{c}n \:=\:a^2 \\ n \:=\:b^2-2 \end{array}$
Then: .$\displaystyle a^2 \:=\:b^2-2 \quad\Rightarrow\quad b^2 - a^2 \:=\:2 \quad\Rightarrow\quad (b-a)(b+a) \:=\:1\!\cdot\!2$
We have a system of equations: .$\displaystyle \begin{array}{ccc}b-a &=& 1 \\ b+a &=& 2\end{array}$
. . which has the solution: .$\displaystyle a \:=\:\tfrac{1}{2},\:b \:=\:\tfrac{3}{2}$
But $\displaystyle a\text{ and }b$ are integers . . . We have reached a contradiction.
Therefore. $\displaystyle n + 2$ is not a perfect square.