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

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- Sep 13th 2009, 06:04 PMzpwnchenhelp me solve this problem
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 - Sep 13th 2009, 08:02 PMSoroban
Hello, zpwnchen!

Quote:

Prove that if $\displaystyle n$ is a perfect square, then $\displaystyle n+2$ isa perfect square.*not*

$\displaystyle n$ is a perfect square: .$\displaystyle n \:=\:a^2\,\text{ for some positive integer }n.$

Suppose $\displaystyle n+2$a perfect square: .$\displaystyle n+2 \:=\:b^2\,\text{ for some positive integer }b.$*is*

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. $\displaystyle n + 2$ isa perfect square.*not*

- Sep 14th 2009, 07:20 AMzpwnchen
Thank you so much!

Can we say that n=a^2 where a is 1/2. therefore, n is not a perfect. we reached a contradiction?

Can we use direct or indirect proof for it?