Prove the following statement: If $n$ and $m$ are nonzero integers, then $n^2-m^2 ≠ 1$.

Question

Here is my attempt.

Proof: By way of contradiction, suppose m and n are nonzero integers and that $n^2-m^2 = 1$. Then, $(n-m)(n+m)=1$.

That is where I get stuck, apparently we are supposed to show $(n-m)=(n+m)$ but I don't see how that would be true. Can anyone help?

$n-m$ and $n+m$ are two integers whose product is $1$. What integers have product $1$? — angryavian, Nov 10 '17 at 07:12

score 1 · Accepted Answer · answered Nov 10 '17 at 07:12

1

$$(n-m)(n+m)=1 $$

$$\implies n- m=n+m=1 \implies (n,m)=(1,0) $$ $$ \;\color{red}{ \text{or}}\; $$$$n-m=n+m=-1 \implies (n,m)=(-1,0) $$ Since, $n$ and $m$ are given to be non-zero, this leads to a contradiction.

answered Nov 10 '17 at 07:12

Jaideep Khare

19,293

Thank you so much, it makes much more sense now. – MANONMARS45 Nov 10 '17 at 07:18
@MANONMARS45 Welcome, Martian! XD – Jaideep Khare Nov 10 '17 at 07:20

Prove the following statement: If $n$ and $m$ are nonzero integers, then $n^2-m^2 ≠ 1$.

1 Answers1