Proof that the difference of two positively squared integers never equals 1

Question

This type of question is usually solved by a proof by contradiction, however I believe I have a direct proof of it, and I would like to know if its correct.

Problem : Prove that there does not exist positive integers m and n, such that $$m^2 - n^2 = 1$$

Proof:

Since $$m, n \in \mathbb{Z^{+}} \implies m \geq 1 \land n \geq 1 $$ Rearranging the first inequality we get. $$m^2 - n^2 \geq 1-n^2$$ To simplify this inequality we can say $$Since :n^2 \in \mathbb{Z^{+}}, \ \ \exists \ k \in \mathbb{Z^{+}} \ | \ k = n^2$$ $$\therefore m^2 - n^2 \geq 1-k $$

And because $$k=n^2 \implies k \geq 1$$ The right hand side of the inequality is never equivalent to 1, as k is never equal to zero. Therefore because the right hand side of this inequality is never equal to one, we have proved that :

$$m^2 - n^2 \neq 1 \ \ \ \forall m,n \in \mathbb{Z^{+}}$$

$$Q.E.D$$

---

Is this direct proof correct? Also if you have any comments or suggestions on my proof-writing, please let me know.

Answer 1

Considering the contrapositive may help to produce clearer proof.

Assume that $\exists m,\;n \in {\mathbb{Z}^ + }$ (with $m > n$) such that$${m^2} - {n^2} = 1.$$ It follows that ($ * $) $m - n \in {\mathbb{Z}^ + }$ and$$(m - n)(m + n) = 1$$so that$$m - n = \frac{1}{{m + n}}$$ But then $m - n \notin {\mathbb{Z}^ + }$ contradicting ($ * $). Therefore no such $m,\;n \in {\mathbb{Z}^ + }$ exist $ \bullet $

Answer 2

(i)...$\;0<m\leq n\implies m^2 \leq n^2$ $\implies m^2-n^2\leq 0\implies m^2-n^2 \ne 1.$

(ii)...$\;m>n>0\implies m>n\geq 1\implies [m\geq n+1\land n\geq 1]\implies$ $\implies [m-n\geq 1\;\land m+n\geq (n+1)+1=2 n +1\geq 3)]\implies$ $\implies m^2-n^2=(m-n)(m+n)\geq 1\cdot 3=3\implies m^2-n^2\ne 1.$