Luz Stokes

2022-07-16

Prove or disprove (discrete math)
Prove or disprove the following statement. The difference of the square of any two consecutive integers is odd
This is working step: let $m,m+1$ be 2 consective integers:
$\left(m+1{\right)}^{2}-{m}^{2}$
${m}^{2}+1+2m-{m}^{2}$
$1+2m$
If m is odd then $2m=\text{even}$, if m is even then $2m=\text{even}$, then adding 1 will make it odd.
Can you please advise me if my working is the right step and could I answer like this?

esbalatzaj

Expert

Step 1
Your proof looks correct. You might want to make it more clear that you are saying $\left(m+1{\right)}^{2}-{m}^{2}=\left({m}^{2}+2m+1\right)-{m}^{2}=2m+1$when you do your arithmetic.
Step 2
You also don't need to consider the cases where m is even and odd separately: since 2m is a multiple of 2, it must be even, and so you can conclude that $2m+1$ is not evenly divisible by 2, so it is odd.

Raynor2i

Expert

Step 1
What does it mean for an integer ℓ to be odd exactly? It means that we may express ℓ in the form $\ell =2n+1$, where $n\in \mathbb{Z}$ (note that n can be any integer).
The difference of the square of any two consecutive integers is odd.
Thus, assume (as you did in your first attempt) that m is an arbitrary integer in Z; then, we are dealing with the difference $\begin{array}{}\text{(1)}& \left(m+1{\right)}^{2}-{m}^{2}.\end{array}$
Step 2
Now expand (1) as Strants did in his answer (except with a slight modification):
$\left(m+1{\right)}^{2}-{m}^{2}=\left({m}^{2}+2m+1\right)-{m}^{2}=2m+1=\ell ,$
where $\ell \in \mathbb{Z}$. We know $\ell \in \mathbb{Z}$ because adding two integers, namely 2m and 1, yields an integer, namely ℓ. Here's the important part: notice what form ℓ takes. We have that $\ell =2m+1,$ where m is an arbitrary integer in Z. Thus, by definition, we can see that ℓ is an odd integer.
Maybe this will clear anything up you did not quite get before.

Do you have a similar question?