prove or disprove: The difference of the squares of any two consecutive integers is

Sep 2013
567
27
Portland
prove or disprove: The difference of the squares of any two consecutive integers is an odd integer

Let m, n and p be any integers such that m and n are consecutive and

Let \(\displaystyle m^2 - n^2 = p\). if m > n, then by definition m=n+1. Because m and n are consecutive integers neither are both even nor odd.


Case (i)

Let m be even and n be odd

because m is even \(\displaystyle m=2k_{1}; k_{1}\in \mathbb{Z}\) and \(\displaystyle n=2k_{2}+1; k_{2} \in \mathbb{Z}\)

\(\displaystyle m^2-n^2=p\)

\(\displaystyle \Rightarrow (2k_{1})^2 - (2k_{2} + 1)^2 = p\)

\(\displaystyle \Rightarrow 4k_{1}^{2} - 4k_{2}^2 - 4k_{2} - 1 = p\)

is this on the right track?
 

topsquark

Forum Staff
Jan 2006
11,578
3,454
Wellsville, NY
Re: prove or disprove: The difference of the squares of any two consecutive integers

prove or disprove: The difference of the squares of any two consecutive integers is an odd integer

Let m, n and p be any integers such that m and n are consecutive and

Let \(\displaystyle m^2 - n^2 = p\). if m > n, then by definition m=n+1. Because m and n are consecutive integers neither are both even nor odd.


Case (i)

Let m be even and n be odd

because m is even \(\displaystyle m=2k_{1}; k_{1}\in \mathbb{Z}\) and \(\displaystyle n=2k_{2}+1; k_{2} \in \mathbb{Z}\)

\(\displaystyle m^2-n^2=p\)

\(\displaystyle \Rightarrow (2k_{1})^2 - (2k_{2} + 1)^2 = p\)

\(\displaystyle \Rightarrow 4k_{1}^{2} - 4k_{2}^2 - 4k_{2} - 1 = p\)

is this on the right track?
You might be able to finish from there, but you do have two mistakes
1) Since we are talking about two consecutive integers then k_2 = k_1.
2) In that case you would want to do (2k + 1)^2 - k^2 > 0. I suppose it doesn't matter because negative numbers can be odd, too. It just looks funny to me.

-Dan
 
Sep 2013
567
27
Portland
Re: prove or disprove: The difference of the squares of any two consecutive integers

okay thanks, I wasn't sure if I could make k_2 = k_1 without sacrificing generality

so if the k's are the same

\(\displaystyle (2k)^2 - (2k+1)^2 = p\)

\(\displaystyle 4k - 4k^2 - 4k - 1 = p\)

\(\displaystyle -4k -1 = p\)

getting hung up again

because if p is odd then presumably \(\displaystyle p=2k+1\) but it may or may not be the same k or would it?

and also \(\displaystyle -4k-1 \neq 2k+1\)

caught myself, clearly did this backwards
 
Last edited:

Prove It

MHF Helper
Aug 2008
12,897
5,001
Re: prove or disprove: The difference of the squares of any two consecutive integers

Since the two integers are consecutive you can call them "x" and "x + 1". Then the difference of their squares is:

$\displaystyle \begin{align*} \left( x + 1 \right) ^2 - x^2 &= x^2 + 2x + 1 - x^2 \\ &= 2x + 1 \end{align*}$

which is an odd integer. Q.E.D.
 
  • Like
Reactions: 2 people