# Next step to take in direct proof or a workaround around current dilemma?

This is a problem from Discrete Math and Its Applications

I used a direct proof to do this proof. I understand the process/idea behind the direct proof, mainly (from https://courses.cs.washington.edu/courses/cse311/14au/slides/lecture07-filled.pdf)

Here is my work so far. Right now I am a mental roadblock. I did a substitution for n in the first equation and ended up with a proof that m - p is an even integer. Is there a workaround where I can get both positive m and p on one side(sum of the two) rather than the difference between the two?

• Since $n$ is common in both $m+n$ and $n+p$ then $m$ and $p$ are either both odd or both even so that $m+p$ is even. This is because for $m+n$ to be even we may either have 2k+2k=4k=2(2k) or 2k+1+(2m+1)=2k+2m+2=2(k+m+1). Try adding an odd and an even and see what happens.
– 123
Jan 11 '15 at 6:40
• I get that idea but showing it via a direct proof Jan 11 '15 at 8:37

Hint: You showed that $m-p$ is an even integer. Now note that $m+p=(m-p)+2p$.

• Thanks, and you go through an easier direct proof to show that the sum of two even numbers is even Jan 11 '15 at 8:37
• Yes, so you know how to complete. Alternately, we could have added $m+n$ and $n+p$, and subtracted $2n$. Jan 11 '15 at 9:17
• thank you. Can you have a look at my latest question as well? This ones a bit tricky. I don't know how to proceed.... Jan 11 '15 at 9:21
• I see it has been answered. Let $n$ be odd. Then there is an integer $k$ such that $n=2k+1$. Now note that $2k+1=(k+1)^2-k^2$. We have explicitly exhibited two squares with difference $n$. Jan 11 '15 at 13:37
• yeah thanks. the k + a and the k + b part was confusing to come up with. Jan 11 '15 at 18:56