Online Course Discussion Forum

Math Challenge II-A Spring 2022 Q8.30

 
 
Picture of Mandy Chang
Math Challenge II-A Spring 2022 Q8.30
by Mandy Chang - Tuesday, May 3, 2022, 5:48 PM
 

Question 8.30:

Find all ordered triples (x,y,z)(x,y,z) of prime numbers satisfying the equation x(x+y)(y+z)=140x(x+y)(y+z)=140


We have that 140=2^2⋅5⋅7. Therefore x is either 2, 5, or 7. Why do we factor it? Are there other ways? 



 
Picture of John Lensmire
Re: Math Challenge II-A Spring 2022 Q8.30
by John Lensmire - Wednesday, May 4, 2022, 9:43 AM
 

I don't want to definitely say there are no other ways to solve, but this is a common technique anytime you're looking for integer solutions.

As a simple example, suppose you're looking at the equation $xy = 20$. In general (over the reals) this has infinitely many solutions. We could even look at the graph of $y = 20/x$ to visualize all the solutions. If we want $x$ and $y$ to be integers, however, we can take advantage of number theory. The key idea is that $x$ and $y$ both need to be factors of $20$, which narrows down the choices.

In the equation given, $x(x+y)(y+z) = 140$, we can similarly say that $x$, $(x+y)$, and$ (y+z)$ must all be factors of $140$. Since we only want prime number solutions, however, we can easily see that $x = 2$, $x = 5$, or $x = 7$ as those are the only prime factors of $140$. This allows us to reduce the problem to two variables and continue from there.

Hope this helps!