My Math Forum  

Go Back   My Math Forum > Math Forums > Math

Math General Math Forum - For general math related discussion and news

Thanks Tree8Thanks
LinkBack Thread Tools Display Modes
April 5th, 2018, 08:00 PM   #21
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,599
Thanks: 2587

Math Focus: Mainly analysis and algebra
Originally Posted by Clonus View Post
if you knew the factors of the number would that help in any "known" way now?
As pointed out above, if you know a pair of factors $m$ and $n$ (with $m \gt n$), of your original number $k=mn$, you write $m=a+b$ and $n=a-b$. You can then solve for $a$ and $b$ $$a=\frac{m+n}2, \; b=\frac{m-n}2$$ and you have $$k=a^2-b^2$$

It's easy to see that, for integer solutions, you need $m$ and $n$ either both odd or both even. This isn't a problem for most numbers as for odd $k$ you may take $(m,n)=(k,1)$ (other pairs may exist); while for $k$ divisible by 4, $(m,n)=(\frac k 2,2)$ works (again other pairs may exist).

Even numbers not divisible by 4 won't have solutions.

Last edited by v8archie; April 5th, 2018 at 08:03 PM.
v8archie is offline  

  My Math Forum > Math Forums > Math

Thread Tools
Display Modes

Copyright © 2019 My Math Forum. All rights reserved.