# A confession and a conjecture $\gcd(a-b,a+b)|2\gcd(a,b)$

A long time ago I studied mathematics at the University of Stockholm. I was mostly interested in algebra and topology but as everybody interested in math I was fascinated by the most famous problems and results in number theory. However, I never took a course in number theory since I considered it as too complicated for me. There was an introductory course ‘Algebra 1’ which included some elementary number theory, as the greatest common divisor, but instead of joining the lectures of ‘Algebra 1’ I made self-studies in ‘Algebra 2’ containing the first half of Hersteins ‘Topics in Algebra’. Today I’m trying to learn elementary number theory from the first two chapters of ‘Introduction to Analytic Number Theory’ by Apostol. But as we say in Sweden “You can’t learn old dogs to sit”.

A couple of years ago I started to develop number theoretical facilities to ANS Forth (which is a wonderful language created with the ambition “Keep it simple”). Then I implemented sets and vectors of numbers, sets and vectors, more general than needed in mathematics, really. Entering

{ 1 2 3 ( 4 5 { 6 7 } 8 ) 9 }

put the set $\{1,2,3,9,(4,5,\{6,7\},8)\}$ on a stack. The set algebra is managed by postfix operations on the items on the stack. Using these implementations I’ve started a semi random search for arithmetical conjectures/theorems. My ignorance make the search more random, but I do understand the language of mathematics and I do have some intuition left for mathematics. And even a blind dog can search using it’s noose.

I guess I have posted almost hundred of conjectures in number theory on MSE, some better and some not. One of my last conjectures could be formulated $$\gcd(a,b)|\gcd(a+b,ab)|\gcd(a,b)^2$$ which has a form reminding of lower and upper bound inequalities. The right part wasn’t to easy to prove and I couldn’t make it on my own. My most recent conjecture may be easy for the experienced MSE users to prove, I guess so but I don’t know. Also this has the form of inequalities: $$\gcd(a,b)|\gcd(a-b,a+b)|2\gcd(a,b)$$ I manage the left side of the chain but want help to prove the right side: $$\gcd(a-b,a+b)|2\gcd(a,b)$$ I have convinced myself there is no counter-examples by testing millions of pseudo random pairs $a,b$.

$a,b$ are positive integers such that $a\ge b$.

I manage the left side of the chain but want help to prove the right side: $$\gcd(a-b,a+b) \mid 2\gcd(a,b)$$
Hint:   write it as $\,\gcd(a+b,a-b)\mid \gcd(2a,2b)\,$ then it reduces to the left side, which you proved.
Show, for positive integers $a$, $b\ge1$, $a\ge b$, that $$\gcd(a,b)\mid\gcd(a-b, a+b)\mid2\gcd(a,b)\tag{1}$$
If $m=\gcd(a,b)$ then $m\ge1$. Let $a=a_1m$, $b=b_1m$, for some positive integers $a_1$, $b_1\ge1$; so $$\gcd(a_1m-b_1m, a_1m+b_1m)=m\gcd(a_1-b_1, a_1+b_1)$$ proving the LHS of $(1)$.
Now for the RHS; note that $\gcd(a_1-b_1, a_1+b_1)=1$, for if not we should have some integer $m_1>1$ s.t. $a_1=a_2m_1$, $b_1=b_2m_1$ and so the mid $\gcd$ of $(1)$ becomes $$\gcd(a_2m_1-b_2m_1, a_2m_1+b_2m_1)=m_1\gcd(a_2-b_2, a_2+b_2)$$
But $m=\gcd(a,b)=\gcd(a_1m,b_1m)=\gcd(a_2m_1m,b_2m_1m)=m_1m\gcd(a_2,b_2)>m$, a contradiction. Hence $\gcd(a_1-b_1, a_1+b_1)=1$ as required.
Now the RHS gives $$\gcd(a-b, a+b)=m\gcd(a_1-b_1, a_1+b_1)=m\mid2\gcd(a,b)=\gcd(2a,2b)=2m$$