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I have discovered some exercise type conjectures which I can't prove and this is one of them:

Given positive integers $a,b$, then
$$ fracgcd(a+b,ab)gcd(a,b) bigg| gcd(a,b)$$

Can this be proved or disproved?

From time to time, when testing my growing math packages BigZ and Forthmath, I recognize some patterns which I can't prove or disprove (or even have the ambition to). I post them here with the hope that it will not annoy too much. I hope you can bear with it.

Write $gcd(a,b)=d$, then $a=da',b=db'$ and thus $fracgcd(a+b,ab)d=gcd(a'+b',a'b'd)$ where $gcd(a',b')=1$. Notice now that $gcd(a'+b',a'b')=1$ since $a'(a'+b')-a'b'=a'^2$ and thus $gcd(a'+b',a'b')|a'^2 implies gcd(a'+b',a'b')|a'$ and $gcd(a'+b',a'b')|a'+b' implies gcd(a'+b',a'b')|a',b' implies gcd(a'+b',a'b')=1$. This means that $gcd(a'+b',a'b'd)=gcd(a'+b',d)$ and thus it divides $d$ by definition. So your conjecture is indeed true.

https://en.wikipedia.org/wiki/P-adic_order

Let $h = gcd(a+b, ab)$ and $g = gcd(a,b).$

For each prime factor $p$ including $2,$ two cases:

(I) $$ a = p^k u, b = p^j v $$
with $k > j$ and $u,v neq 0 pmod p.$
Then $p$-adic valuation $nu_p(g) = j.$ Next $nu_p(ab) = k+j$ while $nu_p(a+b) = j.$ Put together, $nu_p(g) = nu_p(h).$

(II) $$ a = p^k u, b = p^k v $$
with $u,v neq 0 pmod p.$
Then $p$-adic valuation $nu_p(g) = k.$ Next $nu_p(ab) = 2k$ while $nu_p(a+b) geq k.$ Then
$$ k leq nu_p(h) leq 2k $$

Put together, $nu_p(h) leq 2nu_p(g).$

In either case, combining all primes,
$$ h | g^2 $$

Oh, note that we do have $g | h$ and can write $$ frachg ; | ; g $$

Taking $D=(a,b)$, then $a=DA$ and $b=DB$, $(ab,a+b)=D(A+B,ABD)$ and $(A,B)=1$. So you want to know if $(ab,a+b)/D$ divides $D$, i.e. $(A+B,ABD)|D$? well, let's see if the prime common divisors between $A+B$ and $ABD$ are divisor of $D$ as well.

If $p$ is a prime common divisor of $A+B$ and $ABD$, by Gauss lemma, $p|A$ or $p|B$ or $p|D$. If $p|A$ or $p|B$ there is a contradiction with the coprimality of $A$ and $B$ $p$ cause if $p|A$ then $p|(A+B)-A=B$. So $p|D$.

Recall $(a,b) = (a!+!b,rm lcm(a,b)),$ so $,(a!+!b,ab)mid (a,b)^2! = (a!+!b,rm lcm(a,b))^2$ by $,abmid rm lcm(a,b)^2$

Let $m=gcd(a,b)$. Now $mmid a$ and $mmid b$, so $m^2mid ab$.
Let $a=a_1m$, $b=b_1m$, for some positive integers $a_1$, $b_1ge1$, then $a+b=ma_1+mb_1=m(a_1+b_1)$, so $mmid a+b$.

There are three cases to consider:

Case 1:
If $mnmid(a_1+b_1)$ then $gcd(a+b,ab)=m$, and we have
$$fracgcd(a+b,ab)gcd(a,b)=fracmm=1midgcd(a,b)=m$$
and the conjecture is true in this case.

Now we need to check whether $m^2mid a+b=m(a_1+b_1)$, which is equivalent to $mmid(a_1+b_1)$.

Case 2:
If $a_1+b_1=mc$, for some integer $cge1$, then $mmid(a_1+b_1)$, and so $m^2mid a+b=m(a_1+b_1)$. In this case $gcd(a+b,ab)=m^2$ (note we cannot have $gcd(a+b,ab)>m^2$ as $gcd(a,b)=m$, therefore $m^3nmid ab$), and we have
$$fracgcd(a+b,ab)gcd(a,b)=fracm^2m=mmidgcd(a,b)=m$$
and the conjecture is true in this case.

Consider lastly the case:

Case 3:

Let us assume
$$m<gcd(a+b,ab)=gcd(m(a_1+b_1),m^2a_1b_1)=mgcd(a_1+b_1,ma_1b_1)<m^2$$
which simplifies to
$$1<gcd(a_1+b_1,ma_1b_1)<m$$
We cannot have $mmid(a_1+b_1)$, as then $gcd(a_1+b_1,ma_1b_1)=m$, a contradiction. Further simplification gives
$$1<gcd(a_1+b_1,a_1b_1)<m$$
Let $p$ be s.t. $1<p<m$ and $gcd(a_1+b_1,a_1b_1)=p$. Then $pmid a_1+b_1$ and $pmid a_1b_1$, then $pmid a_1$ or $b_1$. If $pmid a_1$ then $pmid a_1+b_1$ implies $pmid(a_1+b_1)-a_1=b_1$, a contradiction since $gcd(a_1,b_1)=1$.

So $gcd(a_1+b_1,a_1b_1)=1$, and it follows $gcd(a+b,ab)$ can take no value in $(m,m^2)$.

The conjecture is true.