1

The problem is correct and I have checked it using multiple prime factor combinations satisfying $ab = cd.$

I tried different ways like considering prime factorization and $gcd$ of $a,b.$ Here is another approach which I figured was closest to solving the problem : $gcd(a,c)=gcd(ad,cd)d=gcd(ad,ab)d=a×gcd(d,b)d$

Similarly

$gcd(a,d)=a×gcd(c,b)c$

Multiplying

$gcd(a,c)×gcd(a,d)=a×gcd(b,c)×gcd(b,d)b$

So the given problem is equivalent to proving

$b×gcd(a,b,c,d)=gcd(b,c)×gcd(b,d)$

Thank you!

Michael Rozenberg
  • 194,933
John Wayne
  • 21
  • 1
    Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Sep 09 '20 at 09:07
  • 2
    @John Wayne Nice problem! Show your attempts and I'll show my solution. – Michael Rozenberg Sep 09 '20 at 09:27
  • 1
    @MichaelRozenberg I tried different ways like considering prime factorization and gcd of a,b. Here is another approach which I figured was closest to solving the problem : $$gcd(a,c) = \frac{gcd(ad,cd)}{d} = \frac{gcd(ad,ab)}{d} = \frac{a\times\gcd(d,b)}{d}$$ Similarly $$gcd(a,d) = \frac{a\times\gcd(c,b)}{c}$$ Multiplying $$gcd(a,c)\times\gcd(a,d) = \frac{a\times\gcd(b,c)\times\gcd(b,d)}{b}$$ So the given problem is equivalent to proving $$b\times\gcd(a,b,c,d)=gcd(b,c)\times\gcd(b,d)$$ Thank you for the consideration – John Wayne Sep 09 '20 at 10:58
  • 1
    @John Wayne I am very busy now. I'll post my solution in evening. Thank you that you showed your attempts. – Michael Rozenberg Sep 09 '20 at 11:20

4 Answers4

3

Induction on $a$ (assume $a,b,c,d$ positive). If $a=1$ the statement is true ($1=1$). Suppose that $a>1$. Let $p$ be a prime dividing $a$. Then $p|cd$, so $p|c$ or $p|d$. Without loss of generality let $p|c$. Then $p|gcd(a,c)$ and $gcd(a,c)=p\cdot gcd(\frac{a}{p},\frac{ b}{p})$. Since $\frac{a}{p}<a$, we can use induction. We have $\frac{a}{p} b= c/p d$. So

$$\frac{a}{p}=gcd(\frac{a}{p},\frac{c}{p}) gcd(\frac{a}{p}, d)/gcd(\frac{a}{p},b,c,d)$$

or $$a=gcd(a,c) gcd(\frac{a}{p}, d)/gcd(\frac{a}{p},b,c,d)$$

Now if $p\not| d$, we have $gcd(\frac{a}{p},d)=gcd(a,d)$ and $gcd(\frac{a}{p},b,c,d)=gcd(a,b,c,d)$, so

$$a=gcd(a,c) \frac{gcd(a, d)}{gcd(a,b,c,d)}$$ as required. And if $p|d$ then $gcd(\frac{a}{p},d)=\frac{gcd(a,d)}{p}$ and $gcd(\frac{a}{p},b,c,d)=\frac{gcd(a,b,c,d)}{p}$, so, again,

$$a=gcd(a,c) \frac{gcd(a, d)}{gcd(a,b,c,d)}.$$ $\Box$

SAGNIK UPADHYAY
  • 436
markvs
  • 19,653
2

Let $a=\prod\limits_{i=1}^np_i^{\alpha_i},$ $b=\prod\limits_{i=1}^np_i^{\beta_i},$ $c=\prod\limits_{i=1}^np_i^{\gamma_i}$ and $d=\prod\limits_{i=1}^np_i^{\delta_i},$ where $\alpha_i$, $\beta_i$, $\gamma_i$ and $\delta_i$ are non-negative integers.

Thus, it's enough to prove that $$\alpha+\min\{\alpha,\beta,\gamma,\delta\}=\min\{\alpha,\gamma\}+\min\{\alpha,\delta\},$$ where $\alpha$, $\beta,$ $\gamma$ and $\delta$ are non-negative integers such that $\alpha+\beta=\gamma+\delta.$

Also, it's enough to assume that $\gamma\geq\delta.$

Id est, we have three cases.

  1. $\alpha\geq\gamma\geq\delta.$

In this case $\beta=\gamma+\delta-\alpha\leq\delta$ and we need to prove that: $$\alpha+\gamma+\delta-\alpha=\gamma+\delta,$$ which is obvious.

  1. $\gamma\geq\alpha\geq\delta.$

Thus, $\beta=\gamma+\delta-\alpha\geq\delta$ and we need to prove that: $$\alpha+\delta=\alpha+\delta,$$ which is obvious again.

  1. $\gamma\geq\delta\geq\alpha.$

Thus, $\beta=\gamma+\delta-\alpha\geq\delta$ and we need to prove: $$\alpha+\alpha=\alpha+\alpha.$$ Done!

Michael Rozenberg
  • 194,933
1

Lets define $ord_m n = k$ to mean that $k$ is the highest integer power (could be $0$) so that $m^k|n$ but for all $j > k$, $m^j\not \mid n$.

Now if we have two naturals $W$ and $U$ and they have unique prime factorizations, then if $W\ne U$ there will be some prime $p$ where $ord_p W \ne ord_p U$. But if $W = U$ then every prime $p$ will be so that $ord_p W= ord_p U$.

I claim that if $ab =cd$ then for any possible prime $p$ then $ord_p a = ord_p \frac {\gcd(a,c),\gcd(a,d)}{\gcd(a,b,c,d)}$, and therefore that $a = \frac {\gcd(a,c),\gcd(a,d)}{\gcd(a,b,c,d)}$.

Pf: Let $p$ be a prime.

Wolog we may assume $ord_p c \le ord_p d$.

Some basic facts about $ord_p$: $ord_p mn = ord_p m + ord_p n$ and $ord_p \gcd(m,n) = \min(ord_p m, ord_p n)$.

Now as $ab = cd$ we have $ord_p a + ord_p b = ord_p c + ord_p d$ or $ord_p b= ord_pc + ord_pd-ord_a$.

And $ord_p \frac {\gcd(a,c)\times \gcd(a,d)}{\gcd(a,b,c,d)} =$

$\min(ord_p a,ord_p c)+ \min(ord_p a,ord_p d) - \min(ord_p a,ord_p b,ord_p c,ord_p d)$.

Case 1: $ord_p a \le ord_p c \le ord_p d$. But that means $ord_p b= ord_d + (ord_c-ord_a) \ge ord_p d$

So $\min(ord_p a,ord_p c)+ \min(ord_p a,ord_p d) - \min(ord_p a,ord_p b,ord_p c,ord_p d)=$

$ord_p a + ord_p a - ord_p a = ord_p a$.

Case 2: $ord_p c < ord_p a \le ord_p d$. But that means $ord_p b = ord_pc + (ord_pd - ord_p a)\ge ord_p c$.

So $\min(ord_p a,ord_p c)+ \min(ord_p a,ord_p d) - \min(ord_p a,ord_p b,ord_p c,ord_p d)=$

$ord_p c + ord_p a - ord_p c = ord_p a$.

Case 3: $ord_p c \le ord_p d < ord_p a$. But that mesn $ord_p b = ord_p c + (ord_p d - ord_p a) < ord_p c$.

So $\min(ord_p a,ord_p c)+ \min(ord_p a,ord_p d) - \min(ord_p a,ord_p b,ord_p c,ord_p d)=$

$ord_p c + ord_p d - ord_p b = ord_p a$.

And that's it.

fleablood
  • 124,253
1

By basic gcd laws (associative, commutative, and distributive) $ $ & $ $ $\color{#c00}{ab = cd}$

$\!\!\begin{align}(a,c)(a,d) &= ((a,c)a,(a,c)d)\\ &= (aa,\,ac,\,ad,\,\color{#c00}{cd})\\ &= \color{#c00}a(a,\ \ c,\,\ \ d,\,\ \ \color{#c00}b) \\[.1em] {\bf Remark}\ \text{ It's a gcd analog of }\ (a\!+\!c)(a\!+\!d) &= aa\!+\!ac\!+\!ad\!+\!\color{#c00}{cd}\\ &= \color{#c00}a(a+c+d+\color{#c00}b)\end{align}$

See here for more on such gcd polynomial arithmetic.

Bill Dubuque
  • 272,048