# Find ℕ a, b, c such that gcd(a,b)=21, gcd(b,c)=21, gcd(a,c)=819

What methods are there to find integers from given GCDs? I need to find the natural numbers a, b, c that satisfy gcd(a,b)=21, gcd(b,c)=21, and gcd(a,c)=819.

I managed to find a set {a=819,b=462,c=819} by some guessing and checking with Mathematica, but I need to be able to do this by hand. I had hoped to reverse engineer some method with a solution and the GCD properties, but so far I'm not having much success. Any help appreciated, thanks.

• Welcome to Mathematics Stack Exchange. Note that $819=21\times39$. $b=21$ would work too – J. W. Tanner Dec 9 '19 at 22:22
• Since $819 = 3^2\cdot 7 \cdot 13$, $a$ and $c$ must both be divisible by $819$, and then $b$ must be divisible by $21$. Any values for all three will work as long as there no additional factors of $3$, $7$, or $13$. – rogerl Dec 9 '19 at 22:24

## 1 Answer

Let $$\dfrac aA=\dfrac bB=21$$ so that $$(A,B)=1$$

Similarly $$b=21D,c=21C,(D,C)=1$$

$$\implies B=D,(B,C)=1$$

Finally we need $$819=(21A,21C)=21(A,C)\implies (A,C)=39$$

WLOG $$\dfrac Ap=\dfrac Cq=39,$$ so that $$(p,q)=1$$

$$(B,39p)=(B,39q)=(p,q)=1$$

where $$b=21B,a=21A=819p,c=819q$$