Solve for natural numbers $\begin{cases}13n-9m=110\\\text {lcm}(n,m)-\text{gcd}(n,m)=3n+12\end{cases}$

Question

I'm going to find $(n,m)\in \mathbb{N}$ such that :

$$\begin{cases}13n-9m=110\\\text {lcm}(n,m)-\text{gcd}(n,m)=3n+12\end{cases}$$

I know that :

$$\text{lcm}(n,m)\cdot\text{gcd} (n,m)=n\cdot m$$ But I don't see any result?

Answer 1

The second equation clearly shows that $\gcd(n,m)|12$, so that there are only six cases for $\gcd(n,m)$. In each of these cases, subsituting $\text{lcm}(n,m)=nm/\gcd(n,m)$ reduces the problem down to a quadratic in $n$ or $m$.

P.S.:

As alex.jordan mentioned, we may further say that $\gcd(n,m)=1,2$.

If $\gcd(n,m)=1$, we have that $nm-1=3n+12$, so that $n(m-3)=13$. Thus, either $n=13$ and $m=4$ or $n=1$ and $m=16$. Neither yields a solution.

If $\gcd(n,m)=2$, we have that $nm/2-2=3n+12$, so that $\frac n2(\frac m2-3)=7$. Thus, either $n=14$ and $m=8$ or $n=2$ and $m=20$. We may check that $n=14$ and $m=8$ is the only solution.

Answer 2

The first equations says that $\gcd(n,m)\mid 110$. Since $\gcd(n,m)\mid lcm(n,m),n$, you can get from the second equation that $\gcd(n,m)\mid 12$, so $\gcd(n,m)\mid 110,12$, and will mean that $\gcd(n,m)\mid 12\cdot10-110=10$. Now, $5\not\mid12$, so $\gcd(n,m)=2$ or $1$.

Now,

• $\gcd(n,m)=1$:

$$lcm(n,m)-1=3n+12$$ $$lcm(n,m)=3n+13$$ Remember that $n\mid lcm(n,m)$, so $n\mid 13$, then $n=1,13$.

1. If $n=1$, then the first equation will be equal to $-9m=97$, and this doesn't have integer solutions for $m$. If $n=13$, then you get $169-9m=110$, so this means $9m=59$, again without solutions.

• If $\gcd(n,m)=2$:

$$lcm(n,m)-2=3n+12$$ $$lcm(n,m)=3n+14$$ Using the Remember argument of the first case, we get $n\mid 14$. Since $n$ is even (or the case wouldn't have sense), $n=2,14$. If $n=2$, we have:

• $13\cdot 2-9m=110$ Obtaining $26-9m=110$, that means $m=84/9$, not an integer.
• $13\cdot 14 -9m =110 $ Obtaining $182-9m=110$, that means $m=8$, so now we just verify the equations: $$13\cdot 14-9\cdot 9=110 $$ $$lcm(8,14)-gcd(8,14)=56-2=54=3\cdot 14+12$$

And we get that the only solution is $(m,n)=(8,14)$.