If $n$ is an integer then $\gcd(2n+3,3n-2)=1\text{ or ?} $

There are two possible Gcd's for integers of the form,

$2n+3$ and $3n-2$

I know the gcd is $1$ if I take the equation modulo $2$. However if I take the equation modulo $3$ I get,

$2n$ and $-2$.

So I think the Gcd is 2. Am I correct?

Solution 1:

We can use a general method: $ $ the linear map $\rm\,(n,1)\overset{\rm A}\mapsto (3n\!-\!2,2n\!+\!3)\,$ has determinant $\rm\, \Delta = {\rm det}\, A= 3^2\!+\!2^2\!=\color{#c00}{13},\,$ so the theorem $\,\Rightarrow\, \gcd(3n\!-\!2,2n\!+\!3)\mid \color{#c00}{13}\gcd(n,1)=\color{}{13}$

Theorem $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y)$

Proof $\ $ Inverting the linear map $\rm\,A\,$ by Cramer's Rule (multiplying by the adjugate) yields

$$\rm \begin{eqnarray} a\ x\, +\, b\ y &=&\rm X\\ \\ \rm c\ x\, +\, d\ y &\ =\ &\rm Y\end{eqnarray} \quad\Rightarrow\quad \begin{array} \rm\Delta\ x\ \ \ =\ \ \ \rm d\ X\, -\, b\ Y \\\\ \rm\Delta\ y\ =\ \rm -c\ X\, +\, a\ Y \end{array}\ ,\quad\ \Delta\ =\ ad-bc\qquad $$

Hence, by RHS system, $\rm\ n\ |\ X,Y\ \Rightarrow\ n\ |\ \Delta\:x,\:\Delta\:y\ \Rightarrow\ n\ |\ gcd(\Delta\:x,\Delta\:y)\ =\ \Delta\ gcd(x,y)\:.$
In particular $\rm\ n = \gcd(X,Y) \mid \Delta\, \gcd(x,y).\ $

Further, by LHS system $\rm\,n\mid x,y\ \Rightarrow\ n\mid X,Y\ \Rightarrow\ n\mid\gcd(X,Y).$
In particular $\rm\ n = gcd(x,y)\mid \gcd(X,Y).\ \ \ $ QED

Solution 2:

You can't do it with remainders.

The gcd divides also $3(2n+3)-2(3n-2)=13$

This can indeed happen, because, for $n=5$, $2n+3=3n-2=13$.

Solution 3:

Apply the euclidean algorithm:

$(3n-2)=(2n+3)+(n-5)$

$(2n+3)=2(n-5)+13$.

So the gcd is either $1$ or $13$.