0
$\begingroup$

$m, n$ are integers, find all integer solutions of the diphantine equation:

$$nx + (n + 2)y = m$$

where $n$ is odd.

I´ve tried with Euclides but i get this:

$$2m = (n + 2)m - nm$$

I need a hint or something.

$\endgroup$
  • 1
    $\begingroup$ Your equation is linear, so you may try and write it in the form $y = \cdots$ and see if that helps. $\endgroup$ – Ruben Jun 20 '19 at 17:41
1
$\begingroup$

The diophantine equation $ax+by=c$ has solutions if and only if $gcd(a,b)|c$.
So for this equation $nx + (n+2)y = m$ to have solution $gcd(n, n+2)|m$ and $gcd(n, n + 2)$ is 1

From the Extended Euclidean Algorithm, given any integers $a$ and $b$ you can find integers $s$ and $t$ such that $$as+bt=gcd(a,b)$$ ($s$ and $t$ may not be unique) where $a = n, b = (n+2)$ and $gcd(a, b)=1$. Now multiply $m$ to both sides. you'll get $n(sm) + (n+2)(tm) = m$, this gives a solution $x=sm$, $y =tm$.

Here is a very similar problem which explains how to get other solutions from this

$\endgroup$
  • $\begingroup$ I'm sorry, I've corrected my post, $n$ is odd. $\endgroup$ – Octavio Berlanga Jun 20 '19 at 18:06
  • $\begingroup$ Hey, as gcd of two consecutive odd integers is 1. So, I changed the part gcd(n, n+2) = 2 to gcd(n, n+2) = 1. And other things a bit as now one solution is x = sm and y = tm $\endgroup$ – wild_fox Jun 20 '19 at 18:15
0
$\begingroup$

Hint if $n=2k+1$ then $n+2=2k+3$.

Then $$1=2k+1-2k=n-k(n+2-n)=n(1+k)-(n+2)k$$

Multiplying by $m$ you get the solution $$x_0=m(1+k)\\ y_0=-km$$

Now solve $$n(x-x_0)+(n+2)(y-y_0)=0$$

$\endgroup$
0
$\begingroup$

Hint $\,\ 1 = n(\overbrace{x\!+\!y}^{\Large z}) + 2y \iff (z,y)\, =\, \overbrace{(1,(1\!-\!n)/2)}^{\rm particular} + \overbrace{k(-2,n)}^{\rm homogeneous}$

$\endgroup$
  • $\begingroup$ $\large {\rm by}\ \bmod n\!:\,\ 2y\equiv 1\equiv 1\!-\!n\iff y\equiv (1\!-\!n)/2 \in\Bbb Z\ \ \, {\rm by}\ n\ \rm odd\ \ $ $\endgroup$ – Bill Dubuque Jun 20 '19 at 19:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.