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$m, n$ 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.

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

3 Answers3

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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

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

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

Bill Dubuque
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  • $\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\ \ $ – Bill Dubuque Jun 20 '19 at 19:04