4
$\begingroup$

I think I'm on the right track, but I can only figure out how to prove for a specific $k$ of my choosing... I don't know how to generalize it for all $k$:

Assume $(3k+2,5k+3)=1$. Therefore, there exists some $x,y\in\mathbb{Z}$ such that $(3k+2)x+(5k+3)y=1$.

\begin{align} (3k+2)x+(5k+3)y&=3kx+2x+5ky+3y\\ &=3kx+5ky+2x+3y\\ &=k(3x+5y)+2x+3y=1 \end{align}

If I let $k=0$, then $2x+3y=1$, and a particular solution is $(x_0,y_0)=(2,-3)$. But this same $(x,y)$ is not valid for other values of $k$... There does appear to exist upon inspection a solution for any value of $k$, but I'm not sure how to prove it.

$\endgroup$
3
$\begingroup$

Your approach is right and is invoking Bezout's Identity which says that gcd$(a,b)$ is the smallest positive integer $c$ such that $ax+by=c$ with $x,y \in \mathbb{Z}$.

Therefore if we can show that there is a pair $(x,y)$ such that

$$(3k+2)x+(5k+3)y=1$$

Then we know that gcd$(3k+2,5k+3)=1$. A quick check shows that $x=5, y = -3$ will do the trick.

$\endgroup$
4
$\begingroup$

$$ 5 \cdot (3k+2) ~-~ 3 \cdot (5k+3) = 1 \qquad \forall k $$

Edit

I'm adding my comment to the answer: in order to remove the $k$s from the linear combination, (multiples of) 5 and 3 as coefficients are the only way. Since they do work, it's done!

$\endgroup$
0
3
$\begingroup$

HINT:

$$5(3k+2)-3(5k+3)=1$$

can you continue?

$\endgroup$
9
  • $\begingroup$ I'm not sure I see where the $5$ and $-3$ came from. $\endgroup$ – Mirrana Oct 4 '13 at 13:47
  • $\begingroup$ OK, I see how the math works out, but I don't see how you came up with those numbers. What procedure did you use? $\endgroup$ – Mirrana Oct 4 '13 at 13:48
  • 1
    $\begingroup$ we want x and y such that the k's get eliminated $\endgroup$ – Shobhit Oct 4 '13 at 13:49
  • $\begingroup$ In general, Euclid's algorithm can work, for finding the greatest common divisor. $\endgroup$ – Berci Oct 4 '13 at 13:49
  • $\begingroup$ @Berci I had considered that, but I couldn't figure out how to use it with the variables... $\endgroup$ – Mirrana Oct 4 '13 at 13:50
3
$\begingroup$

Another, slightly different, proof goes like this: $\gcd(5k+3,3k+2)$ must divide both the sum and difference of the two, i.e.

$$\gcd(5k+3,3k+2) \mid \gcd(8k+5,2k+1).$$

But $8k+5=4(2k+1)+1$, so the gcd must also divide $1$, proving $\gcd(5k+3,3k+2)=1$.

$\endgroup$
2
$\begingroup$

Using the Euclidean algorithm: \begin{align*} 5k+3&=1\times (3k+2)+(2k+1)\\ 3k+2&=1\times (2k+1)+(k+1)\\ 2k+1&=1\times (k+1)+k\\ k+1&=1\times k+1\\ k&= 1\times k+0 \end{align*} Therefore, the GCD of $3k+2$ and $5k+3$ is $1$. Additionally, there exists some $x$ and $y$ such that $(3k+2)x+(5k+3)y=1$. A particular solution is $(x,y)=(5,-3)$.

$\endgroup$

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.