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I could really use some help here:

Prove that $17|2x+3y\iff 17|9x+5y$.

I don't even know how to start. Just pointing me at a similar problem that has a full solution would be helpful enough, but of course I will be extremely thankful if someone could explain how to approach solving this! :)

Bill Dubuque
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saremisona
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2 Answers2

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Note

$$17x+17y = 9x+5y + 4\cdot (2x+3y).$$

The rest follows immediately since $\gcd(17,4)=1$ since $at - au$ is always divisible by $a$. Here $a=17$ $t=x+y$ and $u$ is either ${1\over 17}(9x+7y)$ or ${4\over 17}(2x+3y)$.

Adam Hughes
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Let's see the problem in terms of linear algebra.

Let $u=(2,3)$, $v=(9,5)$, and $w=(x,y)$. These are vectors in $\mathbb F_{17}^2$.

Then, $17\mid 2x+3y\iff 17\mid 9x+5y$ iff $\langle u,w \rangle = 0 \iff \langle v,w \rangle = 0 $, and this happens iff $u^{\perp} = v^{\perp}$, which happens iff $u$ and $v$ generate the same subspace, that is, are linearly dependent.

Therefore, you want to prove that the vectors $(2,3)$ and $(9,5)$ are linearly dependent over $\mathbb F_{17}$. This is easy, because $$ \begin{vmatrix} 2 & 3 \\ 9 & 5 \end{vmatrix} = -17 \equiv 0 \bmod 17 $$

If you want do it explicitly, find $a$ such that $a(2,3)=(9,5)$ by solving $$ 2a \equiv 9 \bmod 17, \quad 3a \equiv 5 \bmod 17 $$ and hope to get the same solution. Indeed, the solution is $a=-4$.

lhf
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