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I want proof that:

If $2x + 3y$ is multiple of $17$, then $9x + 5y$ is multiple of $17$.

My attempt:

By the Bezout's theorem, how $mdc(2, 3) = 1$, exists $r, s \in \mathbb{Z}$ such that $$2r + 3s = 1$$

Then, consider the set

$$\mathcal{A} = \lbrace (r,s) \in \mathbb{Z} : 2r + 3s = 1 \rbrace$$

By the hypothesis, $2x + 3y = 17k$, for some $k \in \mathbb{Z}$. Observe that for all $(r, s) \in \mathcal{A}$, $$(2r + 3s)17k = 2(17kr) + 3(17ks) = 17k$$ Thus, we can consider that for some $(r, s) \in \mathcal{A}$ $$17kr = x$$ and $$17ks = y$$

Hence, $17|9x + 5y = 17(9kr + 5ks)$.

I'd like to know if my attempt is correct. Someone can help me? Thank you in advance

Bill Dubuque
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J.A.G
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5 Answers5

3

$$3(9x+5y)-5(2x+3y) =17x$$

Since $3$and $17$ are relatively prime if $2x+3y$ Is a multiple of $17$ so is $9x+5y$

Mohammad Riazi-Kermani
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As remarked. that type of argument doesn't work. Instead $\rm\color{#0a0}{scale}$ it to match-up $\rm\color{#c00}{coef's}$ of $x$

Hint: $\ \bmod 17\!:\ \ \color{#c00}{9}x\!+\!5y\,\equiv \color{#0a0}{{-}4}\,(\color{#c00}2x\!+\!3y)\equiv -4(0)\equiv 0$

We employ the scaling factor $\,\color{#0a0}{-4}\equiv \color{#c00}{\dfrac{9}2}\equiv \dfrac{\!-8}2\,$ to equalize the coef's on $\,x,\,$ via $\,(9/2)2 = 9$

Remark $ $ It's a perfect scaling since $\,\dfrac{9}2\equiv \dfrac{5}3,\ $ by $\,9\cdot 3\equiv 2\cdot 5,\,$ i.e. the scaling also equalizes the coef's on $\,y,\,$ by $\,(5/3)3 = 5.\,$ Over any field - we can employ such cross products to test for scale equivalence of linear forms - just as we use line slopes to test equivalence of lines (through origin).

Alternatively we could take a linear combination of the forms that eliminates $x$ or $y$ (but this involves scaling both forms so we need to further argue that scaling by an integer coprime to $17$ doesn't affect divisibility by $17$, e.g. see my comment on Mohammad's answer).

Bill Dubuque
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2

Your proof is not correct since $9$ and $5$ never really come into play.
But you may notice that $\frac{9}{2}=9(2)^{-1}$ and $\frac{5}{3}=5(3)^{-1}$ are the same class $\!\!\pmod{17}$, namely $13$.
So $2x+3y\equiv 0\pmod{17}$ implies $13(2x+3y)\equiv 0\pmod{17}$, i.e. $9x+5y\equiv 0\pmod{17}$.

Jack D'Aurizio
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Your proof attempt is not valid. The problem is that the original statement you are trying to prove is a statement about all pairs $x,y$ that satisfy $17 \mid (2x + 3y)$, but your argument just produces a particular pair $x,y$ that satisfies this. It is true that if you make sure $x,y$ are both multiples of $17$, then both $2x+3y$ and also $9x+5y$ will be multiples of $17$, but your argument doesn't say anything about pairs $x,y$ that satisfy that $2x+3y$ is a multiple of $17$ even though $x,y$ are not.

Ben Blum-Smith
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If $2x+3y\equiv 0\pmod{17}$ then $$0\equiv2x+3y\equiv36x+20y=4(9x+5y)\equiv0\pmod{17}$$

Since $(4,17)=1$, the result follows.