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So, basically the title. Prove that $17 | 29x + 18y \implies 17 | 23x + 9y$. Plus, does then $17 | 11x + 8y$ ?

Can you please explain what's the idea for doing these type of problems .. ?

4 Answers4

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For the first part :

$17 \mid 29x+18y \Rightarrow 17 \mid 17x+29x+18y \Rightarrow 17 \mid 46x+18y \Rightarrow 17 \mid 2\cdot (23x+9y) \Rightarrow 17 \mid 23x+9y$

For the second part :

$17 \mid 29x+18y \Rightarrow 17 \mid 17x+17y+12x+y \Rightarrow 17 \mid 12x+y$

On the other hand :

$17 \mid 23x+9y \Rightarrow 17 \mid 12x+y+11x+8y \Rightarrow 17 \mid 11x+8y$

Matko
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For the first part:

Let us first reduce it mod 17. Hence

$$29x+18y \equiv 12x+y \pmod{17}$$ $$23x+9y \equiv 6x+9y \pmod{17}$$

since $17$ is relative ly prime to $2$, just multiply the second equation by $2$. What can you conclude?

Maged Saeed
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    Ok I see it, and for the second just substract $12x + y$ from $23x + 9y$ .. –  Dec 19 '18 at 12:43
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    So basically, theres nothing smart in this, I just need to play around with congruences until I catch the right one, right ? –  Dec 19 '18 at 12:44
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    @someone, it amounts to Gaussian elimination mod $17$. – lhf Dec 19 '18 at 12:45
  • @someone, for the second part, you just have another solution idea. – Maged Saeed Dec 19 '18 at 12:48
  • @someone For the first part, you just reduce numbers to the complete residue system module 17. After that, you will conclude the system is solvable if and only if its determinant is relatively prime to 17. Just like the linear equation $ax+b \equiv 0 \pmod m$ is solvable iff $\gcd(a,m) = 1$ – Maged Saeed Dec 19 '18 at 12:49
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    Got it, thought it was harder. Thanks, again :) –  Dec 19 '18 at 12:52
  • In this case "for the first part", the system is linearly dependent, meaning that these two equations are actually the same equation. – Maged Saeed Dec 19 '18 at 12:55
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    @someone It seems that Maged has misunderstood the 2nd part of your problem. It follows the same way as the first, e.g. see my answer. – Bill Dubuque Dec 19 '18 at 17:06
  • @BillDubuque, you are right. I just deleted that part since its answer is the comments. Thanks for your note. – Maged Saeed Dec 19 '18 at 17:18
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    @Maged It's understandable since the wording is a bit ambiguous (such are the vagaries of natural language) – Bill Dubuque Dec 19 '18 at 17:26
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$\left.\begin{align} \!\!\bmod 17\!:\ \ 0\, \equiv &\, \ 18y + 29x\\ \equiv &\,\ \ \ \ \ y+12x \\ \iff\ \color{#c00}y\,\equiv&\ {-}12x\equiv\color{#c00}{5x} \end{align}\right\}\ $ If so then $\ \left\{ \begin{align} &\,\ 23x + 9\,\color{#c00}y\\ \equiv &\ \ \,\ 6x + 9(\color{#c00}{5x})\\ \equiv &\,\ 51x\,\equiv\, 0 \end{align}\right\}\ $ and $\ \left\{ \begin{align} &\,\ 11x + 8\,\color{#c00}y\\ \equiv &\, \ 11x + 8(\color{#c00}{5x})\\ \equiv &\,\ 51x\,\equiv\, 0 \end{align}\right\}$

Bill Dubuque
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Here is the ansatz:

If $a(29x + 18y)\equiv 23x + 9y \bmod 17$ for some $a \in \mathbb Z$, then the result follows easily.

This has to work for all $x,y$ and so $29a \equiv 23 \bmod 17$, which gives $a \equiv 9 \bmod 17$. This $a$ also works for $18a\equiv 9 \bmod 17$.

This argument actually shows

$17 \mid 29x + 18y \iff 17 \mid 23x + 9y$

Here is an explanation.

Consider the matrix $\pmatrix{ 29 & 18 \\ 23 & 9}$. Its determinant is $-153$, which is $0$ mod $17$. Therefore, its two rows are linearly dependent mod $17$. In particular, the second row is a nonzero multiple of the first row, mod $17$, which implies the result, even without computing an explicit multiplier. (Here it is important that $17$ is prime.)

For the second part, consider the matrix $\pmatrix{ 29 & 18 \\ 11 & 8}$.

lhf
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  • Adapted from https://math.stackexchange.com/a/2947271/589 – lhf Dec 19 '18 at 12:18
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    Thanks but I really prefer doing it using just number theory if possible, as I have no knowledge in linear algebra (even matrices and determinants) .. –  Dec 19 '18 at 12:23
  • @someone, just compute $9(29x + 18y)$ mod $17$. – lhf Dec 19 '18 at 12:25
  • Oh, but whats the trick with 9 ? –  Dec 19 '18 at 12:26
  • @someone $17$ is prime, so $9$ has a multiplicative inverse modulo $17$. – Shaun Dec 19 '18 at 12:29
  • Actually that gives me just $17 | 13x + 9y$ if I've done it right –  Dec 19 '18 at 12:30
  • Sorry, got it now .. But still cant see a way to do these problems in general –  Dec 19 '18 at 12:35
  • Ok, I got it. So this works only if modulo is prime, right ? –  Dec 19 '18 at 12:42
  • @someone, in general if the determinant is invertible mod the modulo. – lhf Dec 19 '18 at 12:43
  • @someone While this works generally it's a bit overkill here because we can easily solve for $,y,$ given that its coef is $\equiv 1\pmod{17},$ see my answer. – Bill Dubuque Dec 19 '18 at 17:03