Understanding a proof that $17\mid (2x+3y)$ iff $17\mid(9x +5y)$

Question

I was studying the number theory and came across this question.

Example 1.1. Let $x$ and $y$ be integers. Prove that $2x + 3y$ is divisible by 17 if and only if $9x + 5y$ is divisible by 17.

Solution. $17 \mid (2x + 3y) \implies 17 | [13(2x + 3y)]$, or $17 \mid (26x + 39y) \implies 17 \mid (9x + 5y)$. Conversely, $17 \mid (9x + 5y) \implies 17 \mid [4(9x + 5y)]$, or $17 \mid (36x + 20y) \implies 17 \mid (2x + 3y)$.

I have a difficulty understanding how $$17\mid(26x+39y)$$ implies $$17\mid (9x+5y)$$ and vice versa. I know this question as already been asked here but I didn't understand from that answer and since I'm new to Math SE, I don't have enough points to add a comment to that post to clarify that answer.

The difference between $26x+39y$ and $9x+5y$ is $17x+34y$, an obvious multiple of $17$. — Hagen von Eitzen, Oct 08 '18 at 15:12
Please link to the prior question so we don't duplicate answers here, and so we know precisely what you are reading there. — Bill Dubuque, Oct 08 '18 at 18:19
@RamshaKhalid Thanks, I added an answer which may prove helpful. — Bill Dubuque, Oct 08 '18 at 22:15
Special case of the linked divisibility mod rule, i.e.
$$,\ \bbox[8px,border:1px solid red] {n\mid A\color{#c00}\iff n\mid (A\bmod n)},\ $$

thus $\ 17\mid 26x+39y\iff 17\mid (26x+39y,\bmod 17)\equiv 9x+5y \ \ $ — Bill Dubuque, Feb 18 '24 at 11:38

nonuser · Answer 1 · 2018-10-08T16:25:14.980

3

From $26x+39y$ you can subtract any multiple of $17$, so you can substract $17x+34y$ to get $9x+5y$.

edited Oct 08 '18 at 16:25

answered Oct 08 '18 at 15:11

nonuser

90,026

Thank you for your answer. It did help me understand but I didn't mark this as answer bcz Robert gave a much detailed explanation.Once again thanks alot. :) – Ramsha Khalid Oct 08 '18 at 19:04

lhf · Answer 2 · 2018-10-08T16:34:24.167

2

For a different take, consider the matrix $\pmatrix{ 2 & 3 \\ 9 & 5}$. Its determinant is $-17$, which is $0$ mod $17$. Therefore, its two rows are linearly dependent mod $17$, which implies the result. (Here it is important that $17$ is prime.)

Explicitly, $a(2x+3y)=9x+5y$ works mod $17$ if $2a \equiv 9 \bmod 17$, that is, for $a=13$, which explains where $13$ came from in the proposed solution.

edited Oct 08 '18 at 16:34

answered Oct 08 '18 at 16:11

lhf

216,483

1

bazooka to kill a fly. – fleablood Oct 08 '18 at 16:34

fleablood · Answer 3 · 2018-10-08T15:32:36.173

1

If $a|b$ and $a|c$ then $a|b \pm c$.

And $26x + 39y = (9x + 5y) + 17(x + 2y)$ and $9x+5y = (26x + 39y) - 17(x+2y)$.

And $17|17(x+2y)$ always.

So $17|26x + 39 \implies 17|(26x + 39y) -17(x + 2y) \implies 17|9x +5y$.

And vice versa.

edited Oct 08 '18 at 15:32

answered Oct 08 '18 at 15:26

fleablood

124,253

score 1 · Accepted Answer · edited Oct 09 '18 at 23:23

I think the "if and only if" (abbreviated "iff" in the title of this question and in the textbook you're studying) is confusing you. It kind of suggests that one requires the other but the other does not necessarily require the one, when in fact the two conditions are mutually dependent: one requires the other and the other requires the one.

Do you remember how to add and subtract binomials from Algebra 101? Align the like terms in columns and then perform arithmetic as usual.

$$\begin{array}{rrrr} & 26x & + & 39y \\ - & 9x & + & 5y \\ \hline = & 17x & + & 34y \\ \end{array}$$

(okay, that's not properly aligned because in the time it takes me to figure out how to do it, twenty other answers will get posted and they'll be like "you copied from me").

Then it's easy to see that $$\frac{17x + 34y}{17} = x + 2y.$$ Since $17x + 34y$ is clearly a multiple of 17, it follows that if you subtract it from another multiple of 17, you will get yet another multiple of 17.

Bonus: if $17 \mid 9x + 5y$, then either $x$ or $y$ may be a square, but not both.

I got how we got this answer but to get $26x+39y$, is there any other method than trial and error method? — Ramsha Khalid, Oct 08 '18 at 18:59
I think there's a way through algebraic number theory, involving complex numbers. But I'm hazy on it and it could turn out to be a major digression. — Robert Soupe, Oct 08 '18 at 19:35

score 0 · Answer 5 · answered Oct 08 '18 at 16:04

0

Note that $26=17+9$ and $39=34+5$

Therefore working in mod (17) you can replace, $26$ with $9$ and $39$ with $5$

answered Oct 08 '18 at 16:04

Mohammad Riazi-Kermani

68,728

Understanding a proof that $17\mid (2x+3y)$ iff $17\mid(9x +5y)$

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