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I am having a lot of trouble with this proof. Here is what I've got:

Let $a,b,c,d,r,p$ be positive.

$p|ra-b$ and $p|rc-d$ imply $$p|r(ad-bc)$$

If $gcd(p,r) = 1$, we are done. So suppose $gcd(p,r) = l$ for $l\not=1$. We then get that $l|p$ and $l|r$ imply $l|b$ and $l|d$ and so $$l|(ad-bc)$$

Notice that $px+ry = l$ for some $x,y$, and thus $pxj+ryj = lj = (ab-bc)$ for some $j$. Thus if $p|r$, we are done.

Now, what if $p\nmid r$, what can I do then? I am very stuck.

note: $p$ doesn't have to be prime.

Bill Dubuque
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Mani
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  • Formatting note: \nmid looks a lot better than "\not |" Also, I prefer to put spaces around the usual division symbol, as I think $2,|,4$ looks better than $2|4$ but that's probably personal taste. – lulu Jul 02 '23 at 17:33
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    Note: if $p,|,r$ then your assumptions tell you that $p,|,b$ and $p,|,d$ which makes the desired claim clear. And: I am assuming that $p$ is prime...though you never said that. – lulu Jul 02 '23 at 17:36
  • correct! I am still stuck on the case where $p\not \nmid r$ though – Mani Jul 02 '23 at 17:39
  • I solved that for you. Well, trusting that you just forgot to say that $p$ was prime, it suffices to address the case $p,|,r$ since, clearly, $\gcd(p,r)\in {1,p}$. – lulu Jul 02 '23 at 17:40
  • @lulu p doesn't have to be prime. If p was prime, this would have been solved. I am assuming, however, that all the integers are positive. I tried the first few thousands using this code: [ import math import random i = 0

    while i < 5: p = random.randint(1, 1000) a = random.randint(1, 1000) b = random.randint(1, 1000) c = random.randint(1, 1000) d = random.randint(1, 1000) r = random.randint(1, 1000)

    if ((r * a - b) % p == 0 and (r * c - d) % p == 0):
     print((a*d - b*c) % p)
i=+1]
    – Mani Jul 02 '23 at 17:50
  • then I would edit your post for clarity. I expect all your readers will assume that $p$ is prime. But, yes. the claim is true even for composites. – lulu Jul 02 '23 at 17:56
  • As in the linked dupe, simply eliminate $,r,$ from $,ra\equiv b,\ rc\equiv d,,$ just like in linear algebra. – Bill Dubuque Jul 02 '23 at 22:33

1 Answers1

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The relation $p |(ra-b)$ gives $ra \equiv_p b$. Likewise, the relation $p|(rc-d)$ gives $rc \equiv_p d$. Applying:

$$ad-bc \equiv_p a(rc) - (ra)c \equiv_p 0.$$

[Indeed, use $d \equiv_p rc$ to get $ad \equiv_p a(rc)$, and use $b \equiv_p ra$ to get $bc \equiv_p (ra)c$.]

Mike
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  • Could you help me understnad why $$ad-bc \equiv_{p} a(rc) - c(ra)$$? Or, to rephrase my question, how do I know that if $\alpha \equiv_{p} \beta$ then $\alpha x \equiv_{p} \beta x$? – Mani Jul 02 '23 at 17:58
  • As per my edits in my answer, plug in $d \equiv_p rc$ to get $ad \equiv_p a(rc)$, and $b \equiv_p ra$ to get $bc \equiv_p (ra)c$. – Mike Jul 02 '23 at 18:01
  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 02 '23 at 22:33