Can $p$ divide both $a_1 − a_2$ and $a_1 + a_2$?

Question

If $a_1^2 ≡ a_2^2 \pmod p$, then $p$ divides $a_1^2−a_2^2$, so $p$ divides the product $(a_1 − a_2)(a_1 + a_2)$.

I read in a chapter related to quadratic residues and nonresidues that

unique prime factorization now tells us that $p$ divides $a_1 − a_2$ or $p$ divides $a_1 + a_2$, and so either $a_1 ≡ a_2 \pmod p$ or $a_1 ≡ −a_2 \pmod p$.

Why? It could be the case where both $p$ divides $a_1 − a_2$ and $p$ divides $a_1 + a_2$.

If $p$ divides both then it divides one of them. And it divides the other. So it is TRUE it divides one or the other. The statement "A or B" includes the possible that both A and B is true. — fleablood, Nov 02 '20 at 17:01
If $p$ divides both then $a_1 \equiv a_2 \pmod p$ and $a_1 \equiv -a_2\pmod p$. That's not a contradiction and nowhere did the author rule out that possibility. (However that will require the $a_2\equiv -a_2\pmod p$ and $a_1\equiv -a_1 \pmod p$. So that means either $p=2$ or $a_1,a_2\equiv 0 \pmod p$ ...[or both]). — fleablood, Nov 02 '20 at 17:05
In general "OR" means one or the other or possibly both. Both is not a contradiction of "OR". If an other means one or the other but not both, the author will usually have to say so. In this case we have $p|a_1 + a_2$ or $p|a_1 + a_2$ or both. And so $a_1\equiv a_2$ or $a_1\equiv -a_2$ *or both*. — fleablood, Nov 02 '20 at 17:13
To beat a dead horse. True table of $A$ or $B$ vs $A$ and $B$. is $A$ or $B$ is true if $A$ and $B$ are both true and true when one or the other is true and only false if both are false. $A$ and $B$ is also true if $A$ and $B$ are both true and false if both are false but are false if only one is true. So $A$ and $B\subset A$ or $B$. And So $A$ and $B\implies A$ or $B$. — fleablood, Nov 02 '20 at 17:18
The question boils down to the meaning of "logical or" (which is not number theory), so I have updated the tags to reflect that. — Bill Dubuque, Nov 03 '20 at 01:20

score 4 · Accepted Answer · answered Nov 02 '20 at 15:14

4

Yes, this could be the case (for instance if $a_1=a_2=p$). But also, they have not said that this cannot be the case. That's how "either ... or ..." is conventionally interpreted in mathematics: unless explicitly stated, that phrase does not exclude the possibility of both.

answered Nov 02 '20 at 15:14

Arthur

199,419

What is the role of unique prime factorization here? – Consider Non-Trivial Cases Nov 02 '20 at 15:16
Nothing. You need Euclid's lemma, and that's it. Of course, unique prime factorisation implies Euclid's lemma, but the logic usually goes the other way: We use Euclid's lemma to prove the fundamental theorem of arithmetic. And we use Euclid's lemma here to go from "$p\mid(a_1-a_2)(a_1+a_2)$" to "$p\mid(a_1-a_2)$ or $p\mid(a_1+a_2)$ (or both)". – Arthur Nov 02 '20 at 15:16
@Cons Euclid's Lemma (a prime $p$ divides a product $\Rightarrow$ $p$ divides some factor) is equivalent to the unuqueness of prime factorizations. – Bill Dubuque Nov 03 '20 at 01:16
@BillDubuque I know what you're trying to say, but it is a fact that any two true statements are equivalent. – Arthur Nov 03 '20 at 06:04
@Arthur They are equivalent properties of integral domains.I can elaborate if that is not clear, – Bill Dubuque Nov 03 '20 at 08:23
@BillDubuque I suspect that we are sitting quite squarely in the particular integral domain of $\Bbb Z$ in this case. Also, Euclid's lemma is the definition of a prime in integral domains (along with not being a unit), so that's trivially always true, while unique factorisation doesn't always hold. – Arthur Nov 03 '20 at 08:33
@Arthur I don't know why you insist on interpreting things in a way that it is clearly not intended, but, again, if you truly don't understand what is meant then I will be happy to elaborate. You can find the general definition of prime and irreducible (in domains) here. – Bill Dubuque Nov 03 '20 at 08:56
@BillDubuque I wasn't planning on taking this further than an innocent quip about people saying that proven theorems are equivalent, but you keep pushing the issue, so I am left here trying to defend my statements seriously. I honestly can't see a single unreasonable thing that I have said. You said they were equivalent, and I pointed out that when people say "two theorems are equivalent" it doesn't mean what they often want it to mean. – Arthur Nov 03 '20 at 09:11
Then (speaking of insisting on interpreting things in a way that was clearly not intended, this question was originally about elementary number theory, not abstract algebra) you said that you meant integral domains in general, rather than $\Bbb Z$ in particular. At which point I commented that in that case, they are not equivalent. If I have said anything untrue, I would love a correction, but otherwise I can't see how my responses are anywhere close to unreasonable, or misinterpreting. – Arthur Nov 03 '20 at 09:12
(I've fallen prey to the "proven theorems are equivalent" misconception myself, most recently my first comment in this thread, responding to the OP, where I claim that one theorem implies the other. Just because I know it's wrong doesn't mean I manage to avoid it all the time.) – Arthur Nov 03 '20 at 09:32
@Arthur In algebra, such claims about equivalences are always relative to some (ambient) base theory (here the theory of integral domain). Usually the base theory is clear from the context. One expects the reader to make intelligent interpretations. – Bill Dubuque Nov 03 '20 at 09:35
@BillDubuque Again, I have no clue how you got the idea that the ambient base theory was that of integral domains, rather than that of elementary number theory on $\Bbb Z$ (at least until you explicitly spelled out that that's what you were doing). That really doesn't seem like the most intelligent interpretation of the original context to me. – Arthur Nov 03 '20 at 09:47
@Arthur If you restricted number theory to $\Bbb Z$ you would greatly hinder your number theoretical education. – Bill Dubuque Nov 03 '20 at 10:16
@BillDubuque Really? When elementary school kids learn about primes for the first time, you think the teachers should go all in with integral domains, the distinction between prime and irereducible and all that? No, you start with $\Bbb Z$, then at some point you switch to the more advanced theory. You could argue that that point should be earlier than it conventionally is today, and I would probably agree. However, I don't think the OP has passed that point yet. And as such, I saw the (original) ambient theory as that of $\Bbb Z$, and phrased my answer and initial comments accordingly. – Arthur Nov 03 '20 at 10:31
@Arthur Red herring, since nothing above concerns "elementary school kids" (unless you happen to be one). – Bill Dubuque Nov 03 '20 at 10:40
@BillDubuque Not a red herring at all, just establishing a completely uncontested lower bound: Elementary school kids do number theory in $\Bbb Z$. At the same time, and upper bound: math professors do number theory on general integral domains. We have now bounded the point in-between, where people swap. The OP and this question is somewhere in-between these two bounds. I think the OP and the question is before the swapping point. You are welcome to disagree, but it is, ultimately, a matter of opinion. I'll let the name-calling slide for now, but please don't do it again. – Arthur Nov 03 '20 at 10:53

score 1 · Answer 2 · answered Nov 02 '20 at 16:22

A longish comment: If $p\mid(a_1-a_2)$ and $p\mid(a_1+a_2)$, then $p$ divides their sums and differences, which are $2a_1,2a_2$. This occurs when $p=2$, or when $p$ divides both of $a_1,a_2$, or when both statements are true, such as when $p=2$ and $a_1,a_2$ are both even.

Can $p$ divide both $a_1 − a_2$ and $a_1 + a_2$?

2 Answers2