-2

I reading the following problem:

Let $a$ and $b$ be coprime integers, both $>1$. Prove that there exists an integer $x$ s.t. $1 \le x \le ab - 1, x \equiv 1 \pmod a$, and $x \equiv -1 \pmod b$. Show that $x^2 \equiv 1 \pmod {ab}$ and $x \not \equiv \pm 1 \pmod {ab}$

To be honest I am not really sure what the question is asking.
My approach so far:
$ab$ is obviously not a prime since it has the factors $a$ and $b$. Since $ab$ is not a prime then the roots of one property does not hold hence there is an $x$ s.t. $x^2 \pmod {ab} \space \And x \not \equiv \pm 1 \pmod {ab}$
This means that we have $x$ that is of the form $x - r = n(ab)$ and $r \not \equiv 0 $
So we would have $(x - r)(x -r) = n(ab)$

Then I am stuck.
Using an example such as $a = 2$ and $b = 5$ I can see that for $x = 9$ we have all the conditions requested satisfied and that $(4(2) + 1)(2(5) - 1) = 8(2\cdot 5) + 1 \equiv1 \pmod {10} $
but I am not sure how I can generalize this. I am also not sure how can I jump between the different modulo

Bill Dubuque
  • 272,048
Jim
  • 1,589

1 Answers1

0

Hint:

Use the Chinese remainder theorem: if $ua+vb=1$ is a Bézout's relation between $a$ and $b$, and if $x\equiv 1\pmod a$, $x\equiv -1\pmod b$, then $$x\equiv-1\cdot ua+1\cdot vb \pmod{ab}.$$ Further, you can rewrite $x$ as $\;x=(ua+vb)-2ua$.

Bernard
  • 175,478
  • The book I am reading has not covered the CRT so far, so I don't think it is expected to be used, I also am not familiar with it (or at least can't recall it) – Jim Jul 25 '21 at 22:30
  • It says that $\mathbf Z/a\mathbf Z\times\mathbf Z/b\mathbf Z\simeq \mathbf Z/ab\mathbf Z$ when $a$ and $b$ are coprime, and I used the explicit form of the isomorphism. – Bernard Jul 25 '21 at 22:37
  • Also I thought the question expects that I have to prove that $x \equiv 1 \pmod a$ and $x \equiv \pmod b$ and not take it as given? – Jim Jul 25 '21 at 22:37
  • "/×/≃/" I am not familiar with this notation – Jim Jul 25 '21 at 22:38
  • The symbol $\simeq$ means ‘ is isomorphic to’. For $x\equiv 1\pmod a, \equiv -1\pmod b$, if you have to prove these congruences, you need hypotheses on $x$ – they're false in general. – Bernard Jul 25 '21 at 22:41
  • 1
    Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 25 '21 at 23:45
  • @Bernard: Neither CRT nor isomporphism has been covered in the book I am reading so far. In fact CRT is covered in the next chapter. So I think something else is expected for the exercise, which to be honest, I am not sure I understand what it is asking – Jim Jul 26 '21 at 15:14
  • At least, did you see Bézout's identity for coprime numbers? – Bernard Jul 26 '21 at 15:16
  • @Bernard: The "$ua+vb=1$" you wrote, reminds me of the diophantine equation, and the name Bezout is mentioned there but the formula you specify not. Also the exercise states: "Prove..$x \equiv 1 \pmod a, and x \equiv −1 \pmod b$ but you seem to take it as a given. Shouldn't that be proven? – Jim Jul 26 '21 at 15:22
  • I took as a starting point the inverse isomorphism of the Chinese remainder theorem. A more elementary approch could be used if you've learnt to solve systems of congruences modulo coprime numbers. – Bernard Jul 26 '21 at 15:29
  • @Bernard: But is the $x \equiv 1 \pmod a$ and $x\equiv -1 \pmod b$ supposed to be proven or used as given? It seems that all the comments take that as a given but I don't understand the exercise question like that – Jim Jul 26 '21 at 15:54
  • For me, as stated, it is given, and you're asked to show, under these hypotheses, that $x^2\equiv 1\bmod ab$. – Bernard Jul 26 '21 at 16:15
  • @Bernard: May be there is something to learn on this. The way I read the statement is literary: "Prove that there exists an integer $x$ s.t. $1 \le x \le ab−1$ ,$ \equiv 1 \pmod a$, and $x \equiv −1\pmod b$". Emphasis mine to show how I read it. – Jim Jul 26 '21 at 23:05
  • @Jim: I see. One may add that it happens the way questions are written are rather loose. – Bernard Jul 27 '21 at 08:30