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I was looking into a couple of exercises:
First exercise:

Prove that if $x \mid (x^2 + 1)$ the $x = 1$ or $x = -1$

My solution was that if $x \mid (x^2 + 1)$ this means that $x | x^2$ and $x | 1$. The only number that is a divisor of $1$ is 1/-1. Hence $x = 1$ or $x = -1$

Second exercise:

Prove that if $6 \mid (x^3 - x)$ for every integer $x$ (hint among $3$ consecutive integers, one must be a multiple of $3$

Now for the second exercise I started with the same approach i.e. that if $6 \mid (x^3 - x)$ then $6 \mid x^3$ and $6 \mid x$ and then I realized that the question is about to prove that it holds for all integers which is not obvious to me that it does. So I plugged in some value e.g. for $x = 7$ we have $7^3 - 7 = 343 - 7 = 336$ which is a multiple of $6$. Also for $x = 5$ we have $5^3 - 5 = 125 - 5 = 120$ again is multiple of $6$.

So the $6 \mid (x^3 - x)$ seems to hold but I notice that for $x = 7$ we indeed have $6 \mid (7^3 - 7)$ but it is not the case that $6 \mid 7$.

So I have two questions here:
a) Is it wrong to consider that if $x | (a + b)$ then $x | a$ and $x | b$? Or are there specific conditions that this formula holds? Does that mean that my solution to the first exercise is wrong?
b) how can I use the hint to solve the exercise? It is not clear to me

Bill Dubuque
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Jim
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  • 2 divides 1+1, but 2 does not divide 1 – Giulio R Jun 09 '21 at 08:47
  • @Giulio: So basically $x | a$ and $x | b$ $\equiv$ $x | (a + b)$ but not the reverse? – Jim Jun 09 '21 at 08:50
  • Yes! If $x$ divides both $a$ and $b$ then $x$ divides $a+b$ – Giulio R Jun 09 '21 at 08:55
  • Your proof is not wrong because $x \mid x^2$. What’s true is that if $x \mid (a + b)$ and $x$ divides one of the summands, then it divides both of them. This question is simply the case where ${a, b} = {x^2, 1}$. – shoteyes Jun 09 '21 at 08:56
  • @shoteyes: my proof is not wrong you say, but it seems that the structure I used to show the proof is wrong since it is only by coincidence that it is correct? – Jim Jun 09 '21 at 09:08
  • Does this helps you? https://math.stackexchange.com/q/859910 –  Jun 09 '21 at 09:11
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    But it’s not a coincidence. Because $x \mid x^2$ and $x \mid (x^2 + 1)$, then $x$ must divide both $x^2$ and $1$. One way to see why is that $1 = (x^2 + 1) - x^2$ which is a difference of two integers that are both divisible by $x$. – shoteyes Jun 09 '21 at 09:13
  • @shoteyes: ah I see what you mean, thanks for clarifying that – Jim Jun 09 '21 at 09:15
  • $$\large \begin{align} \text{This is true:}\ \ \ {\rm if}\ \ d,\mid, a+b\ \ &{\rm then}\ \ d,\mid, a\iff d,\mid, b\[.2em] {\rm i.e.}\ \bmod d!:\ {\rm if}\ \ a+b\equiv 0\ \ &{\rm then}\ \ a\equiv 0\iff b\equiv 0 \end{align}\qquad\qquad$$ – Bill Dubuque Jun 09 '21 at 09:16
  • @BillDubuque: The ⟺ does that mean or? Because I understood based on the counter example $2 \mid (1 + 1)$ but $2$ does not divide $1$ that we can't go the other direction – Jim Jun 09 '21 at 09:22
  • You can't prove the 2nd using the method in the first (i.,e. my prior comment). The 2nd is a FAQ with many tens of dupes, e.g. see those linked, and their links, etc. – Bill Dubuque Jun 09 '21 at 09:25
  • That bidirectional arrow means they are equivalent, if $a\equiv 0$ then $b\equiv 0$, and if $b\equiv 0$ then $a\equiv 0\ \ $ – Bill Dubuque Jun 09 '21 at 09:28
  • @BillDubuque: "This is true: if ∣+ i.e. mod: if +≡0 then ∣⟺∣" so we can equally say that $d | a$ or $d | b$ but not both? I still I am not sure what your comment means – Jim Jun 09 '21 at 09:31
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    It means that if $d$ divides a sum of two integers then if $d$ divides one summand then it also divides the other summand. In congruence language: if the sum of two integers is $\equiv 0$ then if one summand is $\equiv 0$ then the other summand is $\equiv 0,$ too. – Bill Dubuque Jun 09 '21 at 09:39
  • @BillDubuque: Thanks for explaining, I couldn't understand that from the notation – Jim Jun 09 '21 at 09:48

2 Answers2

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The given hint almost gives away the whole proof for the second exercise. This is because $$x^3 - x = x (x^2 - 1) = x (x - 1) (x + 1).$$

Can you conclude from this that $6 \mid (x^3 - x)$?

shoteyes
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  • Is it then: $x^3 - x = 3 \cdot m$ since $x^3 - x$ is divisible by $3$ hence $x^3 - x = 3 \cdot m$ where $m >= 1$ hence $x^3 - x = 3\cdot 2 x = 6\cdot x$ for $x > 2$ so it is divisible by $6$? – Jim Jun 09 '21 at 09:27
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As mentioned by shoteyes in the comments, if $x \mid (a + b)$ and either $x \mid a$ or $x \mid b$, then $x$ divides both $a$ and $b$.

In the first exercise, since it is known that $x \mid x^2$ and it is given that $x \mid (x^2 + 1)$, $x \mid 1$. Therefore, the only values of $x$ for $x \mid 1$ to be true are $-1$ and $1$. Your solution to the first exercise is correct, though the way it was shown was not made entirely clear. It would be best to mention the specific conditions for $x \mid a$ and $x \mid b$ to be true.

In the second exercise, even though it is known that $6 \mid (x^3 - x)$, it does not hold true for $6 \mid x^3$ nor $6 \mid x$ for every integer of $x$.

However, it might be helpful to factorise $x^3 - x$, as follows:

$$x^3 - x = x(x^2 - 1) = x(x - 1)(x + 1) = (x - 1) \cdot x \cdot (x + 1)$$

Since $x$ is an integer, $(x - 1)$, $x$ and $(x + 1)$ are consecutive integers. Therefore, by using the hint given in the question, one of the integers must be divisible by $3$ (take any $3$ consecutive integers and you will find that one will be divisible by $3$). Moreover, since $(x - 1)$, $x$ and $(x + 1)$ are consecutive integers, at least one of the integers must be divisible by $2$. Thus, since there is at least one integer which is divisible by $2$ and one integer which is divisible by $3$, $x(x - 1)(x + 1)$ must be divisible by $6$.

I hope that helps!

anipalur
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  • Is it then: $x^3 - x=3\cdot m$ since $x^3 - x$ is divisible by $3$ hence $x^3 - x=3\codt m$ where $m>=1$ hence $x^3 - x=3\cdot 2x=6 \cdot x$ for $x>2$ so it is divisible by 6? – Jim Jun 09 '21 at 09:34
  • @Jim Yes, since any 3 consecutive integers contain one integer which is divisible by 3 and at least one integer divisible by 2, the product of those three integers will be divisible by 6. I am not too sure whether the variable $x$ can be used in the last statement $x^3 - x = 6 \cdot x$ as it may appear to be an equation. I would opt for a different variable. Either way, $x^3 - x = 6 \cdot n$, where $n$ and $x$ are integers. – anipalur Jun 09 '21 at 10:13
  • I didn't use at all the fact that "at least one integer divisible by 2" though, I missed that part – Jim Jun 09 '21 at 11:03
  • @Jim That is an important part of the proof, as it shows that the product of three consecutive integers is not just divisible by 3, but by 6 as well. I hope it is clear now. :) – anipalur Jun 09 '21 at 11:06
  • So basically for $3$ consecutive integers at least $1$ is divisible by $2$ and at least $1$ is divisible by $3$? How is that proven? – Jim Jun 09 '21 at 15:49
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    @Jim I find it strange that the hint given for the second exercise is a not a known fact by the time that this exercise is given, but yes. You can prove by induction that if $n$ is a nonnegative integer, then every product of $n$ consecutive integers is divisible by $n!$. Thus, the aforementioned products are divisible by every integer between $1$ and $n$, inclusive. – shoteyes Jun 09 '21 at 21:18
  • @Jim For every 3 consecutive integers, at least 1 is divisible by 2 and only one is divisible by 3. For example, the three consecutive integers 4, 5 and 6 have at least 1 integer divisible by 2 (4 and 6) and only one divisible by 3 (6). – anipalur Jun 10 '21 at 01:40
  • @shoteyes: wasn't known explicitly or at least to me. From what I see from here https://math.stackexchange.com/questions/12067/the-product-of-n-consecutive-integers-is-divisible-by-n-without-using-the?rq=1 the proof is fairly complex. I am wondering if that specific conclusion i..e the divisibility by specifically $6$ or $3$ or $2$ have other obvious fundamental explanations that I am not aware of. – Jim Jun 12 '21 at 21:13