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The question goes like this: use the CRT to prove that if an integer $n>1$ is not a power of a prime, then there exists an integer $x$ such that $n|(x^{2}-x)$ but $n$ does not divide $x$ nor $x-1$. I can see this working with a simple example like $n=12$, which lets $x=4,-3$. I got this by using $n|(x^{2}-x)$ to imply $x^{2}-x-nk=0$ but I cannot see to proving this with the CRT in particular. Any help is appreciated.

Bill Dubuque
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Matt R
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2 Answers2

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Suppose that $n$ is not a prime power. Then $n=ab$ for some relatively prime numbers $a$ and $b$, both greater than $1$.

By the Chinese Remainder Theorem, there exists an integer $x$ such that $x\equiv 0\pmod{a}$ and $x\equiv 1\pmod{b}$. Thus $a$ divides $x$ and $b$ divides $x-1$. It follows that $ab$ divides $x(x-1)$. Note that $ab$ does not divide $x$, for $b$ divides $x-1$, and therefore $b$ and $x$ are relatively prime. Similarly, one can show that $ab$ does not divide $x-1$.

André Nicolas
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  • Is it because gcd$(a,b)=1$ that we can construct $x$ to be a multiple of $a$ such that $a|x$ but $b|(x-1)$? Why is is that we can argue of the existence of such an $x$ where $b|(x-1)$? Which part of the CRT says this? – Matt R Apr 25 '14 at 23:56
  • @Matt $\ a\mid x,\ b\mid x-1!\iff! x\equiv 0\pmod a,\ x\equiv 1\pmod b,,$ i.e. $\ x\equiv (0,1)\ $ mod $,(a,b).,$ By CRT that has a (unique) solution mod $,ab.\ \ $ – Bill Dubuque Apr 26 '14 at 00:08
  • Sorry, what I meant was how, with the information that $a$ and $b$ are relatively prime, we can argue that there must exist an $x$ where $x\equiv 0\pmod a$ and $x\equiv 1\pmod b$. – Matt R Apr 26 '14 at 00:36
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    That is a special case of the Chinese Remainder Theorem. The theorem says that if $m_1,m_2,\dots,m_k$ are pairwise relatively prime, then for any $c_1,c_2,\dots, c_k$ there is a solution of the system of congruences $x\equiv c_i\pmod{m_i}$. Moreover, it says that $x$ is unique modulo $m_1m_2\cdots m_k$. Ours is the case $k=2$, $m_1=a$, $m_2=b$, $c_1=0$, $c_2=1$. – André Nicolas Apr 26 '14 at 01:15
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Hint $\bmod n\,$ we seek a solution of $\ x(x-1)\equiv 0\ $ but $\,x\not\equiv 0,1.\, $ If $\,n = p^k\,$ is a prime power we get only trivial solutions: $\,p^k\mid x(x-1)\,$ $\Rightarrow\,p^k\mid x\,$ or $\,p^k\mid x-1\,$ by $\,x,\,$ $\,x-1\,$ coprime. Else $\,n\,$ is not a prime power so we can write $\,n = ab\ $ with coprime $\,a,b>1.\,$ What happens when we CRT lift the solutions $\, x\equiv 0,1\,$ mod $\,a,b\,$ to four solutions mod $\,n\,?$ The solutions $\,x\equiv (0,0),(1,1)\,$ mod $\,(a,b)\,$ map to $0,1\,$ mod $ab,\,$ but the solutions $\,(0,1),(1,0)\,$ mod $\,(a,b)\,$ map to values $\not\equiv 0,1\,$ mod $\,ab.$

Remark $\ $ Elements satisfying $\,x^2 = x\,$ (and $\,\color{#c00}{x\neq 0,1})$ are called $\rm\color{#c00}{(nontrivial)}$ idempotents. They are intimately connected to coprime factorizations (of both elements and rings). As we see above, modulo any non-prime-power composite $\,n,\,$ there are nontrivial idempotents $\,(0,1),(1,0).$

Some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally one can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so any notrivial idempotent or nontrivial square-root will split $\,n,\,$ since it yields a quadratic with $3$ roots).

Bill Dubuque
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  • I am not too strong with Abstract Algebra, but you provide an interesting comparison between it and Number Theory. I'll try my best to understand it. Also, thank you for the link! – Matt R Apr 26 '14 at 00:01
  • @Matt You don't need to understand the remarks to understand the main argument. If anything there is not clear please let me know and I will be happy to elaborate. – Bill Dubuque Apr 26 '14 at 00:05
  • What do you mean when you refer to lifting the solutions? Also, when you use the pairwise notation? Is that just a condensed way of denoting each congruency $x$ has? – Matt R Apr 26 '14 at 00:39
  • @Matt Lifting means to apply CRT to solve $,x\equiv (c,d),$ mod $,(a,b),$ to obtain the value of $,x$ mod $,ab.,$ You can think of the pair/vector notation simply as an abbreviation, but when you study rings you will learn that these pairs have a (product) ring structure given by component-wise operations, i.e. $,(c_1,d_1)+(c_2,d_2) = (c_1+c_2,d_1+d_2),\ $ and $\ (c_1,d_1)(c_2,d_2) = (c_1 c_2, d_1 d_2).,$ – Bill Dubuque Apr 26 '14 at 02:11
  • From this structural view, CRT says this ring of pairs is isomorphic to $,\Bbb Z/ab = $ integers mod $,ab,,$ i.e. $,\Bbb Z/ab = \Bbb Z/a\times \Bbb Z/b,$ is the product of the rings $,Z/a,, \Bbb Z/b,,$ i.e. said ring of pairs with said component-wise operations. For some examples of these pair (product) operations see the computations $,(0,1)+(0,1),$ and $,(1,1)+(0,1),$ in the final two lines of this post. – Bill Dubuque Apr 26 '14 at 02:11