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Someone gave me this problem:

Show that every even number $n$ is the difference of two natural numbers $a,b$ both of which are coprime to a given number $c$.

I have trouble to use the somehow make the information that $a$ and $b$ are coprime to $c$ useful. Any hints for this are appreciated. Thanks in advance I usually come up with something but this time I have no clue. I know we can write $$ \frac{a+b}{2}=m $$ where $n=2m$ but this does not help either. I also constructed a few examples but I found no real pattern.

Bill Dubuque
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calculatormathematical
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  • You seek an integer $,x = ,a,$ such that $,x,$ and $,x+n,$ are coprime to $,c,,$ which is equivalent to their product $,f(x)= x(x+n),$ being coprime to $,c.,$ CRT allows us to easily tackle such problems for any polynomial $f(x)$ with integer coef's - see my comment on Calvin's answer. – Bill Dubuque Jan 29 '23 at 17:27

2 Answers2

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Assume $c>1$ (otherwise the answer is obvious). The problem is equivalent to: find two elements of $\Bbb Z_c^\times$ (the set of invertible elements of $\Bbb Z_c$) which differ by $n\bmod c.$

By the Chinese remainder theorem, we may restrict to the case where $c$ is a power of some prime $p.$

If $p\ne2$ then $\varphi(c)>c/2$ so the injective map $\Bbb Z_c\to\Bbb Z_c,\;t\mapsto t+(n\bmod c)$ cannot send every invertible to a non-invertible, and we are done.

If $p=2,$ the set of differences of two elements of $\Bbb Z_c^\times$ is $2\Bbb Z_c.$ It contains $n\bmod c$ since $n$ is even, so again we are done.

Anne Bauval
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Let $P$ be the set of primes such that $1 + n \not \equiv 0 \pmod{p}$.
Let $Q$ be the set of primes such that $ -1 + n \not \equiv 0 \pmod{p}$.

Claim: Prove that $P \cup Q$ is the set of all primes.
(I'm not requiring that it's a disjoint union.)

Consider the prime decomposition of $c = \prod p_i^{r_i}$.
Solve the modular arithmetic system
$a \equiv \begin{cases} 1 \pmod{p_i} & p_i \in P \\ -1 \pmod{p_i} & \text{otherwise} \end{cases}$

Claim: Prove that $a, b=a+n$ satisfy the conditions of the problem.

Followup: Where did we use that $n$ is even?
Can we modify the proof so that this works for all numbers $n$?

Calvin Lin
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  • More generally, to show $,f(x) = x(x+n),$ has a value coprime to $,c = \prod p_i^{e_i}$ it suffices to show for every $,i,$ that $f,$ has a value coprime to $p_i,,$ i.e. $,f(x_i)\not\equiv 0\pmod{!p_i},,$ since then we get a solution by using CRT to solve the system $,x\equiv x_i\pmod{!p_i}.,$ It's clear if $,p_i > 2 = \deg f,$ since $\Bbb Z_p$ is a field so $f$ has as most $2 < p_i$ roots, so there is a nonroot $,x_i.,$ If some $,p_i = 2,$ then $\bmod 2!:\ n\equiv 0\Rightarrow f = x(x+n)\equiv x^2,$ so $,f(1)\not\equiv 0$. – Bill Dubuque Jan 29 '23 at 06:21
  • The same idea works for higher degree polynomials $,f\in\Bbb Z[x],$ (that don't vanish mod any $p_i,$ i.e. $p_i$ does not divide some coef). As above, we reduce to finding a nonroot for all primes $p_i \le \deg f)$. – Bill Dubuque Jan 29 '23 at 06:21