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Show that $3x+11$ and $5x+18$ are relatively prime for all positive integers $x$.

Hi everyone I've looked around a lot and found similar questions like this here but when trying some of the tips I feel like whenever I get close it doesn't match.

As I understand it there are two approaches I can take for a proof strategy and I can't get them to work (or am doing something wrong and not able to find the next logical step).

For example via the Euclidean Algorithm

So $3x+11$ and $5x+18$ are relatively prime. That means they are not both 0 and $$gcd((3x+11), (5x+18)) = 1$$

Also two integers $a$ and $b$ are relatively prime if and only if there exist integers $s$ and $t$ such that $$as+bt = 1$$

So I first tried dividing and cannot arrive at a way that I get this nice 1 sitting alone at the end.

$$5x+18 = (1) (3x+11) + (2x+7)$$ $$3x+11 = (1) (2x+7) + (x+4)$$ $$2x+7 = (1)(x+4)+(x+3)$$ $$x+4 = (1)(x+3)+1$$ $$x+3 = 1(1)+(x+2)$$ By that last step I think I'm lost...

Another common solution I'm seeing which is no help is people just saying well you know that if $(3x+11)$ and $(5x+18)$ are relatively prime then there there are two integers s, t such that $$as+bt =1$$ and they pull out like $(3x+11)(5) + (-3)(5x+18) = 1 $ which is great and fine and all but I have no idea how to get those two numbers by any sort of method besides just guessing. There must be a way I'm missing or a fundamental step in the Euclidean Algorithm or definitions of Linear Combinations that I'm missing.

backslash
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    You were done in the 2nd last step when you reached the remainder $1.\ $ That's the gcd, so they are indeed coprime. See here for a slicker more general method using Cramer's rule. – Bill Dubuque Nov 04 '18 at 04:16
  • Hi thanks for comment - I see now that there was a remainder of 1 (got so caught up that didn't realize that was happening) however none of the definitions I have regarding coprimes and gcd mention a remainder of 1 being a bonafide proof that two numbers are coprime so I can't make the jump there. It makes sense to me that it would be the case but there must be some law or rule or something involved. – backslash Nov 04 '18 at 04:32
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    Every remainder is an (integer-coef) linear combination of the original gcd arguments (Bezout / extended Euclid). Hence $\ n(3x+1) + k(4x+18) = 1,\ $ so any common divisor of $,3x+1,$ and $4x+18$ divides the LHS, so also the RHS $ = 1\ \ $ – Bill Dubuque Nov 04 '18 at 04:36
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    That's clearest when you do the extended Euclidean algorithm this way, which calculates the linear combinations at each step via an augmented matrix which accumulates the linear operations. – Bill Dubuque Nov 04 '18 at 04:39
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    Or consult the proof of the Euclidean algorithm - the gcd is not changed by any step since $,\gcd(a,b) = \gcd(a,b-na),$ so the initial gcd equals the final $\gcd(x+3,1) = 1\ \ $ – Bill Dubuque Nov 04 '18 at 04:45
  • I am wondering if my proof for Euclidean algorithm is missing something. It describes how we run through it as $b = aq_1 + r_1$ where $0 \le r_1 \lt a$ If $r_1 = 0, gcd(a,b) =a$ If not, then $\exists q_2, r_2 \in \Bbb Z $ such that $ a= r_1, q_2 + r_2$ where $0 \le r_2 \lt r_1$ If $r_2 = 0$ then $gcd(a, r_1) = r_1$ Now it carries on this way, saying $r_1 \gt r_2 \gt$ ... is a decreasing sequence of positive integers it MUST terminate. So there is a smallest n such that $r_n = 0$ It seems to omit any option of $r_n = 1$ and thus proving $a$ and $b$ are coprimes (ie: gcd(a,b) = 1) – backslash Nov 04 '18 at 05:30
  • Because after looking at your links and re-reading the Euclidean algorithm a few times at your suggestion I'm seeing how $r_n-1 = gcd(r_n-2, r_n-1) = gcd(r_n-3,r_n-2) = ... = gcd(a, r_1) = gcd(b,a)$. – backslash Nov 04 '18 at 05:41
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    $r_{n-1}$ is the gcd. i.e. $,\gcd(r_1,r_2) = \gcd(r_2,r_3) =, \cdots, = \gcd(r_{n-1},0) = r_{n-1}\ $ – Bill Dubuque Nov 04 '18 at 05:49
  • @BillDubuque thank you !! – backslash Nov 04 '18 at 06:17

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One simple way to prove these numbers are not relative primes is to show that the following fraction is irriducible: $$\frac{5x+18}{3x+11}=1+\frac{2x+7}{3x+11}$$ Hence, $\,2x+7\,$ and $\,3x+11\,$ can be co-primes iff (if and only if) the following fraction is reducible: $$\frac{3x+11}{2x+7}=1+\frac{x+4}{2x+7}.$$ Now for $\,x+4\,$ and $\,2x+7\,$ to be co-primes this fraction must be reducible: $$\frac{2x+7}{x+4}=1+\frac{x+3}{x+4}\,$$ So, if $\,x+3\,$ and $\,x+4\,$ are co-primes, then the next fraction is reducible: $$\frac{x+4}{x+3}=1+\frac{1}{x+3}\,$$ Now finally, we come to the following fraction which must also be reducible $$\frac{1}{x+3}$$ but it is not. So, the original numbers are not co-primes. Q.E.D.

In order for two numbers to be co-primes, the fraction must be reducible i.e. it must be in the form: $$\frac{xk}{xp}$$ where some number, say, $\,x\,$ is a common factor. If there are no common factor, the fraction is irreducible and its numerator and denominator are co-primes. $$............$$ There's even a simpler way to do it. $$\frac{15x+55}{15x+54}=1+\frac{1}{15x+54}$$ The fraction $$\frac{1}{15x+54}$$ is irreducible. So the initial numbers are co-primes because it is trivial to show that $\,15x+54\,$ is not divisible by $\,5\,$ but divisible by $\,3,$ and $\,15x+55\,$ is conversely not divisible by $\,3\,$but divisible by $\,5$. In other words multiplying one number by $\,5$ and the other by $\,3$ doesn't change the situation because after that we know that the initial numbers will still not be both (at the same time) divisible by $\,5\,$ or by $\,3\,$.

Ken Draco
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