0

How to prove $(11k + 5, 8 + 15k) > 1 \implies (11k + 5, 8 + 15k) = 13 \implies k \equiv 9 \pmod {13}$? I am able to prove the converse, but I'm having trouble with the first implication. Once I can prove that I suspect proving $(11k + 5, 8 + 15k) = 13 \implies k \equiv 9 \pmod {13}$ will not be as hard.

b_pcakes
  • 1,501

3 Answers3

1

\begin{align} \left. \begin{array}{c} g | 11k+5 \\ g|8+15k \\ \end{array} \right\} &\implies g|11(8+15k)-15(11k + 5) \\ &\implies g|13 \\ &\implies g=13 \; \text{or} \; g=1\\ \end{align}

\begin{align} g = 13 &\implies \left\{ \begin{array}{l} 13 |11k+5 \\ 13|8+15k \\ \end{array} \right. \\ &\implies \left\{ \begin{array}{rcl} 11k +5 & \equiv 0 & \pmod{13} \\ 8 + 15k & \equiv 0 & \pmod{13} \\ \end{array} \right.\\ &\implies 2k \equiv 5 \pmod{13} \\ &\implies k \equiv 9 \pmod{13} \end{align}

without mods

\begin{align} g = 13 &\implies \left\{ \begin{array}{l} 13 |11k+5 \\ 13|8+15k \\ \end{array} \right. \\ &\implies 13|-4(11k+5) + 3(8+15k) \\ &\implies 13|k+4 \\ &\implies k+4 = 13 +13n \quad (n \in \mathbb Z)\\ &\implies k = 9 + 13n \quad (n \in \mathbb Z)\\ &\implies k \equiv 9 \pmod{13} \end{align}

Steven Alexis Gregory
  • 26,769
  • 1
    i.e. mod the gcd $ := d:\ \dfrac{8}{15}\equiv -k\equiv \dfrac{5}{11},\Rightarrow 8\cdot 11\equiv 5\cdot 15,\Rightarrow, 0\equiv 88-75\equiv 13.,$ Thus $,d\mid 13,,$ so $,d=13,$ by $,d>1.\ $ – Bill Dubuque Oct 24 '16 at 02:15
  • 1
    Yes, it explains what you did in the language of fractions. – Bill Dubuque Oct 24 '16 at 02:27
  • @BillDubuque - took a minute, but I got it – Steven Alexis Gregory Oct 24 '16 at 04:29
  • It's another way to view the elimination of $,k.,$ Beware that modular fraction arithmetic is valid only for fractions with denominator coprime to the modulus (see here). Here that is true, otherwise there is a prime $,p\mid d,15,$ so $,p\mid d\mid 15k+8,$ $\Rightarrow$ $,p\mid 8,,$ contra $,(15,8)=1.,$ Similarly for $,5/11.\ $ – Bill Dubuque Oct 24 '16 at 14:46
0

$\gcd(11k + 5, 15k + 8) = \gcd(11k + 5, 4k + 3) = \gcd(3k - 1, 4k + 3) = \gcd(3k-1, k + 4) = \gcd(-13, k+4) $

If $\gcd(-13, k+4) \ne 1$ then as $13$ is prime $\gcd(-13,k+4) = 13$ so $k+4 \equiv 0 \mod 13$ so $k \equiv 9 \mod 13$.

fleablood
  • 124,253
0

$11k+5+8+15k=13+26k=13(2k+1)$ suppose that $d=(11k+5,8+13k)$ does not divide $13$, it divides $2k+1$. You have $11k+5=5(2k+1)+k$ so $d$ divides $k$, since $d$ divides $11k+5$ and $8+13k$, it divides $5$ and $8$. This implies $d=1$. Contradiction.

For the second implication check all the value in $Z/13$.

$k=0$ mod $13$, $11k+5=5$ mod 13

$k=1$ mod $13$ $8+15k=10$ mod $13$ .... $k=9$ mod 13 $11k+5=14$ mod $13$ = $13\times 8$ mod $13$ $15k+8$ = $26$ mod $13$

....

so $k=9$ is the only value which satisfies the assertion.

Tsemo Aristide
  • 87,475