3
$\begingroup$

Decide whether the equation has solutions and if it does, find them all.

$10x \equiv 14 \pmod {17}$

Since $17\mid 10x-14$, $x$ must have the following form $x=\frac{17y+14}{10}$. $x$ belongs to natural numbers therefore, $10\mid 17y+14$ and the only solutions are the ones where $y$ finishes in $8$ if $y>0$ or in $2$ if $y<0$.

So $y=\{8,28,38,\ldots\}$. Then $x= \{15,49,66,\ldots\}$ $y=\{\ldots,-32,-22,-12\}$. Then $x=\{\ldots,-66,-49,-15\}$

Isn't there an infinity of solutions? I am not sure if my solution is correct. Any ideas?

$\endgroup$
  • $\begingroup$ There are infinitely many solutions, but they all belong to the same congruence class. You want to approach this problem with first checking if $\gcd(10, 17) = 1$ (guaranteeing a unique solution), and then using the Euclidean Algorithm to find the inverse of $10$ modulo $17$. $\endgroup$ – apnorton Feb 12 '15 at 0:48
  • $\begingroup$ What if $10$ is replaced with, say, $7$? $\endgroup$ – Bernard Feb 12 '15 at 0:49
1
$\begingroup$

Hint $\,\ {\rm mod}\ 17\!:\,\ 10x\,\equiv\, 14\iff x\,\equiv\,\dfrac{14}{10}\,\equiv\, \dfrac{14}{-7}\,\equiv\, -2$

Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus. See here for further discussion.

$\endgroup$
  • $\begingroup$ You shouldn't use fractions in modular arithmetic. $\endgroup$ – Bernard Feb 12 '15 at 1:10
  • $\begingroup$ @Bernard That's a matter of opinion, on which I strongly disagree. $\endgroup$ – Bill Dubuque Feb 12 '15 at 1:12
  • $\begingroup$ I meant it's confusing for beginners. $\endgroup$ – Bernard Feb 12 '15 at 1:17
  • $\begingroup$ @Bernard Only if not taught correctly. $\endgroup$ – Bill Dubuque Feb 12 '15 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.