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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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