1

Solve the linear congruence for $x$ : $$34x ≡ 51( \text{mod}\; 85)$$

I found using the Euclidean Algorithm that the GCD is $17$. Because the GCD evenly divides $51$, there equivalence should be solvable. I made a Diophantine equation to solve: $34p - 85q = 17$

From using the Euclidean Algorithm I have:

$$85 = 2 \cdot34+17$$

$$34=2 \cdot 17+0$$

$$17=85-2 \cdot34$$

I do not know where to go from here in order to make this model the Diophantine equation in order to solve for $p$, and I'm not entirely sure what to do with the solution when I get it, because I am solving for $x$.

Chinnapparaj R
  • 11,589
user52640
  • 105
  • Note that once you have one solution, $x$ say, then $x+5m$ are solutions for all integers $m$, because $34\cdot 5\equiv 0$ mod $85$. – TonyK Dec 10 '18 at 10:53

5 Answers5

3

In this case, it may be easier to reduce by dividing by $17$. You know that $$ 34x\equiv51\pmod{85}. $$ This equivalence is the same as $$ 34x-51=85k $$ for some integer $k$. Since all of $34$, $51$, and $85$ are divisible by $17$, we may divide through by $17$ to get $$ 2x-3=5k. $$ In other words, the original equivalence has the same solutions as $$ 2x\equiv3\pmod{5}. $$ Can you finish it from here?

Michael Burr
  • 32,867
  • Thank you very much. you saved my day. But I have a question for you. what should I do if the problem asks it like "solve x for 9x ≡ 2 (mod 21)" where you can't use the logic you posted. – user800956 Dec 26 '20 at 08:32
  • @user800956 It would be better to ask this as a separate question. The equivalence you posed has no solutions, (if $9x-2=21k$, then you can conclude that $3\mid 2$, which is false). – Michael Burr Dec 26 '20 at 19:54
1

We want to solve for $x$ where $$34x-51=85q$$ for some $q \in \mathbb{Z}$.

$$17(2x-3)=17(5q)$$

$$2x \equiv 3 \pmod{5}$$

Multiply both sides by $3$,

$$x \equiv 9 \equiv -1 \equiv 4 \pmod{5}$$

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
  • Thank you very much. you saved my day. But I have a question for you. what should I do if the problem asks it like "solve x for 9x ≡ 2 (mod 21)" where you can't use the logic you posted. – user800956 Dec 26 '20 at 08:46
  • argue that there is no solution by contradiction, by arguring that combination of multiples of $3$ give us multiple of $3$. – Siong Thye Goh Dec 26 '20 at 09:41
0

$\implies34x=51+85y$ where $y$ is an arbitrary integer

$\iff2x=3+5y$

$\iff2(x+1)=5(y+1)$

$\implies\dfrac{2(x+1)}5= y+1$ which is an integer

$\implies5|2(x+1)\iff5|(x+1)$ as $(2,5)=1$

lab bhattacharjee
  • 274,582
  • See also : My answers in https://math.stackexchange.com/questions/407478/solving-a-linear-congruence OR in https://math.stackexchange.com/questions/361336/how-to-solve-the-diophantine-equation-8x-13y-1571 – lab bhattacharjee Dec 10 '18 at 10:34
0

$$\begin{eqnarray*} 34x & \equiv & 51 \mod 85 \stackrel{34 = 85-51}{\Leftrightarrow}\\ -51x & \equiv & 51 \mod 85 \stackrel{:17}{\Leftrightarrow}\\ -3x & \equiv & 3 \mod 5 \stackrel{:(-3)}{\Leftrightarrow}\\ x &\equiv &-1 \equiv 4 \mod 5 \end{eqnarray*}$$

trancelocation
  • 32,243
0

Theorem: If $\gcd\{a,m\}=d$ and suppose $d \vert b$, then the linear congruence $$ax \equiv b \;(\text{mod}\;m)$$ has exactly $d$ solutions modulo $m$. These are given by $$t,t+\frac{m}{d},t+\frac{2m}{d},\cdots,t+\frac{m(d-1)}{d}$$ where $t$ is the solution , unique modulo $\frac{m}{d}$, of the linear congruence $$\frac{a}{d}x \equiv \frac{b}{d}\;(\text{mod}\;\frac{m}{d})$$

In your case, $a=34$, $b=51,m=85$ and $\gcd\{34,85\}=17$. so $t$ is the solution of $$2x \equiv 3\;(\text{mod 5})$$ Now since $\gcd\{2,5\}=1$, we have $$t \equiv3\cdot2^{\phi(5)-1}(\text{mod}\;5)$$That is $$t \equiv 24\;(\text{mod}\;5) \equiv 4$$

So the $17$ solutions are $$4, 4+\frac{85}{17},4+\frac{2 \cdot 85}{17}, \cdots,4+\frac{16\cdot 85}{17}$$

Reference : Tom M. Apostol : Introduction to Analytic number theory

Chinnapparaj R
  • 11,589