I'm stuck with this exercise:

Find all $(a,b) \in \Bbb Z^2$ such that $b \equiv 2a \pmod 5$ and $28a+10b=26$

It's from my algebra class, we are looking into diophantic and congruence equations.

I started by looking to the $(a,b)$ that would solve the equation: $14a+5b=13$, those would be of the form $a=-13+5s$ and $b=39-14s$. Here is where I don't understand what I should do.

I've looked into $a$'s congruence mod 5 and got $5s \equiv 3 (5)$.

If $b$ should be two times $a$ then it would be: $10s\equiv1(5)$. Right?

So if I'm on the right path I still don't see how should I make to combine the first solution for b and this congruence requirement. What should I try? Thanks a lot.

  • $\begingroup$ Is this not modulo 2? $\endgroup$ – mvw May 31 '16 at 17:21
  • $\begingroup$ Sorry! I forgot to write down mod 5. $\endgroup$ – jrs May 31 '16 at 17:28

$$28a+10b=26\iff 14a+5b=13$$

Now use mod $14$ and mod $5$: $$14a\equiv 13\pmod{5}\iff -a\equiv -2\pmod{5}$$

$$\stackrel{:(-1)}\iff a\equiv 2\pmod{5}$$

$$5b\equiv 13\pmod{14}\iff 5b\equiv 55\pmod{14}$$

$$\stackrel{:5}\iff b\equiv 11\pmod{14}$$

Therefore it's necessary that $a=5k+2$ and $b=14t+11$ for some $k,t\in\mathbb Z$. $$14a+5b=13\iff 14(5k+2)+5(14t+11)=13$$

$$\iff 70(k+t)+83=13\iff k+t=-1$$

Therefore all the solutions are given by $$(a,b)=(5k+2,14(-k-1)+11)$$

$$=(5k+2,-14k-3),\, k\in\mathbb Z$$

You're given $b\equiv 2a\pmod{5}$, i.e. $$-14k-3\equiv 2(5k+2)\pmod{5}$$

$$\iff k+2\equiv 4\pmod{5}\iff k\equiv 2\pmod{5}$$

$$\iff k=5r+2,\, r\in\mathbb Z$$


$$=(25r+12,-70r-31),\, r\in\mathbb Z$$

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