Find $b \in \Bbb Z$ for which exists $a \equiv 4 \pmod 5$ such that $6a+21b=15$

Question

I'm starting to study diophantic equations and congruence and I have found this problem that I don't know how to solve:

Find $b \in \Bbb Z$ for which exists $a \equiv 4 \pmod 5$ such that $6a+21b=15$.

I can see that the equation is equivalent to $2a + 7b = 5$ and that all $a$ would be of the form $a = -1 + 7s$ and similarly $b = 1-2s$. But I don't understand how to pick $b$ such that $a \equiv 4 \pmod 5$. Any thoughts on how to solve it? Thanks

Answer 1

Notice that if $a\equiv 4 \pmod 5$ then

$$6a+21b \equiv a+b \equiv 0 \pmod 5,$$

so $b \equiv 1 \pmod 5$. In this case, using the form of the general answer you yourself found, we have that $-2s \equiv 0 \pmod 5$, which implies $s=5k$ for some $k \in \mathbb{Z}$. It follows that any solution respecting the modular condition on $a$ is of the form $b = 1-10k$ and $a=-1+35k$ for some $k \in \mathbb{Z}$.

Finally, observe that for any choice of $k$, we have $a \equiv -1 \equiv 4 \pmod 5$ and $b \equiv 1 \pmod 5$, so all of those are solutions. Even $k=0$ ($a=-1, b=1$) is a solution

Answer 2

Write $a = 5k+4$. The we may solve for $b$: \begin{align} 6(5k+4) + 21 b &= 15 \\ 21b &= 15-24-30k \\ b &= (-9-30k)/21 \\ b &= -(10k+3)/7 \text{.} \end{align}

For $b \in \Bbb{Z}$, we must have $10k+3 \cong 0 \pmod{7}$, which simplifies to $k \cong 6 \pmod{7}$.

As a check:

• $k=-1 \implies a = -1 \implies b = 1$ works,
• $k=6 \implies a = 34 \implies b=-9$ works,
• and none of $b=0, -1, -2, \dots, -8$ allow an integer $a$.