For every $n \in \Bbb Z$, prove there exists $a, b \in \Bbb Z$ such that $n = 5a + 2b$

Question

I was thinking of breaking the $n$ into two cases: one where $n$ is even and one where $n$ is odd. Can anyone give me a hint on how to proceed from here?

Thanks.

Answer 1

Not too hard:

We don't need to look at the cases of even and odd $n$ separately; we have

$\gcd(2, 5) = 1 \Longrightarrow \exists x, y \in \Bbb Z, \; 5x + 2y = 1, \tag 1$

which is known to some as Bezout's identity, though it is not always referred to as such; then

$n = n(1) = n(5x + 2y) = 5(nx) + 2(ny); \tag 2$

now take

$a = nx, \; b = ny, \tag 3$

and write

$n = 5a + 2b. \tag 4$

As pointed out by Ethan Bolker in his comment, a solution to

$5x + 2y = 1 \tag 5$

is

$x = 1, \; y = -2, \tag 6$

which is easy enough to guess; knowing that $x$, $y$ exist motivates us to do so.