Prove by induction that every integer n, n >=0, can be represented by 5a + 7b for a,b in Z

Question

I'm having trouble approaching this by induction.

I prove the base case in that if n = 0, 5a + 7b = 5(0) + 7(0) = 0

I assume that 5a + 7b = n is true, then I try to show that it holds true for n+1. I'm assuming that I can't use the same a and b for the second equality, so I wrote it as 5a₂ +7b₂ = n+1.



From here, I substituted 5a + 7b for n, simplified and got to 5(a₂ - a) + 7(b₂ - b) = 1. I'm not sure if that's right or where to go from here

Answer 1

Note 5(-4) + 7(3) = -20 + 21 = 1. So if 5a + 7b = n, then

n + 1 = 5a + 7b + 5(-4) + 7(3) = 5(a - 4) + 7(b + 3)

So the result is true for n + 1.