Prove that every integer greater than 27 can be written as 5a+8b, where a,b in ZZ^+

1.Statement: P(n)For ${\displaystyle n\in \mathbb{N}}$ there exist a,b > 0 such that
n+27 = a * 5 + b * 8
We use mathematical induction to show that every integer greater than 27 can be written as
a*5+b*8, for a,b>0
Base Case: Let n = 1. Note that
28 = 4 * 5 + 1 * 8
Thus base case is true.
Induction Hypothesis: Let P(n) is true for all ${\displaystyle n\le k}$. That is there exist a,b > 0 such that
k+27=a*5+b*8
We have to find a'b' > 0 such that
(k+1)+27=a*5+b*8
2.Inductive Steps: First note that if n = a* 5 + b*8 for n > 27 then either ${\displaystyle a\ge 3\text{or}b\ge 3}$. Otherwise note that if a < 2 and b < 2 then
a*5+b*8 < 2*5+2*8=26, a contradiction.
Case 1:First assume ${\displaystyle b\ge 3}$ Now by using induction hypothesis we get
(k+1)+27=(k+27)+1
=(k+27)+(5*5-3*8)
=(a*5+b*8)+(5*5-3*8)
=(a+5)*5+(b-3)*8
In this case we get a' = a+5>0 and b' = b - 3>0.
Case 2: Now assume ${\displaystyle a\ge 3}$. Then we consider following derivation
(k+1)+27=(k=27)+1
=(k+27)+(-3*5=2*8)
=(a*5+b*8)+(-3*5+2*8)
=(a-3)*5+(b+2)*8
In this case we consider a' = a-3>0 and b' = b + 2>0.
Result: For ${\displaystyle n\in \mathbb{N}}$ there exist a,b > 0 such that
n+27 = a*5+b*8