Can you check if my proof is right?

Theorem. ∀x≥8,x can be represented by 5a+3b where a,b∈N.

Base case(s): x=8=3⋅1+5⋅1✓x=9=3⋅3+5⋅0✓x=10=3⋅0+5⋅2✓

Inductive step:

n∈Na1=8,an=a1+(x−1)⋅3b1=9,bn=b1+(x−1)⋅3=a1+1+(x−1)⋅3c1=10,cn=c1+(x−1)⋅3=b1+1+(x−1)⋅3=a1+2+(x−1)⋅3S={x∈N:x∈ax∨x∈bx∨x∈cx}

Basis stays true, because 8,9,10∈S

Lets assume that x∈S. That means x∈an∨x∈bn∨x∈cn.

If x∈an then x+1∈bx,

If x∈bx then x+1∈cx,

If x∈cx then x+1∈ax.

I can’t prove that but it’s obvious. What do you think about this?

**Answer**

**Proof by induction.**

For the base case n=8 we have 8=5+3.

Suppose that the statement holds for k where k>8. We show that it holds for k+1.

There are two cases.

1) k has a 5 as a summand in its representation.

2) k has no 5 as a summand in its representation.

**For case 1**, we delete “that 5” in the sum representation of k and replace it by two “3“s ! This proves the statement for k+1.

**For case 2**, since k>8, then k has at least three “3“s in its sum representation. We remove these three 3‘s and replace them by two fives! We obtain a sum representation for k+1. This completes the proof.

**Attribution***Source : Link , Question Author : 88sandvich , Answer Author : Fermat*