Are the two statements concening number theory correct? Statement 1: any integer no less than four can be factorized as a linear combination of two and three.
Statement 2: any integer no less than six can be factorized as a linear combination of three, four and five.
I tried for many numbers, it seems the above two statement are correct. For example, 
4=2+2; 5=2+3; 6=3+3+3; ...
6=3+3; 7=3+4; 8=4+4; 9=3+3+3; 10=5+5; ...
Can they be proved?
 A: The question could be generalized, but there is a trivial solution in the given cases.  


*

*Any even number can be written as $2n$. For every odd number $x > 1$, there exists an even number $y$ such that $y+3 = x$.  

*Likewise, numbers divisible by $4$ can be written as $4n$. All other numbers over $2$ are $4n + 5$, $4n + 3$ or $4n + 3 + 3$.
A: Statement 1: Either $b$ is even or $b-3$ is even. 
Statement 2: Either $b$ is divisible by 3, or $b-4$ is or ... now complete the argument.
A: Hints:


*

*Use the division algorithm: $n=2q+r.$ What are the values of $r$? If $n≥4,$ then $q≥2,$ and we can re-write $n$ as: $$n=2(q−1)+2+r.$$ What are the values of $2+r$?

*Again use the division algorithm: $n = 3q + r,$ and with $n ≥ 6,$ we have $$n = 3(q-1) + 3 + r.$$ What are the possible values for $3 + r$?
A: Read part 1.2 of this book:
http://shoup.net/ntb/ntb-v2.pdf
..on Ideals.  An Ideal is the set of integers that can be expressed as multiples of some constant k.  Let this be $I_k$
Your question is whether $I_2 + I_3 = \mathbb{Z}$
It can be shown that $I_a + I_b = I_{\gcd(a,b)}$
And easily shown that $I_1 = \mathbb{Z} $
