# Find all even natural numbers which can be written as a sum of two odd composite numbers.

Find all even natural numbers which can be written as a sum of two odd composite numbers.

Let $n\geq 100$ an even number. Consider the quantities $n-91$, $n-93$ and $n-95$, one of these is a multiple of 3, and not exactly 3 cause $100-95>3$, then is a composite odd number. Observing thet 91,93 and 95 are composite, you conclude that every $n\geq 100$ works. Now check directly the remaining numbers, and you have the solution.
• +1 To finish this (without further optimization) requires processing around $\color{#c00}{50}$ or more possible exceptional cases. $\$ Much faster is to use $\,40\,$ vs. $\,100,\,$ which employs only $\,\color{#c00}7\,$ cases - see my answer. Commented Jan 19, 2014 at 19:01
$n\ge 40\,\Rightarrow\, 3$ divides one of $\,n\!-\!a,\ a \in\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{\{9,25,35\}}^{\large\qquad\ \equiv\ \{0,\ \ 1,\ \ 2 \}\,\pmod{\! 3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ so $\ n = (n-a)+ a\,$ works, by $\,3\mid n-a > 3.\,$ Else $\,39 \ge n = a + b,\,$ wlog $a \le b\,$ so $\,a< 20\,$ so $\,a = \color{#c00}9$ or $\color{#0a0}{15}\,$ (the only odd composites $< 20).$ Hence the only $\,n\le 39\,$ that work are
$$\begin{eqnarray}\color{#c00}9+\!\!\!\!\!\!\!\!\!\!\overbrace{\{9,15,21,25,27\}}^{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#c00}9,\ 39-\color{#c00}9{\bf ]}}\!\!\!\!\!\!\!\!\!\! &=& \{18,24,30,34,36\}\\ \\ \color{#0a0}{15}+ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\{15,21\}}_{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#0a0}{15},\ 39-\color{#0a0}{15}{\bf ]}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &=& \{30,36\}\quad\ \ \, {\bf QED}\end{eqnarray}$$
• @anorton Yes, we may choose notation so that $\,a\,$ denotes the least summand. Commented Jan 19, 2014 at 18:25