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For each even number $n$ greater than $2$, there exists a $3$-regular graph with $n$ nodes.

Attempt: I already noticed that it's true for $n=4$ and $n=6$, so it would be true for any even number because we can generate all even numbers with the generators $4$ and $6$.

Example:

$12 = 4+4+4$ or $6+6$

$8 = 4+4$

$10 = 6+4$

Proposition: $\forall p \in 2 \mathbb{N}$, $\exists x,y \in \mathbb{N}$ such that $4x+6y=p$ where $p>3$

Proof by Strong induction

base case: p=4

4(1)+6(0)=4

Inductive hypothesis: Assume $\forall i \leq p \in 2 \mathbb{N}$, $\exists x,y \in \mathbb{N}$ such that $4x+6y=i$ where $i>3$

Show that the above claim is true for $p+2$

So $4t+6r=p-2$ by inductive hypothesis where $t$ and $r$ are the desired numbers for $p-2$. Therefore, $$4+4t+6r=p-2+4$$ $$4(t+1)+6r=p+2$$ which implies that $t+1$ and $r$ are the desired numbers for the even number p+2 and therefore the claim is proven

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    Right, so you're done, since the disjoint union of 3-regular graphs is a 3-regular graph. – 2017-01-10
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    Yes, but I still don't know if I have to prove my claim by using some number theory trick. $\forall p \in 2\N$, $\exists x,y \in \N$ such that $4x+6y=p$ – 2017-01-10
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    Prove by induction that every integer greater than $1$ can be expressed as $2x + 3y$, where $x,y$ are nonnegative integers (at least one of which is positive). Then multiply by $2$. – 2017-01-10
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    I think I got it. I will update the thread with the proof – 2017-01-10
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    You can get a connected graph by arranging the vertices in a circle and connect each vertex to its two neighbours and the vertex diametrically opposite. But if you don't need the graph to be connected, your graph works. – 2017-01-10
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    Yes I was also trying to understand that proof, but I just came up with mine. I tried to prove my claim by induction. I hope it's right. – 2017-01-10

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$Alternate$: Let there exists a $3-$regular graph with an arbitrary say $n$ vertices and $e$ edges. By Handshaking theorem;

$\sum_{i=1}^{n} deg(v_i)=2e$

$\implies deg(v_1)+deg(v_2)+.....+deg(v_n)=2e$

$\implies 3+3+.....+3 (n times)=2e$

$\implies 3n=2e$

$\implies e=\frac {3n}{2}\in \mathbb N$ iff $n$ is even

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    @Arthur...but see the word 'IFF' in the last line so this is a biconditional (not a conditional) which is true in both directions. – 2017-01-11