Determine all $a$ and $b$ natural numbers such that $\frac {a^2+2b} {b^2-2a}$ and $\frac {b^2+2a} {a^2-2b}$ are whole numbers. I proceeded in the following way:  
It is clear that $a \ne 0$ and $b \ne 0$.  
Let $\frac {a^2+2b} {b^2-2a} = k, k \in \mathbb{Z} \tag 1$
and $\frac {b^2+2a} {a^2-2b} = m, m \in \mathbb{Z} \tag 2$ 
1. If $a = b$:
Let $a = b = n$. Then $k = m = \frac {n^2+2n} {n^2-2n} = \frac {n(n+2)} {n(n-2)} = \frac {n+2} {n-2} = p, p \in \mathbb{Z}$. From here we get that $n \in \{1, 3, 4, 6\}$. I don't know if these are all so I can't prove that there are no others.
2. If $a \ne b$:
For $k$ and $m$ to be whole numbers the numerators have to be greater than or equal to the denominator in $(1)$ and $(2)$.
From $(1)$ we get that $a^2+2b \ge b^2-2a \implies a+2 \ge b \tag {3}$
From $(2)$ we get that $b^2+2a \ge a^2-2b \implies b+2 \ge a \tag {4}$
Here I am stuck again... WolframAlpha gives me these solutions to $(3)$ and $(4)$.

So we have $b\in \{a-2,a-1,a,a+1,a+2\}$ 
Case 1:  $b= a-2$, then $$(a-2)^2-2a\mid a^2+2(a-2)$$
so $$a^2-6a+4\mid a^2+2a-4$$ 
and thus  $$a^2-6a+4\mid  (a^2+2a-4)-(a^2-6a+4)=8a-8$$ Now if $a=1$ then $b=-1$ which works. So $a>1$ and now we have $$a^2-6a+4\leq 8a-8\implies a^2-14a+12\leq 0$$
so $$ (a-7)^2\leq 37\implies |a-7|\leq 6...$$
We do similary for all other cases...
