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Find all positive integers $a$ and $b$ such that $\frac {a^2 + b} {b^2 − a}$ and $\frac {b^2 + a} {a^2 − b}$ are both integers.

WLOG, Let, $a\geq b$ So,
\[b^2+a\geq a^{2}-b\Rightarrow a+b\geq (a+b)(a-b)\Rightarrow b+1\geq a\]
Case 1: $a=b$
\[b-1|b+1\Rightarrow b-1|2\]
So, $a=b=2$ or $3$ in this case.
Case 2: $a=b+1$
\[b^{2}-b-1|(b+1)^{2}+b\Rightarrow b^2-b-1|4b+2\]
but $b^{2}-b-1\equiv 1$(mod $2$)
So, \[b^{2}-b-1|2b+1\]
But$b^{2}-b-1> 2b+1$ for $b\geq 4$
$b=3$ doesn't hold the divisibility. But true for b=1,2

So in this case (a,b)=(2,1) or (3,2)

A question: How to show "not divisible" sign in Latex

Use \not command.
Example $\not |$ $\not = \not \leq$