My first problem - Are any of the numbers equal to each other?

Notice that I called the problem "Are any of the numbers equal to each other?" as a kind of a hint, to eliminate the possibility of falsely deducing that there are 24 permutations, since if any of the numbers are by any chance equal to each other, there will be duplicates. This is the case here. Let's start by simplifying the inequality $\frac{a}{cd}$ > $\frac{1}{b}$ . We can multiply the equation by the denominators if and only if they are both strictly greater than $0$ . We can see from there that if both $c$ , and $d$ are negative, $cd$ must be positive. So, by multiplying, we get $ab>cd$ . Now, let $ab=x$ and $cd=y$ . We know that $x, y>0$ , $x>y$ , and that they both are integers. Substituting that into the fraction from the exercise condition, we get $\frac{x^{x}+y^{x}}{x+y}$ . Since it mustn't be an integer, the numerator mustn't be divisible by the denominator. If $x=1$ , the numerator is equal to the denominator, which isn't allowed. If $x$ is greater or equal to 3, the numerator can be factored to $(x+y)(x^{x-1}-yx^{x-2}+...\pm y^{x-1})$ , allowing it to be cancelled with the denominator, which also isn't allowed. So $x$ must be equal to 2. Since $x>y$ and both are greater than 0, y must be equal to 1. Substituting back in, we have to solve the diophantine equations $ab=2$ and $cd=1$ . The possible solutions for the first equation are $(a,b)=(\pm 1,\pm 2$ ), and using the conditions from earlier - $x, y>0$ , $x>y$ we end up with $a=2, b=1$ . Likewise, the possible solutions for the second equations are $(c,d)=(\pm 1,\pm 1$ ), and by eliminating we get $c=-1$ , and $d=-1$ . So now we see that c and d are equal to each other. Because of that, the possible number of permutations is $\frac{4!}{2!}$ , or 12.