<s>[INST] Triplets of distinct integers > 1 that return integer values. If $(A, B, C)$ are distinct integers $> 1$, and $$f(A, B, C) = \frac{\frac{A^2-1}{A} + \frac{B^2-1}{B}}{\frac{C^2-1}{C}},$$ then for what (if any) triplets $(A, B, C)$ is $f(A, B, C)$ an integer?
UPDATE: I've done some more work and have come up with the following: Based on the equations I combined to arrive at $f(A,B,C)$, it follows that: $$f(A,B,C)>2\Rightarrow A>B>C>1$$ and $$f(A,B,C)=2\Rightarrow A>C>B>1$$
My further work: 
Let $f(A,B,C)=k$ where $k>1$ is an integer. Rewrite $f(A,B,C)$ as $$k=\frac{A-\frac{1}{A}+B-\frac{1}{B}}{C-\frac{1}{C}}$$
Thus $$A-\frac{1}{A}+B-\frac{1}{B}-k(C-\frac{1}{C})=A-\frac{1}{A}+B-\frac{1}{B}-kC+\frac{k}{C}=0$$
Note that $\frac{1}{A}+\frac{1}{B}<1$ for all integer $A,B$ such that $A>B>1$.
Now we consider three cases: $1)$ $k<C$, $2)$ $k=C$, $3)$ $k>C$.
Case $1$: $k<C$
Since $\frac{k}{C}<1$ it follows that these equations hold:$$A+B=kC$$ $$\frac{1}{A}+\frac{1}{B}=\frac{A+B}{AB}=\frac{k}{C}$$ 
By substitution $$\frac{kC}{AB}=\frac{k}{C}\Rightarrow AB=C^2$$
This is only true if $A=np^2$, $B=nq^2$, and $C=npq$ for integers $n,p,q\geq 1$ and $p>q$. From this we have $$p^2>pq; pq>q^2\Rightarrow p^2>pq>q^2\Rightarrow np^2>npq>nq^2\Rightarrow A>C>B>1\Rightarrow k\leq2$$ This last condition, $k\leq2$, follows from the conditions stated above and leaves us with two possibilities: $k=2$ or $k=1$.
If $k=1$ we have $A+B=C$ which violates the condition $A>C$.
If $k=2$ we have $A+B=2C\Rightarrow \frac{A+B}{2}=C$. By substitution we have $$\frac{A+B}{AB}=\frac{2}{\frac{A+B}{2}}$$ $$4AB=(A+B)^2\Rightarrow A^2-2AB+B^2=0\Rightarrow (A-B)^2=0$$
However, this leads to the result that $A=B$ which violates that stated conditions.
Case $1$, $k<C$, fails.
Case $2$: $k=C$
Since $\frac{k}{C}=1$ it follows that $$A+B+1=C^2$$ $$\frac{1}{A}+\frac{1}{B}=0$$
This later equation is clearly false and thus Case $2$, $k=C$ fails.
Case $3$: $k>C$
Let $p,q$ be integers with $C>q\geq0$ and $p>1$. From this we can write $k$ as $k=pC+q$.
Note that $q=0 \Rightarrow A+B+p=C^2$ and $\frac{1}{A}+\frac{1}{B}=0$ (as above) which is clearly a contradiction, so $q\geq1$.
Substituting in $k$ above we have $$A-\frac{1}{A}+B-\frac{1}{B}-pC^2-qC+p+\frac{q}{C}=0$$ 
Applying the same logic as above we have the following equations $$\frac{A+B}{AB}=\frac{q}{C}$$ $$A+B=pC^2+qC-p$$
By substitution we have the following system of equations (one cubic and one quadratic) $$pC^3+qC^2-pC=qAB\Rightarrow pC^3+qC^2-pC-qAB=0$$ $$pC^2+qC-p-A-B=0$$
Since at least one $C$ that works must solve both equations we denote the roots of the cubic as $x_1,x_2,x_3$ and the roots of the quadratic as $x_1,y_2$.
By Vieta's formulas we have $$x_1+x_2+x_3=-\frac{q}{p}$$ $$x_1 x_2+x_1 x_3+x_2 x_3=-1$$ $$x_1 x_2 x_3=\frac{qAB}{p}$$
and $$x_1+y_2=-\frac{q}{p}$$ $$x_1 y_2= -\frac{p+A+B}{p}$$
From the first equations (setting $x_1+x_2+x_3=x_1+y_2$) we have $y_2=x_2+x_3$.
Thus $$x_1 y_2=x_1 x_2+x_1 x_3=-\frac{p+A+B}{p}\Rightarrow x_2 x_3-\frac{p+A+B}{p}=-1\Rightarrow x_2 x_3=\frac{A+B}{p}$$ $$x_2 x_3=\frac{A+B}{p}\Rightarrow x_1\frac{A+B}{p}=\frac{qAB}{p}\Rightarrow x_1=\frac{qAB}{A+B}$$
Since $x_1=C$ and $B>C$ we have $$B>\frac{qAB}{A+B}\Rightarrow B^2>(q-1)AB$$
Since $A>B$ it follows $AB>B^2$ and so $q-1<1\Rightarrow q<2$. Since $q\neq0$ (see above) we have $q=1$ and so $C=\frac{AB}{A+B}$.
Substituting for $C$ in the quadratic (although the cubic has the same result) we have $$0=p(\frac{AB}{A+B})^2+\frac{AB}{A+B}-p-A-B$$ Which, after a fair bit of rearranging, becomes $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ Thus $f(A,B,C)$ is an integer (with $A,B,C$ constrained as above) iff $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ has integer solutions $p>0$ and $A>B>2$.
Any help with solving this last bit would be greatly appreciated. So far guesswork and Wolfram Alpha have failed to produce results.
 [/INST] $f(A,B,C)$ has no integral solutions for distinct integers $A,B,C$.
Let $f(A,B,C)=k$ as in the question. $k<C$ and $k=C$ have been shown above to lead to contradictions. $k>C$ has been shown to reduce to solving $$p=\frac{(A+B)^3-AB(A+B)}{(AB)^2-(A+B)^2}$$ $$C=\frac{AB}{A+B}$$ for positive integers as restricted above. 
Start with: $$C=\frac{AB}{A+B}$$ This equation is identical to $$\frac{1}{A}+\frac{1}{B}=\frac{1}{C}$$ Let $A=C+s$ and $B=C+t$ with $s,t$ positive integers with $s>t$. We now have  $$\frac{1}{C+s}+\frac{1}{C+t}=\frac{1}{C} \Rightarrow (C+s)(C+t)=C(C+s+C+t)$$ $$C^2+C(s+t)+st=2C^2+C(s+t)\Rightarrow C^2=st$$
The solution set for this last equation is $s=na^2, t=nb^2, C=nab$ for positive integral $n,a,b; a>b$. From this we have $A=na^2+nab$ and $B=nb^2+nab$. Substituting into our equation for $p$ we have: $$p=\frac{(na^2+2nab+nb^2)^3-(na^2+nab)(nb^2+nab)(na^2+2nab+nb^2)}{((na^2+nab)(nb^2+nab))^2-(na^2+2nab+nb^2)^2}\Rightarrow$$ 
$$p=\frac{(n(a+b)^2)^3-n^2ab(a+b)^2(n(a+b)^2)}{(n^2ab(a+b)^2)^2-(n(a+b)^2)^2}\Rightarrow$$
$$p=\frac{n^3(a+b)^6-n^3ab(a+b)^4}{n^4a^2b^2(a+b)^4-n^2(a+b)^4}\Rightarrow$$
$$p=\frac{n(a+b)^2-nab}{n^2a^2b^2-1}=\frac{na^2+nab+nb^2}{n^2a^2b^2-1}\Rightarrow$$
$$p=\frac{na^2}{n^2a^2b^2-1}+\frac{nab}{n^2a^2b^2-1}+\frac{nb^2}{n^2a^2b^2-1}\Rightarrow$$
$$p\approx\frac{1}{nb^2}+\frac{1}{nab}+\frac{1}{na^2}<1|a,b,n\geq2$$
Thus $n,a,b$ must be small. In fact, a quick computer search reveals that $p=\frac{na^2+nab+nb^2}{n^2a^2b^2-1}$ is only greater than $1$ if $n=b=1$. Rewriting for $p$ we have $$p=\frac{a^2+a+1}{a^2-1}$$ which is clearly asymptotic to $1$.
Testing a few values, if $a=2$ then $p=2.\overline{3}$, if $a=3$ then $p=1.625$. 
Since as $a$ gets larger $p$ gets smaller, and $p$ is asymptotic to $1$, clearly no integer $a\geq2$ results in an integral $p$.
Since there exist no integral $p$'s that can solve $k=pC+q$ (see the question for how to derive this expression) clearly $k$ is not greater than $C$ and Case $3$, $k>C$ fails.
$k=f(A,B,C)$ is neither less than, greater than, nor equal to $C$, which is clearly a contradiction and so,
$f(A,B,C)$ has no integral solutions for distinct integers $A,B,C$.
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