Question: For simplicity let $\Bbbk$ be a field of characteristic $0$ and let $A$ be a finitely generated unital associative $\Bbbk$-algebra.

Is it true that for any two simple $A$-modules $S\_1, S\_2$, we have that $\operatorname{Ext}^1\_A (S\_1, S\_2)$ is finite-dimensional?

If not, what would be a simple counterexample, and what sort of conditions do we need to ensure this?

(As far as I understand, the Weyl algebra $\Bbbk \langle x, y \rangle / (xy - yx - 1)$ has only infinite-dimensional simple modules, but their first extension groups are still finite-dimensional.)

Answer: 
$$797a+50b=292c$$
Here we see that $a$ must be even, and if $a\ge4$ then the LHS is too big for the RHS to match in magnitude (remember that $0\le c\le 9$). Hence $a=2$:
$$1594+50b=292c$$
Now we see that $c$ must be at least $6$ so that the RHS can match the LHS in magnitude, and must be $2$ or $7$ to match in the last digit (the LHS's last digit must be $4$). Hence $c=7$, from which we get $b=9$ and the unique solution as $\overline{abc}=297$.
> 
> Not a 'real' answer, but it was too big for a comment.
> 
> 
> 

I wrote and ran some [Mathematica](https://en.wikipedia.org/wiki/Wolfram_Mathematica)-code:

```
In[1]:=ParallelTable[
  If[TrueQ[8*(a*100 + b*10 + c*1) == 3*(100*c + 10*b + 1*a)], {a, b, 
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

```

Running the code gives:

```
Out[1]={{{{2, 9, 7}}}}

```

So, we can see that you're right!

---

When we expand this, with different values of the constants before the numbers we get:

```
In[2]:=ParallelTable[
  If[TrueQ[4*(a*100 + b*10 + c*1) == 7*(100*c + 10*b + 1*a)], {a, b, 
    c}, Nothing], {a, 1, 9}, {b, 1, 9}, {c, 1, 9}] //. {} -> Nothing

Out[2]={{{{2, 3, 1}}}, {{{4, 6, 2}}}, {{{6, 9, 3}}}}

```