A triple of integers $(a,b,c)$ satisfies $a+bc=2017$ and $ b+ca=8$. Find all possible values of $c$
I have applied subtraction and addition to these 2 equations which resulted $2c= \frac {2025}{a+b} - \frac{2009}{a-b} $. Since $a+b$ has 15 distinct values and $a-b$ has 6 distinct values , if I do the combination, the value of $2c$ will be 80 values which is so much to write. So, is there any method?
Add the 2 equations to get $(a+b)(c+1)=2025$, from which you can then get the factors and hence the possible values for $c+1$.
Similarly, subtract the 2 equation and you'll get $(c-1)(b-a)=2009$, and do the same here to get another set of possible values of $c-1$. Here, the values of $c$ we get are $-2008,-286,-48,-40,-6,0,2,8,42,50,288,2010$.
Now any value of $c$ that is admissible, must satisfy both the above equations.Thus,of these values of $c$ we just found, only $c=-6,0,2,8$ satisfy the first equation. Hence these are the only values of $c$ satisfying both the equation.