Solve $\frac {(x-a)(x-b)} {(x-c)}+1 = \frac {(x-a+1)(x-b+1)} {(x-c+1)}$

Question

I don't know how to solve $\frac {(x-a)(x-b)} {(x-c)}+1 = \frac {(x-a+1)(x-b+1)} {(x-c+1)} $ with $x,a,b,c \in \mathbb N$

I suspected that it's right for $a=b \lor a=c$ what actually is true. But I can't tell if there are more solutions.

Answer 1

There are no more solutions, after doing some algebra you will get $(c-b)\cdot(c-a) = 0$.

$\frac {(x-a)(x-b)} {(x-c)} + 1 = \frac {(x-a)(x-b)} {(x-c+1)} + \frac {(x-b)} {(x-c+1)} + \frac {(x-a)} {(x-c+1)} + \frac 1 {(x-c+1)}$ (first and last term to the left)

$\frac {(x-a)(x-b)}{(x-c+1)(x-c)} + \frac {(x-c)}{(x-c+1)} = \frac {(x-a)}{(x-c+1)} + \frac {(x-b)} {(x-c+1)}$ Multiply both sides by (x-c)(x-c+1)

$(x-a)(x-b) + (x-c)(x-c) = (x-c)(x-a) + (x-c)(x-b)$ (group them)

$(x-a)(c-b) - (x-c)(c-b) = 0$ (group them)

$(c-b)(c-a) = 0$

Answer 2

Let $$X=x-a \qquad Y=x-b \qquad Z=x-c $$ your equation becomes: $$ \frac{XY}{Z}=\frac{(X+1)(Y+1)}{Z+1} $$ that, for $Z \ne0$ and $Z+1 \ne 0$, is: $$ XY(Z+1)+Z(Z+1)=(X+1)(Y+1)Z \quad \iff \quad (Y-Z)(X-Z)=0 $$ that has solutions: $$ Y=Z \qquad X=Z $$ that is: $$ x-b=x-c \qquad x-a=x-c $$

so the starting equation is an identity if $a=c$ or $ b=c$ and has no solutions otherwise.