A number theory problem by Aquilino Madeira

$\frac{(n+1)!-3(n-1)!}{(n+1)!+3(n-1)!}=\frac{(n-1)!}{(n-1)!}\times \frac{n(n+1)-3}{n(n+1)+3}=\frac{53}{59}$

$59(n(n+1)-3)=53(n(n+1)+3)$

$6(n(n+1))=3(53+39)$

$n^{2}+n=\frac{53+59}{2}$

$n^{2}+n-56=0$

$n=7,-8$

$n-1\ge0 \rightarrow n=\boxed{7}$

$\boldsymbol {\frac{(n+1)!-3(n-1)!}{(n+1)!+3(n-1)!}=\frac{53}{59}}$ $\boldsymbol {\Rightarrow \frac{(n+1)!-3(n-1)!+(n+1)!+3(n-1)!}{(n+1)!-3(n-1)!-(n+1)!-3(n-1)!}=\frac{53+59}{53-59}}$ $\boldsymbol {\Rightarrow \frac{(n+1)!}{(n-1)!} = \frac {112}{-6}\times(-3)=56}$ $\boldsymbol {\Rightarrow n(n+1)=56}$ $\boldsymbol {\Rightarrow n^2+n-56=0}$ $\boldsymbol {\Rightarrow (n+8)(n-7)=0}$ $\boldsymbol {\Rightarrow n=-8,7}$

Since, factorial of a negative number is undefined the only answer will be $\color{#69047E} {\boxed 7}$ .