Am I Evil ?

Let $x=dx_{1}$ and $y=dy_{1}$ where $d=gcd(x,y$ and $gcd(x_{1},y_{1})=1$

Hence, $\frac{13}{d^{2}x_{1}^{2}}+$ $\frac{1996}{d^{2}y_{1}^{2}}=\frac{z}{1997}$

or, $1997(13)y_{1}^{2}+1997(1996)x_{1}^{2}=d^{2}zx_{1}^2y_{1}^2$

Hence, $x_{1}^{2}|1997\times13$ and $y_{1}^{2}|1997\times1996$

Now, note that $1997$ is square-free and $1996=2^{2}\times499$

Hence, $(x_{1},y_{1})=(1,1) or (1,2)$

Case 1

$(x_{1},y_{1})=(1,1)$

Therefore, $d^{2}z=1997\times13 + 1997\times1996$

or, $d^{2}z=1997(13+1996)=1997.7^{2}$

Hence, $d=1$ or $7$

$(x,y,z)=(1,1,4011973),(7,7,81877)$

Case 2

$(x_{1},y_{1})=(1,2)$

$d^{2}z=(13+499)1997=1997.2^{9}$

These give $(x,y,z)=(1,2,1022469),(2,4,255616),(4,8,63904),(8,16,15976),(16,32,3994)$

$frac{13}{x^2}$+$frac{1996}{y^2}$=$frac{z}{1997}$ now cross multiplying you get $x^2y^2z$=$1997(13y^2+1996x^2)$ =$1997(13y^2+499(2x)^2$ now 1997,13,499 all are primes , therefore as x,y,z are integers z=1997 k where k is nothing but $(13y^2+499 4 x^2)/(x^2 y^2)$ now both $x^2$ and $y^2$ are factors of the numerator , therefore notice that there is a factor of x^2 in the numerator , therefore y=kx, now as there is a factor of y^2 as well k can only be 1 or 2 as 4 is present in the term having x^2 , now check for both cases , i.e $y=x$ & $y=2x$ in the original equation you get 7