WakeUp wrote: Determine all triples $(x,y,z)$ of positive integers such that \[\frac{13}{x^2}+\frac{1996}{y^2}=\frac{z}{1997} \] Wlog, $x>0, y>0$. The given one implies $13\cdot 1997\cdot y^2+1996\cdot 1997\cdot x^2=zx^2y^2.$ And since $zx^2y^2$ is divisible by $x^2$ and $y^2$, $13\cdot 1997\cdot y^2$ is divisible by $x^2$ and $1996\cdot 1997\cdot x^2$ is divisible by $y^2$. These ones imply $x^2 | 13\cdot 1997\cdot y^2$ and $y^2 | 1996\cdot 1997\cdot x^2$, imply $x | 13\cdot 1997\cdot y^2$ and $y | 1996\cdot 1997\cdot x^2$, imply ($x =1$ or $x| y^2$) and ($y=1 $ or $y= 2$ or $y|x^2$), imply $(x,y)=(1,1), (x,y)=(1,2)$ or $x=y$.

Expanding, we have $\frac{13y^{2}+1996x^{2}}{x^{2}y^{2}}=\frac{z}{1997}\Longleftrightarrow 1997\left ( 13y^{2}+1996x^{2} \right )=zx^{2}y^{2}$ Note that $1997\mid zx^{2}y^{2}$. As $1997$ is a prime number, then it must divide $x^2 $, $y^2 $ or $z$. Suppose it does not divide $z$, then ${1997\mid x^2 y^2=\left (xy \right )^2\Rightarrow 1997^2\mid zx^2y^2=1997\left (13y^2+1996x^2 \right )\Rightarrow 1997\mid 13y^2+1996x^2}$ This can be written in terms of congruences as $13y^2+1996x^2\equiv 0\pmod {1997}\Longleftrightarrow 13y^2\equiv x^2\pmod {1997}$ So $13y^2$ is a quadratic residue modulo $1997$. But the Law of Quadratic Reciprocity gives us an easy contradiction: ${\left ( \frac{13y^{2}}{1997} \right )=\left ( \frac{13}{1997} \right )\left ( \frac{y^{2}}{1997} \right )=\left ( \frac{13}{1997} \right )=\left ( \frac{1997}{13} \right )=\left ( \frac{8}{13} \right )=\left ( \frac{2}{13} \right )^{3}=\left ( \frac{2}{13} \right )=-1}$ Therefore $1997\mid {z}$. Let $z=1997w$ for some natural $w$, then $13y^2+1996x^2=wx^2y^2$ Now, it is easily seen that $x^2\mid 13y^2$ and $y^2\mid 1996x^2$. So there exists natural numbers $k$ and $h$ such that $\left\{\begin{matrix} 13y^{2}=kx^{2}\\ 1996x^{2}=hy^{2} \end{matrix}\right.$ These naturals are not equal to $1$, otherwise $13y^2=x^2$ or $\left ( 2x \right )^{2}\cdot 499=y^{2}$. In any case we have an expression of the form $pm^2=n^2$, with $m$ and $n$ integers and $p$ prime. This is impossible because $p\mid n^2\Rightarrow p^2\mid n^2=pm^2\Rightarrow p\mid m^2$ And consequently in the left hand side $p$ has an odd exponent, while in the right hand side it appears with an even exponent, a contradiction. Finding $x^2$ on the first expression and replacing it on the second one, we have $1996\frac{13y^{2}}{k}=hy^{2}\Longleftrightarrow 2^{2}\cdot 499\cdot 13=kh$ Therefore $k\in \left \{ 2,4,13,26,52,499,998,1996,6487,12974 \right \}$. It can not equal $2^2\cdot 499\cdot 13$ because it would imply $h=1$. By a similar argument exposed on the case $pm^2=n^2$ it can be proven that, given two prime numbers $p$ and $q$, there are no two integers $m$ and $n$ such that $pm^2=qn^2$. As we have $13y^2=kx^2$, we can put away $k=2$ and $k=499$. We have eight cases to analyze: Case 1: $k=4$ Then $13y^2=4x^2$ whence $13\mid x$ and $y^{2}=\frac{4}{13}x^{2}$, so if we replace that on $13y^2+1996x^2=wx^2y^2$ we have $13\frac{4}{13}x^2+1996x^2=wx^2\frac{4}{13}x^2\Longleftrightarrow 13\cdot 2^{2}\cdot 5^{3}=wx^{2}$ As $13$ does not divide any perfect square that divides $13\cdot 2^{2}\cdot 5^{3}$, there is no solution in this case. Case 2: $k=13$ Then $y^2=x^2$, whence $x=y$, so $13x^2+1996x^2=wx^2x^2\Longleftrightarrow 2009=wx^2$ The only perfect squares that divide $2009$ are $49$ and $1$. Let us take $49$. Then $x=7$ and $w=41$, consequently $z=1997\cdot 41=81877$ and we found the solution $\left (7,7,81877 \right )$. Let us take now $x=1$, then $w=2009$ and we found the solution $\left (1,1,4011973 \right )$. Case 3: $k=26$ Then $13y^2=26x^2\Longleftrightarrow y^2=2x^2$, a contradiction since it is $pm^2=n^2$. Case 4: $k=52$ Then $13y^2=52x^2\Longleftrightarrow y^2=4x^2$, whence $y=2x$. Replacing it on $13y^2+1996x^2=wx^2y^2$ we have $13\cdot 4x^2+1996x^2=wx^2\cdot 4x^2\Longleftrightarrow 512=wx^2$ So $x^2\in \left \{ 1,4,16,64,256 \right \}$, and we found the solutions $\left (1,2,1022464 \right ),\left (2,4,255616 \right ),\left (4,8,63904 \right ),\left (8,16,15976 \right ),\left (16,32,3994 \right )$ Case 5: $k=998$ Then $13y^2=998x^2\Longleftrightarrow y^2=\frac{998}{13}x^2$, whence $13\mid x$ and replacing it we have $998x^2+1996x^2=wx^2\frac{998}{13}x^2\Longleftrightarrow 39=wx^{2}$ A contradiction since none perfect square divisible by $13$ divides $39$. Case 6: $k=1996$ Then $13y^2=1996x^2\Longleftrightarrow y^2=\frac{1996}{13}x^2$, whence $2$, $13$ and $499$ divide $x$. Replacing it we have $1996x^2+1996x^2=wx^2\frac{1996}{13}x^2\Longleftrightarrow 26=wx^2$ This is an evident contradiction. Case 7: $k=6487$ Then $13y^2=6487x^2\Longleftrightarrow y^2=499x^2$, a contradiction since it is $pm^2=n^2$. Case 8: $k=12974$ Then $13y^2=12974x^2\Longleftrightarrow y^2=998x^2$. Therefore $499\mid x$ and replacing it we have $12974x^2+1996x^2=wx^2\cdot 998x^2\Longleftrightarrow 15=wx^2$ And we have an evident contradiction again. Conclusion: The solutions $\left ( x,y,z \right )$ are $\left ( 7,7,81877 \right )$ $\left ( 1,1,4011973 \right )$ $\left ( 1,2,1022464 \right )$ $\left ( 2,4,255616 \right )$ $\left ( 4,8,63904 \right )$ $\left ( 8,16,15976 \right )$ $\left ( 16,32,3994 \right )$