Let $x, y, z$ be positive integers such that $$\frac{13}{x^2}+\frac{28}{y^2}=\frac{z}{85}$$ Find all $x+y+z$.
I observe that $(x, y)=(1, 1)$ or $(1, 2)$ will give some solutions. However I can neither find any more solutions nor prove that there are only these solutions. Can anyone help me out.
Here's a solution that doesn't need computer assistance:
Multiplying with $85x^2$, the equation is equivalent to $$\frac{28\cdot85x^2}{y^2}=zx^2-13\cdot85.$$ Since the right-hand side is an integer, so is the left-hand side and $y^2\mid 28\cdot85x^2=2^2\cdot5\cdot7\cdot17\cdot x^2,$ hence $y\mid 2x$. Similarly, $x^2\mid 13\cdot85y^2$, hence $x\mid y$. This only leaves two options: $y=x$ or $y=2x$.
If $y=x$, the equation becomes $$\frac{41}{x^2}=\frac z{85}\quad\Longleftrightarrow\quad x^2z=41\cdot85.$$ But $41\cdot 85$ is square-free, so the only solution is $x=1$ and $z=41\cdot85$.
If $y=2x$, the equation simplifies to $$\frac{20}{x^2}=\frac z{85}\quad\Longleftrightarrow\quad x^2z=20\cdot 85=2^2\cdot5^2\cdot17.$$ For $x^2$ to divide this product, we must have $x=1,2,5$ or $10$. Therefore, the complete list of solutions is $$(1,1,41\cdot85),\ (1,2,2^2\cdot5^2\cdot17),\ (2,4,5^2\cdot17),\ (5,10,2^2\cdot17),\ (10,20,17).$$
This doesn't looks short, but here's a hint that reduces the problem to a finite search:
Write the equation as $(x^2z-1105)(y^2z-2380)=2629900$. This means that both $x^2z-1105$ and $y^2z-2380$ are among the divisors of $2629900$, which gives only a finite number of possibilities.
We giver further details of the method of completing a product / rectangle, suggested in jjagmath's prior answer.
The equation can be written as: $$\frac{13y^2+28x^2}{x^2y^2}=\frac{z}{85}$$ $$x^2y^2z=85(13y^2+28x^2)$$ $$x^2y^2z^2-1105y^2z-2380x^2z=0$$ $$x^2z(y^2z-2380)-1105(y^2z-2380)-2629900$$ $$(y^2z-2380)(x^2z-1105)=2629900$$ Let say, we have found two factors to be $a$ and $b$, then we can say: $$x^2z-1105=a$$ from which we can say:$$z=\frac{a+1105}{x^2}$$ and $$y^2z-2380=b$$ and then: $$z=\frac{b+2380}{y^2}$$ Combining these, we get: $$\frac{a+1105}{x^2}=\frac{b+2380}{y^2}$$ There seems to be a number of constraints here, first of all, both the sides need to be an integer, second, they need to be equal to each other. Third, $a$ and $b$ can only be factors of $2629900$, for which we can set $b=\frac{2629900}{a}$ and write: $$\frac{a+1105}{x^2}=\frac{\frac{2629900}{a}+2380}{y^2}$$ $$ x=y\sqrt{\frac{a+1105}{\frac{2629900}{a}+2380}} $$ This produces some of the values, through algebraic brute force, setting value of $a$ starting from $1$ to $2629900$, a total of $108$ factors, for which the term under the radical is a perfect square fraction [need not to be an integer], which is: $$x=\frac{y}{2}$$ for values of $a=595$ and again, $$x=y$$ for $a=2380$. For $x=y$, we can write our first equation as: $$ \frac{41}{x^2}=\frac{z}{85}\Rightarrow z=\frac{3485}{x^2}$$ Now $3485=1\times5\times17\times41$, since $1$ is the only perfect square here, only $x=1$ can produce an integer value of $z$. Therefore, one of our solution is $(x,y,z)=(1,1,3485)$
Considering, $x=\frac{y}{2}$, we can write our first equation as: $$\frac{13}{x^2}+\frac{7}{x^2}=\frac{z}{85}$$ $$z=\frac{1700}{x^2}$$ As before $1700=1\times2^2\times5^2\times17$, therefore, $x$ can take the value of $x=1,2, 5, 2\times 5$ Therefore, the only values we can find here is: $$ (x,y,z)=(1,1,3485),(1,2,1700), (2,4,425), (5,10,68), (10,20, 17) $$ And that's it. There are no other values.
