# Finding all $x+y+z$ for positive integers satisfying $\frac{13}{x^2}+\frac{28}{y^2}=\frac{z}{85}$ [closed]

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.

• There is also $(x,y)=(2,4)$, $(x,y)=(5,10)$, $(x,y)=(10,20)$. Commented Aug 4 at 9:44
• @lulu This is NOT a finite problem. If you set for example, $y=1$, there are infinitely many values of $x$ that satisfy $\frac{13}{x^2}+\frac{28}{y^2} \ge \frac{1}{85}$. Commented Aug 4 at 12:40
• @jjagmath Absolutely right, my error. Will delete prior comment.
– lulu
Commented Aug 4 at 12:41

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 $$(1,1,41\cdot85),\ (1,2,2^2\cdot5^2\cdot17),\ (2,4,5^2\cdot17),\ (5,10,2^2\cdot17),\ (10,20,17).$$

• Uses a few times that if $\,n\in \Bbb Z\,$ is squarefree then $\, y^2\mid nz^2\,\Rightarrow\, y\mid z,\,$ which is proved here. $\ \$ Commented Aug 4 at 19:16

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.

• @readers This is the standard method of completing a product (or rectangle, a generalization of completing a square. Commented Aug 4 at 17:25

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.

• What about $(1,1,3485)$? Commented Aug 4 at 15:17
• And credit to jjagmath? (for the main idea $(y^2z-2380)(x^2z-1105)=2629900$) Commented Aug 4 at 15:29
• Done both the edits. Thanks @jjagmath Commented Aug 4 at 15:58