Unwind the equation

Question

Let $x, y, z, t$ be positive integers. Given that $$68(xyzt+xy+zt+xt+1)=157(yzt+y+t)$$ Find the value of the product $xyzt$. I couldn't even start with the problem. I just know that the expression n the left bracket is a multiple of $157$ and that in the right bracket is a multiple of $68$. Then what do I do?

Answer 1

Thanks to individ. His approach made me thinking and here is the solution.From the given equation, $$\frac{157}{68}=\frac{xyzt+xy+zt+xt+1}{yzt+y+z}$$$$=\frac{x(yzt+y+t)+zt+1}{yzt+y+z}$$$$=x+\frac{zt+1}{yzt+y+z}$$$$=x+\frac{1}{\frac{yzt+y+z}{zt+1}}$$$$=x+\frac{1}{\frac{y(zt+1)+z}{zt+1}}$$$$=x+\frac{1}{y+\frac{z}{zt+1}}$$$$=x+\frac{1}{y+\frac{1}{\frac{zt+1}{z}}}$$$$=x+\frac{1}{y+\frac{1}{t+1/z}}$$ Thus, we want to write the fraction $\frac{157}{68}$ as a continued fraction: $\frac{157}{68}=2+\frac{1}{3+\frac{1}{4+1/5}}$ Therefore, $x=2, y=3, z=4, t=5$. Hence $xyzt=120$.

Answer 2

This equation is written differently.

$$x=\frac{157(yzt+t+y)-68(zt+1)}{68(yzt+y+t)}$$

$$x=\frac{157}{68}-\frac{zt+1}{yzt+y+t}$$

Means $x=1$ and then. $$89(yzt+y+t)=68(zt+1)$$ $$y=\frac{68(zt+1)-89t}{89(zt+1)}=\frac{68}{89}-\frac{t}{zt+1}$$ So no positive solutions.

For the case when $x=2$ I made a mistake.

The procedure is the same. $$21(yzt+y+t)=68(zt+1)$$

$$y=\frac{68}{21}-\frac{t}{zt+1}$$ $$\frac{a}{21}=\frac{t}{zt+1}$$

$a=5 ; 26 ; 47 $ - And as a consequence it is necessary to iterate.

$$t=\frac{a}{21-az}$$

Never saw that $21-5z=1$ That means $z=4$