# Find $x,y,z \in Z$ , $1<x<y<z$ and $xyz - 1 = t(x-1)(y-1)(z-1)$

Find $x,y,z \in \Bbb Z, 1<x<y<z$ and $xyz - 1 = t(x-1)(y-1)(z-1)$.

Help me!

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What values does $t$ take? – T. Eskin Feb 25 '13 at 15:07
If OP's $t$ is an integer it has to range between 1 to 3 given the conditions. – Vincent Tjeng Feb 25 '13 at 15:50
@VincentTjeng please show us how you deduced that. – franklin Mar 7 '13 at 2:27
Hi @franklin, my bound can be found below as an answer. It was a bit too long to write as a comment. – Vincent Tjeng Mar 8 '13 at 3:34
I've solved the problem over integer $t$ after some work. – Vincent Tjeng Mar 8 '13 at 5:59

Rearranging, we have

$$xyz-1=zt(x-1)(y-1)-t(x-1)(y-1)$$ $$z\left(xy-t(x-1)(y-1)\right)=1-t(x-1)(y-1)$$ $$z=\frac{1-t(x-1)(y-1)}{xy-t(x-1)(y-1)}=\frac{t(x-1)(y-1)-1}{t(x-1)(y-1)-xy}=\frac{t-\frac{1}{(x-1)(y-1)}}{t-\frac{x}{x-1}\frac{y}{y-1}}$$

Now, since $z>y>x>1$, we have $z \geq 4$, and we must have

$$\frac{t-\frac{1}{(x-1)(y-1)}}{t-\frac{x}{x-1}\frac{y}{y-1}}\geq 4$$

Looking at the terms in the denominator, we have $\frac{x}{x-1} \leq \frac{2}{1}, \frac{y}{y-1} \leq \frac{3}{2}$, since $x\geq 2, y\geq 3$

Thus $\frac{x}{x-1}\frac{y}{y-1} \leq 3$. If $t>3$, the denominator is positive, and we have

$${t-\frac{1}{(x-1)(y-1)}}\geq4\left({t-\frac{x}{x-1}\frac{y}{y-1}}\right)$$ $${-\frac{1}{(x-1)(y-1)}}\geq\left({3t-4\frac{x}{x-1}\frac{y}{y-1}}\right)$$

However, if $t>3$, the RHS is strictly nonnegative, while the LHS is strictly negative. Thus, we reach a contradiction.

Edit: In fact, you can improve the bound slightly and go some way to solving the problem for integer $t$ by considering the following cases.

Firstly, for $t=3$, you can show that

$$3t-4\frac{x}{x-1}\frac{y}{y-1}$$

for $x\geq3$. Therefore, for $t=3$, it suffices to check the cases where $x=2$.

Now when we have $t=3, x=2$, the equation simplifies to $(y-3)(z-3)=5$, corresponding to the solution $(x,y,z,t)=(2,3,8,3)$.

Similarly, for $t=2$, it suffices to check the cases $x=2, 3, 4$ or $(x,y)=(5,6)$.

Considering the equation modulo $2$, we can eliminate the cases $x=2,4$. For $x=3$, the equation simplifies to give $(y-4)(z-4)=11$, corresponding to the solution $(x,y,z,t)=(3,5,15,2)$. Finally, we substitute $(x,y)=(5,6)$ into the original equation and obtain $30z-1=40(z-1)$, which has no integer solutions.

Finally, for $t=1$, we can simplify the original equation to the following

$$xy+yz+zx=x+y+z$$.

But we have $$xy+yz+zx\geq3x+4y+2z>x+y+z$$ Thus there are no solutions to the equation when $t=1$.

To summarize, the two integer solutions for integer t are $(x,y,z,t)=(2,4,8,3) or (3,5,15,2)$.

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