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Let $x,y,z \in \mathbb{Z^+}$ and $x \neq y \neq z$.

Given the conditions above, find when $x$, $y$, $z$ satisfy below:

$$ (x^2-1)(y+1)=\frac{z^2+1}{y-1}\,.$$

What I did was I factored the numerator to

$$(x+1)(x-1)(y+1)=\frac{z^2+1}{y-1}\,,$$

but I am having trouble figuring out how to isolate the variables. I tried some values with trial and error and wasn't able to get any.

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3 Answers 3

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I solve this Diophantine equation with the assumption that $x$, $y$, and $z$ can be any integers (not necessarily positive and not necessarily unequal). Rewrite the equation as $$\left(x^2-1\right)\,\left(y^2-1\right)=z^2+1\,.$$ Since the largest power of $2$ that divides $z^2+1$ is $2$ (i.e., $4\nmid z^2+1$ for all $z\in\mathbb{Z}$) and $8\mid t^2-1$ for any odd integer $t$, we conclude that both $x$ and $y$ must be even.

If either $x$ or $y$ is nonzero, say, $x\neq 0$, then $x^2-1$ is a positive integer congruent to $3$ modulo $4$. Therefore, a prime natural number $p\equiv 3\pmod{4}$ divides $x^2-1$. This means $p\mid z^2+1$ as well, but we know this is impossible as $-1$ is not a quadratic residue modulo $p$. Hence, the only possible solution in $\mathbb{Z}$ must come from $(x,y)=(0,0)$, which then produces $(x,y,z)=(0,0,0)$.

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Multiply both sides by $y-1$, expand, and simplify. You then have several relatively easy ways to attack the question. Is that enough to go on?

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  • $\begingroup$ After doing what you said, i get $z=\sqrt{x^2y^2-y^2-x^2}$ but im not sure how will that help. $\endgroup$
    – kero
    Nov 30, 2015 at 3:41
  • $\begingroup$ 1. Instead of taking the square root, factor. For example, the equation implies both $$ y^2+x^2 = (xy)^2 - z^2 = (xy-z)(xy+z) $$ and $$ y^2+z^2 = x^2(y^2-1). $$ By using well-known theorems regarding numbers which are the sum of two squares, you determine things about each of the resulting factors. 2. Instead of taking the square root, use a parameterization (esp. for the resulting sums-of-squares [2.1.3] equation $$(xy)^2 = x^2+y^2+z^2$$ to figure out the form of $x$, $y$, and $z$. $\endgroup$ Nov 30, 2015 at 22:19
  • $\begingroup$ A quick computer check shows there are no positive integer solutions to $$\sqrt{(xy)^2-x^2-y^2} = \mathbb{Z}$$ with $x,y<1000$ and $x\neq y$. A more rigorous approach will probably show there are no solutions at all. $\endgroup$ Dec 1, 2015 at 15:45
  • $\begingroup$ @TitoPiezasIII: My intuition told me there are none. But I haven't actually gone further than my hints. Maybe I will… $\endgroup$ Dec 1, 2015 at 17:40
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The only solution is $x = y = z = 0$.

Suppose there were a solution with x, y, and z not all zero. Choose the maximum $k$ with $2^k | gcf(x, y, z)$. So $x = 2^k m$, $y = 2^k n$, and $z = 2^k p$, and $m$, $n$, and $p$ are not all even.

$(xy)^2 = x^2 + y^2 + z^2$

$(2^k m 2^k n)^2 = (2^k m)^2 + (2^k n)^2 + (2^k p)^2$

$(4^k mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$

$16^k (mn)^2 = 4^k m^2 + 4^k n^2 + 4^k p^2$

$4^k (mn)^2 = m^2 + n^2 + p^2$

But the last equation only has solutions (mod 4) when m, n, and p are all even, contradicting the choice of k.

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