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 If $$x,y,z \in \mathbf{N}^+$$, when is $$(xy+1)(yz+1)(zx+1)$$ a perfect square?
04-19-2010, 09:54 AM
Post: #1
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
If $$x,y,z \in \mathbf{N}^+$$, when is $$(xy+1)(yz+1)(zx+1)$$ a perfect square?

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05-08-2010, 11:29 PM
Post: #2
 elim Moderator Posts: 577 Joined: Feb 2010 Reputation: 0
Theorem For $$x,y,z \in \mathbf{N}^+$$,$$(xy+1)(yz+1)(zx+1)$$ is a perfect square iff $$\{p,q,r\}$$ is $$P_1$$-set
Let's see the complete proof.
If the theorem is not true, there are $$p,q,r \in \mathbb{N}^+$$ such that
$$p \le q \le r$$ and $$(pq+1)(qr+1)(rp+1)$$ is a perfect square but not all of
$$pq+1, qr+1, rp+1$$ are perfect squares. We assume $$p,q,r$$ also make
$$p+q+r$$ the smallest among such choices as counterexamples of the theorem.
Consider the equation
$$s^2-2(p+q+r+2pqr)s + (p^2+q^2+r^2-2(pq+qr+rp)-4)=$$
$$p^2+q^2+r^2+s^2-2(pq+qr+rp+ps+qs+rs)-4pqrs-4=0$$
Let $$s=p+q+r+2pqr-\sqrt{(p+q+r+2pqr)^2-(p^2+q^2+r^2-2pq-2qr-2rp-4)}$$
$$\quad \quad\quad =p+q+r+2pqr-2\sqrt{(pq+1)(qr+1)(rp+1)}$$
Clearly $$s$$ is a integer and
$$(p+q-r-s)^2=p^2+q^2+r^2+s^2+2(pq-pr-ps-qr-qs+rs)=4pq+4rs+4pqrs+4$$
$$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad=4(pq+1)(rs+1)$$ and symmetrically
$$(p+r-q-s)^2=4(rp+1)(qs+1)$$, $$(p+s-q-r)^2=4(ps+1)(qr+1)$$ and so
$$(pq+1)^2(p+r-q-s)^2(p+s-q-r)^2=16(pq+1)^2(pr+1)(qs+1)(qr+1)(ps+1)$$
therefore $$(pq+1)(qs+1)(ps+1)$$ is a perfect square. And since $$qs+1$$ and $$rp+1$$
must be or not be perfect squares the same time, and the same thing is true for $$ps+1$$ and $$qr+1$$, we see that $$pq+1, qs+1, ps+1$$ are not all perfect squares either.
Moreover, $$rs+1=(p+q-r-s)^2/(4(rp+1)) \ge 0 \Rightarrow s \le -1/r$$
If $$r=1$$ then $$p=q=r=1 \Rightarrow (pq+1)(qr+1)(rp+1)=8$$ not a perfect square. So $$r>0, \quad s \ge 0$$
If $$s=0$$, then $$(p+q-r)^2=4(pq+1),(p+r-q)^2=4(rp+1),(p-q-r)^2=4(qr+1)$$ thus all of $$pq+1, qr+1, rp+1$$ are perfect squares, contradicts to the assumption and we see that $$s > 0$$
Let $$s' = p+q+r+2pqr+2\sqrt{(pq+1)(qr+1)(rp+1)}$$, then
$$s^2<ss'=p^2+q^2+r^2-2(pq+qr+rp)-4<r^2-p(2r-p)-q(2r-q)<r^2$$ thus
$$0<s<p+q+s<p+q+r$$ which contradicts the minimum property of $$p,q$$ and $$r$$. $$\square$$
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