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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
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
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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