SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Suppose that $abcd=n^2$ for some $n\in {\mathbb{Z}}^{+}$. We can suppose that $a<c<d<b$. From this, we have $bd+ac-ad-bc=(b-a)(d-c)>0$, thus $$ad+bc<\frac{1}{2}(ad+bc+bd+ac)=\frac{(a+b)(c+d)}{2}=\frac{p^2}{2}.$$ Denote $\gcd (ad,bc)=k\in {\mathbb{Z}}^{+}$ then $ad=k{u}^2$, $bc=k{v}^2$ with $u,v\in {\mathbb{Z}}^{+}$, and $u<v$, $\gcd (u,v)=1$. Since $a,b,c,d$ are coprime to $p$ then $\gcd (k,p)=1$. We have $$k\left(v^2-u^2\right)=bc-ad=(p-a)c-a(p-c)=p(c-a).$$ So $p\mid v^2-u^2$ since $\gcd (k,p)=1$. Hence, $p\mid v-u$ or $p\mid v+u$. In both cases, we always have $u+v\ge p$. From this, we can conclude that $$ad+bc=k\left(u^2+v^2\right)>k\frac{(u+v{)}^2}{2}\ge \frac{p^2}{2}$$ by AM-GM inequality. But this contradicts the inequality we stated above, then $abcd$ cannot be a perfect square.