Team Selection Test for JBMO

a. The answer: permutations of $(2,3,11)$. Let $p+q+r=x^2$ and $pq+qr+rp+3=y^2$ where $x$, $y$ are integers. Let us show that one of the primes $p$, $q$, $r$ is $2$. If all primes $p$, $q$, $r$ are odd all possibilities up to permutations are: $(p,q,r)=(1,1,1)$, $(1,1,3)$, $(1,3,3)$, $(3,3,3)$ (mod $4$). We get a contradiction in the cases $(1,1,1)$, $(1,3,3)$ since $x^2=p+q+r\equiv 3(\mathrm{mod}4)$ and in the cases $(1,1,3)$, $(3,3,3)$ since $y^2-3=pq+qr+rp\equiv 3(\mathrm{mod}4)$. Therefore, at least one of $p$, $q$, $r$ is equal to $2$. W.l.o.g. $p=2$ and $q\le r$. Then $q+r=x^2-2$, $qr=y^2-2x^2+1$. Now if $3\mid y$, then $(q+2)(r+2)=y^2+1\equiv 1(\mathrm{mod}3)$. Thus, either $q\equiv r\equiv 2(\mathrm{mod}3)$ or $q\equiv r\equiv 0(\mathrm{mod}3)$. But for $q\equiv r\equiv 0(\mathrm{mod}3)$ we get a contradiction: $x^2-2\equiv 0(\mathrm{mod}3)$. For $q\equiv r\equiv 2(\mathrm{mod}3)$ we get $x^2-2\equiv 1(\mathrm{mod}3)$ and $3\mid x$, but by assumption $3\nmid x$. Thus, $3\mid y$ is not possible. Now since $3\nmid x$ we get $x^2\equiv y^2\equiv 1(\mathrm{mod}3)$ and consequently $qr=y^2-2x^2+1\equiv 0(\mathrm{mod}3)$. Thus, $q=3$. Now $r=x^2-5$ and $3r=y^2-2x^2+1$. Therefore $5r=y^2-9=(y-3)(y+3)$. For $r=2,3,5$ $x$ is not an integer number. Therefore, $r>5$. Since $y-3=1$ yields no solution $y-3=5$, $r=y+3$

and $r=11$. For $x=4$, $y=8$ we get $(p,q,r)=(2,3,11)$.

b. $(p,q,r)=(2,11,23)$ satisfies the conditions for $(x,y)=(6,18)$.