62nd Czech and Slovak Mathematical Olympiad

We show that the only solution is $a=1$, $b=4$ and $c=3$. Let $s=b+c\ge 7$. Substituting $a=5-b$ and $c=s-b$ the first condition gives $$(5-b{)}^2+b^2+(s-b{)}^2=26,$$ thus $$3b^2-2(s+5)b+s^2-1=0.$$ The equation has a real solution iff the discriminant $4(s+5{)}^2-12(s^2-1)\ge 0$. This yields $s^2-5s-14\le 0$, or $(s+2)(s-7)\le 0$. Since $s\ge 7$, there must be $s=7$. If we substitute to the previous equation we get $$3b^2-24b+48=0;$$ with the only solution $b=4$. Then $a=1$ and $c=3$.