Japan Mathematical Olympiad

$8,9,10,13,17,19,27$

Let $a$ be the sum of two numbers the player $A$ wrote, and $b$ be the sum of the two numbers $B$ wrote. If we let $t,s$ be the two numbers $A$ wrote, then $t+s=a$ and $ts=2b$, so $t,s$ must be the solutions of the quadratic equation $x^2-ax+2b=0$. Similarly, the two numbers $B$ wrote down are the solutions of the equation $x^2-bx+2a=0$. From this we obtain the fact that the discriminants $a^2-8b$ and $b^2-8a$ of these equations must be perfect squares. Conversely, suppose that $a^2-8b$ is a perfect square, equal to $k^2$ for positive integer $k$, say. Then, we see that from $(a+k)(a-k)=8b$ that $a>k$ and that $a$ and $k$ have the same even-odd parity. Therefore, the two roots $\frac{a+k}{2}$, $\frac{a-k}{2}$ of the equation $x^2-ax+2b=0$ are positive integers. Similarly, if $b^2-8a$ is a perfect square, we see that the two roots of the equation $x^2-bx+2a=0$ are positive integers. Thus, we see that the conditions for the problem are equivalent to the following: $a$ and $b$ are positive integers and both $a^2-8b$ and $b^2-8a$ are perfect squares, and $a\ge b$. In the sequel, we determine the pairs $(a,b)$ satisfying these conditions. Since $a\ge b$, we see that if $a>9$, then $$a^2>a^2-8b\ge a^2-8a>(a-6{)}^2$$ must hold. As $a^2-8b$ and $a$ have the same even-odd parity, we must have either $a^2-8b=(a-2{)}^2$ or $a^2-8b=(a-4{)}^2$. From the former we get $b=\frac{a-1}{2}$ and from the latter $b=a-2$. Now, suppose $b=\frac{a-1}{2}$ is satisfied. Then, we have $b^2-8a=b^2-8(2b+1)=b^2-16b-8<(b-8{)}^2$. Since $b^2-8a$ and $b$ have the same even-odd parity, we see that we can write $$b^2-16b-8=(b-2\ell {)}^2,$$ where $\ell$ is an integer greater than 4. Then, we get $$b=\frac{{\ell }^2+2}{2}=(\ell +4)+\frac{18}{\ell -4}$$ and we conclude that $\ell =5,6,7,10,13,22$ must hold and these values of $\ell$ yield $b=17,19,27$ with corresponding values for $a=35,39,55$ respectively. Next suppose $b=a-2$ is satisfied. Then, we have $b^2-8a-16<(b-4{)}^2$. Since $b^2-8a$ and $b$ have the same even-odd parity, we can write $$b^2-8b-16=(b-2j{)}^2,$$ where $j$ is an integer greater than 2. Then, we get $$b=\frac{j^2+4}{j-2}=(j+2)+\frac{8}{j-2},$$ from which we get $j=3,4,6,10$, which yield $b=10,13$ and $a=12,15$ respectively. Finally, let us consider the case where $a\le 9$. If $b<8$, then we get $b^2-8a\le b^2-8b<0$, which does not satisfy our requirement. So, we must have $b=8$ or $9$. It is easy to check that $(a,b)=(9,8)$ does not satisfy the requirement, while $(a,b)=(8,8),(9,9)$ do. So, we conclude that the possible values that $b$ can take to satisfy the requirement of the problem are $8,9,10,13,17,19,27$.

Alternate Solution:

Let us denote by $x,y$ and $z,w$ the two numbers that the players $A$ and $B$ wrote down, respectively. We may assume that $x\ge y$ and $z\ge w$ are satisfied. The conditions of the problems can be stated as $$xy=2(z+w),zw=2(x+y),x+y\ge z+w.$$

Adding the respective sides of the first two equations, we obtain $xy+zw=2(x+y+z+w)$, which can be transformed into the form $$(x-2)(y-2)+(z-2)(w-2)=8.$$ First, let us consider the case where $y=1$. Since $x=2z+2w$, we get $zw=2(x+y)=4z+4w+2$, which is transformed to $(z-4)(w-4)=18$. From this we obtain $(x,w)=(22,5)$, $(13,6)$, $(10,7)$. We find the corresponding values of $x$ are $54,38,34$, respectively, and the requirement $x+y\ge z+w$ is also satisfied in all of the three cases. We also find that $z+w=27,19,17$, respectively. When $w=1$, we can argue in the same way as above, and find the other two are no solutions satisfying the conditions of the problem. Next, we assume that $x\ge y\ge 2$ and $z\ge w\ge 2$ are satisfied. Then, we have $(x-2)(y-2)\ge 0$ and $(z-2)(w-2)\ge 0$, and since $(x-2)(y-2)+(z-2)(w-2)=8$, we must have $(x-2)(y-2)\le 8$ and $(z-2)(w-2)\le 8$. If $y\ge 5$, then we get $(x-2)(y-2)\ge (y-2{)}^2\ge 9$. So, we must have $y\le 4$. If $y=4$, then from $(x-2)(y-2)\le 8$, we conclude that $x=4$ or $5$ or $6$. When $x=4$: From $zw=2(x+y)=16$, $z+w=\frac{xy}{2}=8$, we get $(x,w)=(4,4)$. When $x=5$: We see there are no pairs $(x,w)$ satisfying $zw=18$ and $z+w=10$. When $x=6$: We get $zw=20$ and $z+w=12$, which yield $(x,w)=(10,2)$. We see that among the possibilities found for $(x,w)$ above only $(x,w)=(4,4)$ satisfies the condition $x+y\ge z+w$. So, the only solution we get for the case $y=4$ is $(x,y,z,w)=(4,4,4,4)$ for which $z+w=8$. If $y=3$, then from $xy=2(z+w)$ we see that $x$ must be even, and from $(x-2)(y-2)\le 8$, $x$ must be one of $4,6,8,10$. When $x=4$: There are no $(x,w)$ satisfying $zw=14$, $z+w=6$. When $x=6$: Only $(x,w)=(6,3)$ satisfies $zw=18$, $z+w=9$. When $x=8$: There are no $(x,w)$ satisfying $zw=22$, $z+w=12$. * When $x=10$: Only $(x,w)=(13,2)$ satisfies $zw=26$, $z+w=15$. Among the possibilities found above, we see that only $(x,y,z,w)=(6,3,6,3)$ satisfies the additional condition $x+y\ge z+w$, and we have $z+w=9$ in this case. Finally, when $y=2$, we see that from $(z-2)(w-2)=8$, we must have $w=3$ or $w=4$. When $w=3$, then $z=10$ and $x=\frac{2(z+w)}{y}=13$, and $(x,y,z,w)=(13,2,10,3)$ satisfies the requirements, and we get $z+w=13$ in this case. If $w=4$, then we have $z=6$, $x=\frac{2(z+w)}{y}=10$, and $(x,y,z,w)=(10,2,6,4)$ satisfies the requirements, and we get $z+w=10$. Summarizing what we considered above, we get the possible values for the sum of the two numbers $B$ wrote to be $8,9,10,13,17,19,27$.