Baltic Way 2024 · Problem 20

(a) Rearranging terms in the first equation gives

$$a^{2} b^{2}-2 a b=c\left(a^{2}+b^{2}\right)+a b$$

By substituting $a b=a^{2}+b^{2}-c^{2}$ into the right-hand side and rearranging the terms we get

$$a^{2} b^{2}+c^{2}=(c+1)\left(a^{2}+b^{2}\right)+2 a b$$

By adding $2 a b c$ to both sides and factorizing we get

$$(a b+c)^{2}=(c+1)(a+b)^{2}$$

Now it is obvious that $c+1$ has to be a square of an integer. (b) Let us say $c+1=d^{2}$, where $d>1$ and is an integer. Then substituting this into the equation (10) and taking the square root of both sides (we can do that as all the terms are positive) we get

$$a b+d^{2}-1=d(a+b)$$

We can rearrange it to $(a-d)(b-d)=1$, which immediately tells us that either $a=b=d+1$ or $a=b=d-1$. Note that in either case $a=b$. Substituting this into the second equation of the given system we get $a^{2}=c^{2}$, implying $a=c$ (as $\left.a, c>0\right)$.

• If $a=b=d-1$, then $a=c$ gives $d-1=d^{2}-1$, so $d=0$ or $d=1$, neither of which gives a positive $c$, so cannot be a solution.
• If $a=b=d+1$, then $a=c$ gives $d+1=d^{2}-1$, so $d^{2}-d-2=0$. The only positive solution is $d=2$ which gives $a=b=c=3$. Substituting it once again into both equations we indeed get a solution.

(a) Substituting $a^{2}+b^{2}$ from the second equation to the first one gives

$$(a b-1)^{2}=c\left(c^{2}+a b\right)+a b+1$$

Rearranging terms in the obtained equation gives

$$(a b)^{2}-(c+3) a b-c^{3}=0$$

which we can consider as a quadratic equation in $a b$. Its discriminant is

$$D=(c+3)^{2}+4 c^{3}=4 c^{3}+c^{2}+6 c+9=(c+1)\left(4 c^{2}-3 c+9\right)=(c+1)\left(4(c-1)^{2}+5(c+1)\right)$$

To have solutions in integers, $D$ must be a perfect square. Note that $c+1$ and $4(c-1)^{2}+5(c+1)$ can have no common odd prime factors. Hence $\operatorname{gcd}\left(c+1,4(c-1)^{2}+5(c+1)\right)$ is a power of 2 , so $c+1$ and $4(c-1)^{2}+5(c+1)$ are either both perfect squares or both twice of some perfect squares. In the first case, we are done. In the second case, note that

$$4(c-1)^{2}+5(c+1) \equiv 4(c-1)^{2}=(2(c-1))^{2} \quad(\bmod 5)$$

so $4(c-1)^{2}+5(c+1)$ must be $0,1,4$ modulo 5 . On the other hand, twice of a perfect square is $0,2,3$ modulo 5 . Consequently, $c+1 \equiv 0(\bmod 5)$ and $4(c-1)^{2}+5(c+1) \equiv 0(\bmod 5)$, the latter of which implies $c-1 \equiv 0(\bmod 5)$. This leads to contradiction since $c-1$ and $c+1$ cannot be both divisible by 5 . (b) By the solution of part (a), both $c+1$ and $4 c^{2}-3 c+9$ are perfect squares. However, due to $c>0$ we have $(2 c-1)^{2}=4 c^{2}-4 c+1<4 c^{2}-3 c+9$, and for $c>3$ we also have $(2 c)^{2}=4 c^{2}>4 c^{2}-3 c+9$. Thus $4 c^{2}-3 c+9$ is located between two consecutive perfect squares, which gives a contradiction with it being a square itself. Out of $c=1,2,3$, only $c=3$ makes $c+1$ a perfect square. In this case, the quadratic equation $(a b)^{2}-(c+3) a b-c^{3}=0$ yields $a b=9$, so $a=1, b=9$ or $a=3, b=3$ or $a=9, b=1$. Out of those, only $a=3, b=3$ satisfies the equations. Remark: Part (a) of the problem can be solved yet another way. By substituting $a^{2}+b^{2}$ from the second equation to the first one, we obtain

$$(a b-1)^{2}=c\left(c^{2}+a b\right)+a b+1=c^{3}+1+a b c+a b=(c+1)\left(c^{2}-c+1+a b\right)$$

Whenever an integer $n$ divides $c+1$, it also divides $a b-1$. Therefore $c \equiv-1(\bmod n)$ and $a b \equiv 1$ $(\bmod n)$. But then $c^{2}-c+1+a b \equiv 4(\bmod n)$, i.e., $n$ divides $c^{2}-c+1+a b-4$. Thus the greatest common divisor of $c+1$ and $c^{2}-c+1+a b$ divides 4 , i.e., it is either 1,2 or 4 . In the first and third case we are done. If their greatest common divisor is 2 , then clearly all three of $a, b, c$ are odd, so $a^{2} \equiv b^{2} \equiv c^{2} \equiv 1(\bmod 8)$. Thus from the second equation we have $a b=a^{2}+b^{2}-c^{2} \equiv 1$ $(\bmod 8)$, so $(a b-1)^{2}$ is divisible by 64 . Now, if $c \equiv 1(\bmod 4)$, then $c^{2}-c+1+a b \equiv 2(\bmod 4)$, which means that $(c+1)\left(c^{2}-c+1+a b\right) \equiv 4(\bmod 8)$, giving a contradiction with $(a b-1)^{2}$ being divisible by 64 . If instead $c \equiv 3(\bmod 4)$, then $c^{2}-c+1+a b \equiv 0(\bmod 4)$. This means that both $c+1$ and $c^{2}-c+1+a b$ are divisible by 4 , giving a contradiction with the assumption that their greatest common divisor is 2 . Therefore $c+1$ must be a perfect square.