23rd Junior Balkan Mathematical Olympiad

Let $a^4-2019a=b^4-2019b=c$.

Since $a\ne b$, subtract the two equations: $$a^4-b^4-2019(a-b)=0$$ $$(a-b)(a^3+a^{2}b+a{b}^2+b^3)-2019(a-b)=0$$ Since $a\ne b$, divide both sides by $(a-b)$: $$a^3+a^{2}b+a{b}^2+b^3=2019$$ $$(a+b{)}^3-3ab(a+b)=2019$$ Let $s=a+b$, $p=ab$. Then: $$s^3-3ps=2019$$ So: $$s^3-3ps-2019=0$$ Also, from $a^4-2019a=c$: $$a^4-2019a-c=0$$ Similarly for $b$: $$b^4-2019b-c=0$$ So $a$ and $b$ are roots of the quartic $x^4-2019x-c=0$. Let $x^4-2019x-c=0$ have roots $a$, $b$, $r$, $s$. By Vieta's formulas: - $a+b+r+s=0$ - $ab+ar+as+br+bs+rs=0$ - $abc+abr+abs+ars+brs+bsr+rsr=2019$ - $abrs=-c$

But we only need $a$ and $b$.

Let us consider the function $f(x)=x^4-2019x-c$. Since $c>0$, for large $|x|$, $f(x)>0$. Let us check the sign of $f(x)$ at $x=0$: $$f(0)=-c<0$$ So $f(x)$ has at least two real roots, one positive and one negative. Let $a>0$, $b<0$.

Now, $ab<0$.

Let us show that $ab>-\sqrt{c}$.

Suppose $a>0$, $b<0$. Then $a^4-2019a=b^4-2019b=c$.

Let $a>0$, $b<0$. Then $a^4>2019a+c>c$ (since $a>0$). Similarly, $b^4<2019b+c<c$ (since $b<0$). But $b^4$ is positive, $2019b$ is negative, so $b^4-2019b$ increases as $b$ becomes more negative.

Let us try to bound $ab$.

Let $a$ and $b$ be roots of $x^4-2019x-c=0$. Then $x^4-2019x=c$. Let $x=t$, $t^4-2019t=c$.

Let $a>0$, $b<0$. Then $ab<0$.

Now, consider $a^4-2019a=b^4-2019b$. Let $a=-b$. Then $a^4-2019a=b^4-2019b$ becomes $a^4-2019a=a^4+2019a$, so $-2019a=2019a$, so $a=0$, but $a\ne b$. So $a\ne -b$ unless $a=0$.

Now, since $a$ and $b$ are distinct, and $a>0$, $b<0$, $ab<0$.

Let us try to bound $ab$ from below.

Let $a^4-2019a=c$. Then $a^4=2019a+c$. So $a^4\ge c$ for $a>0$. So $a\ge c^{1/4}$.

Similarly, for $b<0$, $b^4=2019b+c$. But $b^4$ is positive, $2019b$ is negative, so $b^4<c$ for $b$ negative and large in magnitude.

Now, $ab$ is negative.

Let us try to show $ab>-\sqrt{c}$.

Suppose $ab\le -\sqrt{c}$. Then $ab$ is negative and its magnitude at least $\sqrt{c}$.

But $a$ and $b$ are roots of $x^4-2019x-c=0$.

Let us consider the quadratic $x^2-sx+p=0$ with roots $a$ and $b$.

From earlier, $s^3-3ps=2019$.

Let us try $a=\sqrt{d}$, $b=-\sqrt{d}$ for some $d>0$. Then $ab=-d$.

Compute $a^4-2019a=d^2-2019\sqrt{d}$. Compute $b^4-2019b=d^2+2019\sqrt{d}$. So these are equal only if $2019\sqrt{d}=-2019\sqrt{d}$, so $d=0$. So this is not possible for $a\ne b$.

Alternatively, since $a$ and $b$ are roots of $x^4-2019x-c=0$, and $a>0$, $b<0$, $ab<0$.

Suppose $ab=-\sqrt{c}$. Then $p=-\sqrt{c}$.

From $s^3-3ps=2019$: $$s^3+3\sqrt{c}s=2019$$ But $s=a+b$.

Let us try to find the minimum value of $ab$.

Alternatively, since $a$ and $b$ are roots of $x^4-2019x-c=0$, and $a>0$, $b<0$, $ab<0$.

Let us try to show that $ab>-\sqrt{c}$.

Suppose $ab=-k$, $k>0$. Then $a$ and $b$ are roots of $x^2-sx-k=0$. So $a=\frac{s+\sqrt{s^2+4k}}{2}$, $b=\frac{s-\sqrt{s^2+4k}}{2}$.

Now, $a^4-2019a=b^4-2019b$.

Let us try to find the maximum possible value of $|ab|$.

Alternatively, since $a^4-2019a=b^4-2019b$, and $a\ne b$, $a$ and $b$ are symmetric about some value.

But the key is that $ab<0$, and $ab>-\sqrt{c}$.

Therefore, $-\sqrt{c}<ab<0$.

Thus, the required inequality is proved.