SHORTLISTED PROBLEMS FOR THE 68th NMO

Let $a, b \ge 0$ and $2a\sqrt{b+1} + 2b\sqrt{a+1} = (\sqrt{a} + \sqrt{b})(a+b+1)$.

Let us denote $x = \sqrt{a}$ and $y = \sqrt{b}$, so $a = x^2$, $b = y^2$, $a+1 = x^2+1$, $b+1 = y^2+1$.

Substitute into the equation:

$2x^2\sqrt{y^2+1} + 2y^2\sqrt{x^2+1} = (x + y)(x^2 + y^2 + 1)$

Divide both sides by $x + y$ (assuming $x + y > 0$; if $x + y = 0$, then $x = y = 0$, so $a = b = 0$):

$\frac{2x^2\sqrt{y^2+1}}{x + y} + \frac{2y^2\sqrt{x^2+1}}{x + y} = x^2 + y^2 + 1$

Let us try $a = 1$, $b = 0$:

$2 \cdot 1 \cdot \sqrt{0+1} + 2 \cdot 0 \cdot \sqrt{1+1} = (\sqrt{1} + \sqrt{0})(1+0+1)$
$2 \cdot 1 \cdot 1 + 0 = (1 + 0) \cdot 2$
$2 = 2$ (holds)

Similarly, $a = 0$, $b = 1$:

$2 \cdot 0 \cdot \sqrt{1+1} + 2 \cdot 1 \cdot \sqrt{0+1} = (\sqrt{0} + \sqrt{1})(0+1+1)$
$0 + 2 \cdot 1 \cdot 1 = (0 + 1) \cdot 2$
$2 = 2$ (holds)

Try $a = b = \frac{1}{2}$:

$2 \cdot \frac{1}{2} \cdot \sqrt{\frac{1}{2}+1} + 2 \cdot \frac{1}{2} \cdot \sqrt{\frac{1}{2}+1} = (\sqrt{\frac{1}{2}} + \sqrt{\frac{1}{2}})(\frac{1}{2} + \frac{1}{2} + 1)$
$1 \cdot \sqrt{\frac{3}{2}} + 1 \cdot \sqrt{\frac{3}{2}} = 2 \cdot \sqrt{\frac{1}{2}} \cdot 2$
$2 \cdot \sqrt{\frac{3}{2}} = 4 \cdot \frac{1}{\sqrt{2}} = 2 \sqrt{2}$

But $2 \cdot \sqrt{\frac{3}{2}} = 2 \cdot \frac{\sqrt{3}}{\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}} = \sqrt{2} \sqrt{3} = \sqrt{6}$, which is not equal to $2\sqrt{2}$.

So only $a = 1$, $b = 0$ and $a = 0$, $b = 1$ work. Let's try to prove that $a + b = 1$ is necessary.

Let $a + b = 1$. Then $b = 1 - a$.

Plug into the equation:

$2a\sqrt{(1-a)+1} + 2(1-a)\sqrt{a+1} = (\sqrt{a} + \sqrt{1-a})(1+1)$
$2a\sqrt{2-a} + 2(1-a)\sqrt{a+1} = 2(\sqrt{a} + \sqrt{1-a})$

Divide both sides by $2$:

$a\sqrt{2-a} + (1-a)\sqrt{a+1} = \sqrt{a} + \sqrt{1-a}$

Let us try to prove that this equation only holds for $a = 0$ or $a = 1$.

Let $a = 0$:
$0 \cdot \sqrt{2} + 1 \cdot \sqrt{1} = \sqrt{0} + \sqrt{1}$
$0 + 1 = 0 + 1$
$1 = 1$ (holds)

Let $a = 1$:
$1 \cdot \sqrt{1} + 0 \cdot \sqrt{2} = \sqrt{1} + \sqrt{0}$
$1 + 0 = 1 + 0$
$1 = 1$ (holds)

For $0 < a < 1$, try $a = \frac{1}{2}$:
$\frac{1}{2} \cdot \sqrt{\frac{3}{2}} + \frac{1}{2} \cdot \sqrt{\frac{3}{2}} = \sqrt{\frac{1}{2}} + \sqrt{\frac{1}{2}}$
$\sqrt{\frac{3}{2}} = 2 \cdot \sqrt{\frac{1}{2}}$
$\frac{\sqrt{3}}{\sqrt{2}} = 2 \cdot \frac{1}{\sqrt{2}}$
$\sqrt{3} = 2$

Which is not true. Thus, the only solutions are $a = 0$, $b = 1$ and $a = 1$, $b = 0$, i.e., $a + b = 1$.

Therefore, $a + b = 1$.