Mualpha7 wrote: Consider all pairs of positive integers $(a,b)$, with $a<b$, such that $\sqrt{a} +\sqrt{b} = \sqrt{2.160}$ Determine all posible values of $a$. Because $a,b \geq 1$ So $\sqrt{a} +\sqrt{b} \geq 2$ So $\sqrt{2.160} \geq 2$ which is impossible SO $\text{NO SOLUTION}$

Mualpha7 wrote: Consider all pairs of positive integers $(a,b)$, with $a<b$, such that $\sqrt{a} +\sqrt{b} = \sqrt{2.160}$ Determine all posible values of $a$. Consider the lowest possible values $a=b=1$ then $2>\sqrt {2.160}.$ Since we can't go smaller, there are no such positive integer $a,b.$

In many Spanish speaking countries, they have . and , swapped from America/other countries, so it's most likely 2,160

Actually, I looked it up and Paraguay actually uses "," as their decimal separator. So the problem is actually Quote: Consider all pairs of positive integers $(a,b)$, with $a<b$, such that $\sqrt{a} +\sqrt{b} = \sqrt{2160}$ Determine all posible values of $a$.

$$\sqrt{a} +\sqrt{b} = \sqrt{2160}$$$$\sqrt{b} = \sqrt{2160}-\sqrt{a}$$$$b = 2160+a-2\sqrt{2160a}$$$$24\sqrt{15a} = 2160+a-b$$Hence $\sqrt{15a}$ is a rational number, but $15a$ is an integer so $\sqrt{15a}$ is also an integer. There exists $k\in\mathbb{Z}_+$ such that $15a=(15k)^2$. $$24\cdot 15k=2160+a-b< 2160$$$$k< 6$$$$b=2160+a-24\sqrt{15a}=15(12-k)^2$$The solutions are $$(a,b)\in\left\lbrace \left(15k^2, 15(12-k)^2\right): k\in\lbrace1,2,3,4,5\rbrace\right\rbrace$$verified by direct substitution to the initial equation.