1) Solve the equation $y^3 = x^3 + 8x^2 - 6x + 8$ for positive integer x and y.

2) Consider two positive integer a and b which are such that $a^{a}b^{b}$ is divisible by 2000. What is minimum value of ab? Note by Dev Sharma
5 years, 9 months ago

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Solution for $2^{nd}$ question :

$2000|a^{a}b^{b}\Rightarrow 2|a\text{ or }b \text{ and } 5|a \text{ or } b. \\ \text{Therefore, in every case } 10|ab. \\ \text{Thus, the minimum value of }ab \text{ is obviously }\boxed{10}\text{, possible cases }10^{10}1^{1} \text{ and }1^{1}10^{10}.$

- 5 years, 9 months ago

Done very well!!!

- 5 years, 9 months ago

Can you post solution of the first one ?

- 5 years, 8 months ago

1)x=0,y=2

- 5 years, 8 months ago

x=9,y=11

Can you provide its solution ?

- 5 years, 8 months ago

Here is the solution: $y^3=x^3+8x^2-6x+8.Now,let us see when $y^3$ is greater than or equal to $(x+2)^3$ $\Longrightarrow x^3+6x^2+12x+8 \leq x^3+8x^2-6x+8\\ \Longrightarrow 9 \leq x$.Now,we have that for only one value of $x$ can the given expression be a cube and that happens when,$x=9,y=(x+2)=11$.And done!

- 5 years, 8 months ago

The proof why $y < (x+3)$.

Consider $(x+3)^3- y^3 = x^2 +33x +19$. Discriminant of this quadratic $\Delta < 0$ and since the coefficient of $x^2$ is positive, it implies that $(x+3)^3 -y^3$ is always positive. So, $y^3 < (x+3)^3$. So, $y < x+2$.

- 5 years, 8 months ago

Actually I have left an integral part,I haven't proved that for $x \leq 8$,there are no solutions.

- 5 years, 8 months ago

Sorry ,0 is not an positive integer!!

- 5 years, 8 months ago