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?

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## Comments

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

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Done very well!!!

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Can you post solution of the first one ?

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1)x=0,y=2

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x=9,y=11

Can you provide its solution ?

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Here is the solution: $y^3=x^3+8x^2-6x+8<x^3+9x^2+27x+27=(x+3)^3 \text{because}\\ x^3=x^3\\ 8x^2<9x^2\\ -6x<27x\\ 8<27$.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!

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$y < (x+3)$.

The proof whyConsider $(x+3)^3- y^3 = x^2 +33x +19$. 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, $y < x+2$.

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$x \leq 8$,there are no solutions.

Actually I have left an integral part,I haven't proved that forLog in to reply

Sorry ,0 is not an positive integer!!

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