Question

Find all integers \(n\) that satisfy the equation
\[(n-1)^3 + n^3 + (n+1)^3 = (n+2)^3\]

[Hint: Work with the equation obtained by replacing \(n\) by \(k+4\).]

Answer 1

\(\large \begin{align} \color{black}{
(n-1)^3+n^3+(n+1)^3=(n+2)^3 \hspace{.33em}\\~\\
\implies \cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
\implies \cdot\cdot\cdot\cdot \hspace{.33em}\\~\\
2n^3-6n^2-6n-8=0 \hspace{.33em}\\~\\
n^3-3n^2-3n-4=0 \hspace{.33em}\\~\\
\color{red}{1}n^3-3n^2-3n-\color{blue}{4}=0 \hspace{.33em}\\~\\
\normalsize \text{by rational roots theorm} \hspace{.33em}\\~\\
\normalsize \text{possible roots are factors of 4 by factors of 1} \hspace{.33em}\\~\\
n=\{ \pm\dfrac{\color{blue}{4}}{\color{red}{1}},\pm\dfrac{\color{blue}{2}}{\color{red}{1}},\pm\dfrac{\color{blue}{1}}{\color{red}{1}}\}\hspace{.33em}\\~\\
\normalsize \text{by synthetic division} \hspace{.33em}\\~\\
4|1 ~~-3~~-3~~-4\hspace{.33em}\\~\\
\underline{~~~- ~~~~4~~~~~~4~~~~~~~4}\hspace{.33em}\\~\\
~~~~~1~~~~~1~~~~~~1~~~~~~0\hspace{.33em}\\~\\
\normalsize \text{n=4 satisfies it} \hspace{.33em}\\~\\
}\end{align}\)

Answer 2

Nice :) i didn't think of rational root thm before xD

Answer 3

Please note that, if I compute the third powers, i get this:
\[n ^{3}-1-3n ^{2}+3n+n ^{3}+n ^{3}+1+3n ^{2}+3n=\]
\[=n ^{3}+8+6n ^{2}+12n\]
and after a simplification, I get:
\[n ^{3}-3n ^{2}-3n-4=0\]
which can be re-written as below:
\[n ^{3}-1-3(n ^{2}+n+1)=0\]
then we can write:
\[[(n-1)-3](n ^{2}+n+1)-3(n ^{2}+n+1)=0\]
now I factor out the quantity \[n ^{2}+n+1\]
and I get:
\[(n-4)(n ^{2}+n+1)=0\]
which admits the only real solution n=4

Answer 4

Wow! that looks great xD

Answer 5

I'm thinking why the textbook gave that hint lol

Answer 6

both above methods look neat to me and they didnt use the hint, so that makes me feel the textbook hint is kinda unnecessary..

Answer 7

here is the actual problem
http://www.chegg.com/homework-help/find-integers-n-satisfy-equation-hint-work-equation-obtained-chapter-mp-problem-25p-solution-9780077418120-exc

Answer 8

@Michele_Laino 's solution looks creative !

Answer 9

Thanks! @mathmath333

Answer 10

http://math.stackexchange.com/questions/120254/sum-of-three-consecutive-cubes

Answer 11

Thanks for your appreciation @ganeshie8

Answer 12

hehehe well that hint was to reduce work :3
\((n+3)^3+(n+4)^3+(n+5)^3-(n+6)^3=0\\ \downarrow \\

(n^3+9n^2+27n+27)+(n^3+12n^2+48n+64)+\\(n^3+15n^2+75n+125)-(n^3+18n^2+108n+216)=0\\ \downarrow \\

27+64+125-216=0\\
\downarrow \\

(n^3+9n^2+27n)+(n^3+12n^2+48n)+\\(n^3+15n^2+75n)-(n^3+18n^2+108n)=0\\
\downarrow \\

n[(n^2+9n+27)+(n^2+12n+48)+(n^2+15n+75)-(n^2+18n +108)]=0\\
n[(2n^2+18n+42)]=0\\
2n(n^2+9n+21)\\
n=0 \text{ the real solution }\\
n^2+9n+21 \text{ has no real solution }\\\)

Answer 13

and one thing i dint find this question in my book xD
what is CHMP is it the appendix ?