With a³+b³+c³-3abc=m (m-random integer)
And
a³+b³+c³-3abc=n(n-another integer)
How to find
a³+b³+c³-3abc=mn(m and n are Co prime)
I came across this in an online math contest.
Hint was given as properties of determinants.
We have to find integer values of a, b, c which satisfy the equation
a³+b³+c³-3abc=mn
(All three cases have different values for a,b,c)
(in online question numbers where used instead of m and n)
Pls help
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Use http://math.stackexchange.com/questions/475354/how-to-show-that-a3b3c3-3abc-abcab-omegac-omega2ab-omega2 – lab bhattacharjee Jul 15 '16 at 14:57
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Welcome to MathSE. You might want to read this informative tutorial on formatting and edit your post accordingly. – Théophile Jul 15 '16 at 15:15
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How the expression $a^3+b^3+c^3-3abc$ has three values: $m$, $n$ and $mn$? – Ng Chung Tak Jul 15 '16 at 15:15
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1 St one different a,b,c . – Jul 15 '16 at 15:38
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2 nd one different a,b,c – Jul 15 '16 at 15:38
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3 Rd one different a,b,c – Jul 15 '16 at 15:39
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http://www.artofproblemsolving.com/community/c3046h1049581___ – individ Jul 15 '16 at 16:23
1 Answers
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There is an easy way to look at this. Take a three by three matrix $$ T = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) $$ Note that $$ T^2 = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) $$ and $$ T^3 = I$$
$$ aI + bT + cT^2 = \left( \begin{array}{rrr} a & b & c \\ c & a & b \\ b & c & a \end{array} \right) $$
Using $T^3 = I, \; T^4 = T,$ given $(a,b,c)$ and $(d,e,f),$ it is easy to calculate $(j,k,l)$ in $$ (aI + b T + c T^2)(dI + eT + fT^2) = jI + kT + lT^2 $$ Meanwhile, we are using determinants, in that $$ \det(aI + b T + c T^2) = a^3 + b^3 + c^3 - 3abc $$
Will Jagy
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