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Prove that a number \(10^{3n+1}\), where \(n\) is a positive integer, cannot be represented as the sum of two cubes of positive integers.

Note by Sharky Kesa 2 years, 10 months ago

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By Fermat's Little Theorem either \(a\equiv 0\pmod{7}\) or

\(a^6\equiv 1\pmod{7}\iff 7\mid \left(a^3+1\right)\left(a^3-1\right)\). By Euclid's Lemma:

\(\iff \left(7\mid a^3+1 \text{ or } 7\mid a^3-1\right)\iff a^3\equiv \pm 1\pmod{7}\).

\(10^{3n+1}\equiv \left(10^3\right)^n\cdot 10\equiv (-1)^n\cdot (-3)\equiv \pm 3\pmod{7}\)

\(a^3+b^3\equiv \{-2,-1,0,1,2\}\pmod{7}\). Therefore \(10^{3n+1}\neq a^3+b^3\).

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TopNewestBy Fermat's Little Theorem either \(a\equiv 0\pmod{7}\) or

\(a^6\equiv 1\pmod{7}\iff 7\mid \left(a^3+1\right)\left(a^3-1\right)\). By Euclid's Lemma:

\(\iff \left(7\mid a^3+1 \text{ or } 7\mid a^3-1\right)\iff a^3\equiv \pm 1\pmod{7}\).

\(10^{3n+1}\equiv \left(10^3\right)^n\cdot 10\equiv (-1)^n\cdot (-3)\equiv \pm 3\pmod{7}\)

\(a^3+b^3\equiv \{-2,-1,0,1,2\}\pmod{7}\). Therefore \(10^{3n+1}\neq a^3+b^3\).

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