Not to do with taxicab numbers

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.

#NumberTheory #Sharky

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $10^{3n+1}\neq a^3+b^3$ .

mathh mathh - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$