Prove that $512^3 + 675^3 + 720^3$ is a composite number.

I want to solve this problem be without computer assisted solution.

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

Note first that $a = 512 = 2^{9}, b = 675 = 3^{3}5^{2}$ and $c = 720 = 2^{4}3^{2}5,$ and thus $2c^{2} = 3ab.$

Next, note that

$a^{3} + b^{3} + c^{3} = a^{3} + b^{3} - c^{3} + 2c^{2}c = a^{3} + b^{3} - c^{3} + 3abc = (a + b - c)(a^{2} + b^{2} + c^{2} - ab + ac + bc).$

Thus $a + b - c = 512 + 675 - 720 = 467$ divides the given expression, i.e., the given expression is composite.

Brian Charlesworth - 6 years, 1 month ago

Nice observation with $2c^3 = 3ab$ !

Calvin Lin Staff - 6 years, 1 month ago

512^3-512+675^3-675+720^3-720+(512+675+720) by fermats little theoram it is divisible by 3

Shiwang Gupta - 6 years, 1 month ago

Hm, can you explain in detail? I'm pretty sure that the number is not a multiple of 3.

Calvin Lin Staff - 6 years, 1 month ago

Yes you are right it not divisible by 3

Tanu Choudhary - 11 months, 2 weeks ago

You can watch a very elegant solution in video by clicking down the link https://youtu.be/hbYVwoSan90

Chase Your Challenges :- Yashaswi Raj - 10 months, 4 weeks 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}`	$\boxed{123}$

Prove that 5123+6753+7203512^3 + 675^3 + 720^3 5123+6753+7203 is a composite number.

Comments

Prove that $512^3 + 675^3 + 720^3$ is a composite number.