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

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

Note by Chung Kevin
3 years, 1 month ago

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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.

- 3 years, 1 month ago

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

Staff - 3 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

- 3 years, 1 month ago

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

Staff - 3 years, 1 month ago

The problem here is the fact that hcf is one. So, I feel that fermats little theorem must be used

- 3 years, 1 month ago