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October 14

Sum of three cubes

Is the sum of cubes of three positive integers divisible by the sum of a, b, c?--109.166.137.3 (talk) 21:59, 14 October 2019 (UTC)[reply]

If you mean is it ever the case, then yes:

Is 3³ + 5³ + 7³ divisible by (3+5+7) ?
Is 27 + 125 + 343  divisible by 15 ?
Is 495 divisible by 15 ?
Yes

Do you mean to ask if it is always the case ? SinisterLefty (talk) 22:06, 14 October 2019 (UTC)[reply]

Yes, I do wonder about always situation! I asked this question starting from the case of 2 cubes a³ + b³ which can be seen easily to be divisible by the sum a + b. I wondered whether the property of divisibility for the cases of 2 cubic terms in the sum can be extended to 3 terms and this could be proven by polynomial factorisation for sum of 3 odd powers, as easily as the sum of 2 odd powers terms.--109.166.137.3 (talk) 22:28, 14 October 2019 (UTC)[reply]

In other words what is the quotient of a³ + b³ + c³ divided by a + b + c?--109.166.137.3 (talk) 22:36, 14 October 2019 (UTC)[reply]

No. An easy counterexample occurs for

(a,b,c)=(1,1,2).

You can write

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

so

a+b+c

divides

a^{3}+b^{3}+c^{3}

precisely when it divides

3abc.

–Deacon Vorbis (carbon • videos) 22:49, 14 October 2019 (UTC)[reply]

In the case of 1, 2, and 4 as the values of a, b, and c the result is that the sum is 7 but 3 times the product is 24, and 7 does not divide 24. The sum of the cubes is 1 + 8 + 64 = 73, which is not divisible by 7. Georgia guy (talk) 23:19, 14 October 2019 (UTC)[reply]

FWIW, the solutions to (a+b+c)|(a³+b³+c³), barring degenerate solutions, are parameterized by b, c arbitrary, u any factor of 3bc(b+c), a = u-b-c. --RDBury (talk) 12:17, 15 October 2019 (UTC)[reply]

As a simple example, (1³ + 1³ + 1³) is divisible by (1+1+1)... :) --CiaPan (talk) 17:58, 17 October 2019 (UTC)[reply]