Wikipedia:Reference desk/Archives/Mathematics/2019 October 14

= 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)


 * 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)


 * 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)
 * 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)
 * 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 &bull; videos) 22:49, 14 October 2019 (UTC)
 * 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)


 * 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)


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