A377739 Array of positive integer triples (x,y,z) where the sum of their cubes equals another cubic number.

3, 4, 5, 1, 6, 8, 6, 8, 10, 2, 12, 16, 9, 12, 15, 3, 10, 18, 7, 14, 17, 12, 16, 20, 4, 17, 22, 3, 18, 24, 18, 19, 21, 11, 15, 27, 15, 20, 25, 4, 24, 32, 18, 24, 30, 6, 20, 36, 14, 28, 34, 2, 17, 40, 6, 32, 33, 21, 28, 35, 16, 23, 41, 5, 30, 40, 3, 36, 37, 27, 30, 37, 24, 32, 40, 8, 34, 44, 29, 34, 44, 6, 36, 48, 12, 19, 53, 27, 36, 45, 36, 38, 42

Offset: 1

Author

Luke Voyles, Nov 05 2024

Keywords

• nonn

Comments

The Shiraishi theorem demonstrated that there were an infinite number of cubic number triples whose sum equaled a cubic number (A226903). Through an analysis of the cubic number triples found by Russell and Gwyther, another way to prove that there are an infinite number of cubic number triples who sum equals a cubic number appeared. For any triple, one can add a zero to the end of the three numbers. The new three numbers will also equal a cubic number. For example, 3^3+4^3+5^3=6^3 can be transformed into 30^3+40^3+50^3=60^3. The number of zeros that are consistently applied to each of the numbers who cubic numbers will always create new cubic numbers. For example, 30^3+40^3+50^3=60^3 can become 300^3+400^3+500^3=600^3 and 3000^3+4000^3+5000^3=6000^3, and so on. Through experiments, the formula holds true for Pythagorean triples and Pythagorean quadruples as well. To apply the method to Pythagorean triples, 3^2+4^2=5^2 can be transformed into 30^2+40^2=50^2, 300^2+400^2=500^2, and so on. For Pythagorean quadruples, 3^2+4^2+12^2=13^2 can be transformed into 30^2+40^2+120^2=130^2 and then to 300^2+400^2+1200^2=1300^2. The property holds even beyond the second and third powers. For example, 3530^4=300^4+1200^4+2720^4+3150^4 just as 353^4=30^4+120^4+272^4+315^4. Additionally, 1440^5=270^5+840^5+1100^5+1330^5 just as 144^5=27^5+84^5+110^5+133^5. Once one set is found, it appears there can be an infinite number of similar sets for any power through this method.

The list of Russell and Gwyther also reveals that the cube of 38 can be represented as the sum of the cubes of nine unique positive integers. This is because 38^3=3^3+4^3+5^3+7^3+14^3+17^3+18^3+24^3+30^3.

Examples

			3^3+4^3+5^3=6^3
1^3+6^3+8^3=9^3
6^3+8^3+10^3=12^3
2^3+12^3+16^3=18^3
9^3+12^3+15^3=18^3
		

Links

• A. Russell and C. E. Gwyther, The Partition of Cubes, The Mathematical Gazette, Vol. 21, No. 242 (Feb., 1937), pp. 33-35 (3 pages).

Crossrefs

The sum of each cubic number triple produce the sequence A023042. The comments produce another method to produce an infinite number of cubic number triples whose sum equals a cube that the method shown by Shiraishi according to A226903. The comments discuss qualities of Pythagorean triples A103606 and Pythagorean quadruples A096907. The title's structure drew inspiration from A291694.

Formula

If a^3+b^3+c^3=d^3, then any specific number k that has a zero as the last digit will make k(d^3) another cubic number through the formula k(a^3)+k(b^3)+k(c^3)=k(d^3)