Replace the numerals 1 through 8 with ever increasing prime numbers, always using the next lowest possible that is available
^{1} to fulfill the criteria on the left. Then do the same for the right.
Present a series for the left, and one for the right.
7 + 8 = Square
6 = Prime
4 + 5 = Square
1 + 2 + 3 = Square


7 + 8 = Cube
6 = Prime
4 + 5 = Cube
1 + 2 + 3 = Cube

If it was required that the Right set required "1" to be the next Prime following on after that used for the "8" in the Left set, what might the Right set read, if indeed it is possible?
^{1.}
Note,
"always using the next lowest possible that is available" means that if it is next on the list it cannot be dismissed unless it is
the last of a group of two or three and will not fulfill the criterion. Only then may you advance to the next.
Under the interpretation that what is sought is a set of primes where P
_{1}< P
_{2 }< P
_{3 }< P
_{4 }< P
_{5} < P
_{6 }< P
_{7 }< P
_{8}, beginning with the lowest possible value of P
_{1}, where all criteria may be met; with the additional criteria being that for each series: P
_{1}+P
_{2}+P
_{3}, P
_{4}+P
_{5} and P
_{7}+P
_{8}, each are perfect powers (squares for the left and cubes for the right), the following is the solution:
Left (2, 3, 11, 13, 23, 29, 31, 131)L_{1} +
L_{2} +
L_{3} is a Square; 2 + 3 + 11 = 16 = 4
^{2}L_{4} +
L_{5} is a Square; 13 + 23 = 36 = 6
^{2}L_{7} +
L_{8} is a Square; 31 + 131 = 144 = 12
^{2}Right (2, 3, 59, 61, 1667, 1669, 1693, 4139)R_{1} +
R_{2} +
R_{2} is a Cube; 2 + 3 + 59 = 64 = 4
^{3}R_{4} +
R_{5} is a Cube; 61 + 1667 = 1728 = 12
^{3}
R_{7} +
R_{8} is a Cube; 1693 + 4139 = 5832 = 18
^{3}Edited on January 7, 2010, 8:26 am

Posted by Dej Mar
on 20100106 14:52:27 