A well-known factorisation involves the sum of two cubes:a3+b3=(a+b)(a2−ab+b2)Assuming that both a and b are positive integers, then this sum of two cubes is a semiprime if a+b and a2−ab+b2 are both prime. It's interesting to explore when this happens.
Let's begin with b=1 and restrict ourselves to semiprimes between 4 and 4000. There are only seven that qualify and they are:9,65,217,4097,5833,10649,21953These numbers form the initial terms of OEIS A237040. Their factorisations are as follows:9=32=23+1365=5×13=43+13217=7×31=63+134097=17×241=163+135833=19×307=183+1310649=23×463=223+1321953=29×757=283+13The next is b=8 which yields six numbers:35,133,737,1339,3383,24397These numbers factorise as follows:35=5×7=33+23133=7×19=53+23737=11×67=93+231339=13×103=113+233383=17×199=153+2324397=31×787=293+23These terms are not listed in the OEIS database. Next we'll let b=27. This yields eight numbers and we see that 35 makes a reappearance as the 2 and 3 swap places. The numbers are:35,91,1027,2771,8027,17603,21979,39331These numbers are not listed in the OEIS database and their factorisations are:35=5×7=23+3391=7×13=43+331027=13×79=103+332771=17×163=143+338027=23×349=203+3317603=29×607=263+3321979=31×709=283+3339331=37×1063=343+33Here is a permalink to an algorithm that allows further exploration using additional values of b. A further avenue for investigation would be to consider semiprimes that are a difference of two cubes since we know that:a3−b3=(a−b)(a2+ab+b2)Using b=−1, we find that the following semiprimes qualify:26,215,511,1727,2743,7999,13823These numbers form the initial members of OEIS A242262. Their factorisations are as follows:26=2×13=33−13215=5×43=63−13511=7×73=83−131727=11×157=123−132743=13×211=143−137999=19×421=203−1313823=23×601=243−13
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