Numbers of the form (p^2)q, p and q different primes, are "special".
arithmetic/consecutive.p2q New
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Problem
Numbers of the form (p^2)q, p and q different primes, are "special". Lyrians are students until their age reaches the latter of two consecutive special numbers, masters until it reaches the last of three consecutive special numbers, and sages until death at the last of four. At what ages do Lyrians become masters and sages, and how long do they live? Spring Bent, Lyrian Ages *** To the inhabitants of Planet Lyre III, which circles the planet Betelgeuse, numbers of the form (p^2)q, where p and q are different prime numbers, are called special numbers because they have special significance. Lyrians are students until their ages in Lyrian years reach the latter of two consecutive numbers that are special. They are then masters until their ages reach the last of three consecutive numbers that are special. They are sages until death, which occurs at the ripe old age of the last of four consecutive numbers that are special. At what ages do Lyrians become masters and sages, and how long do they live? Follow-on question: in the year 1, when the society was founded, the infallible Grand Sage pronounced that the society would only fall on a year number that was the last of five consecutive special numbers. Can it be proven that the society will never fall? If not, what is the first year in which it might fall? -- TRM <diogenes@mail.tds.net>
Show solution
The puzzle has already been answered. Everybody stopped posting once they became satisfied with the answers. The following summary is a concatenation of parts of previous postings. A student becomes a master at 45. 44 = 11*2^2. 45 = 5*3^2 A master becomes a sage at age 605. 603 = 67*3^2 604 = 151*4^2 605 = 5*11^2 [ Editor: 604 = 151*2^2; 4^2 is not a prime squared. ] Of four consecutive numbers, one must be a multiple of four, and another must also be even, but not a multiple of four. To be "special" (the product of a prime and a prime squared), the latter must be twice the square of an odd prime. Odd squares are congruent to 1 mod 8, so this number, 2p^2, is 2 mod 8, and the multiple of four, 4q, must be 2p^2 + 2. Neither p nor q is 3, so the odd number between these two even numbers must be a multiple of three; either 3r^2 or 9r for some prime r. One can check quickly up to large values by choosing p, seeing if 2p^2 + 2 is four times a prime, and if 2p^2 + 1 is 3r^2 or 9r. In the few cases that remain, check if 2p^2 - 1 or 2p^2 + 3 is special. The first case that works is p = 92311: 17042641441 = 7^2 347809009 17042641442 = 2 92311^2 17042641443 = 3^2 1893626827 17042641444 = 2^2 4260660361 (but 17042641440 = 2^5 3 5 11 17 181 1049, and 17042641445 = 5 52561 64849, so the streak ends at four) So Lyrians live for 17,042,641,444 years. On the follow-on question: No. The society might fall in year 10093613546512325: 10093613546512321 = 7^2 205992113194129 10093613546512322 = 2 71040881^2 10093613546512323 = 3^2 1121512616279147 10093613546512324 = 2^2 2523403386628081 10093613546512325 = 5^2 403744541860493 I made some assumptions to speed up the search, so I don't know if the example above is the smallest one. [ Editor: the follow-on answer is quoted, unsigned, in Rhoads's posting. ] -- Glenn C. Rhoads <rhoads@paul.rutgers.edu>