mersenneforum.org 31*10^2-13
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 2020-03-20, 10:23 #1 enzocreti   Mar 2018 10138 Posts 31*10^2-13 Consider the example (31*10^2-13)/9=7^3 are there other primes p and q (with p the reverse of q) besides 31 and 13, such that (p*10^2-q)/9=c^3 for c>0?
 2020-03-20, 18:35 #2 Dylan14     "Dylan" Mar 2017 33×19 Posts Up to the ten millionth prime the only combinations in which ((p*10^2-q)/9)^(1/3) is a positive integer and p and q are emirps are Code: 31 13 124981 189421 126653 356621 96139019 91093169 96179059 95097169 124430101 101034421 124448201 102844421 124486601 106684421 124494701 107494421 124538111 111835421 124540211 112045421 124550311 113055421 124633121 121336421 124643221 122346421 124668421 124866421 124670521 125076421 124687621 126786421 124755331 133557421 124780631 136087421 124782631 136287421 124794731 137497421 124838141 141838421 124853341 143358421 124856341 143658421 124990751 157099421 126481603 306184621 126553313 313355621 126569413 314965621 126572513 315275621 126627023 320726621 126751333 333157621 126770533 335077621 126782633 336287621 126784633 336487621 126795733 337597621 126949253 352949621 126951353 353159621 126972553 355279621 126986653 356689621 The first pair you already have found. The next two yield the number 1367631, which is 111^3. I'll leave it to you to determine the others. The Mathematica code I used to get these: Code: For[n = 1, n <= 10000000, n++, {p = Prime[n]; q = IntegerReverse[p]; If[PrimeQ[q] && IntegerQ[((100*p - q)/9)^(1/3)], Print[p, " ", q]]}]