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Old 2020-03-20, 10:23   #1
enzocreti
 
Mar 2018

17·31 Posts
Default 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?
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Old 2020-03-20, 18:35   #2
Dylan14
 
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"Dylan"
Mar 2017

24·3·11 Posts
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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]]}]
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