## Sunday, 15 July 2012

### Twin Prime Proof Appendix

Here’s a bit more detail on how the first hypercuboid was constructed – this is the simple version before we added the complication of residues of squares. This is a version that is looking for pn and pn+2 numbers (eg twin primes) starting from the point where we have narrowed the possible numbers down to the 2n+1 planes and the 3n+1 planes:

 1 7 13 19 25 31 37 43 49 55 61 67 73 79 85 91 97 103 109 115 121 127 133 139 145 151 157 163 169 175 181 187 193 199 205

Numbers coloured grey are either 5N or 5N +2.
Numbers coloured blue are either 7N or 7N+2.

We turn this into a box by adding the following five slices starting from each of the five columns:

 1 31 61 91 121 151 181 211 241 271 301 331 361 391 421 451 481 511 541 571 601 631 661 691 721 751 781 811 841 871 901 931 961 991 1021 1051 1081 1111 1141 1171 1201 1231 1261 1291 1321 1351 1381 1411 1441 1471 1501 1531 1561 1591 1621 1651 1681 1711 1741 1771 1801 1831 1861 1891 1921 1951 1981 2011 2041 2071 2101 2131 2161 2191 2221 2251 2281

 7 37 67 97 127 157 187 217 247 277 307 337 367 397 427 457 487 517 547 577 607 637 667 697 727 757 787 817 847 877 907 937 967 997 1027 1057 1087 1117 1147 1177 1207 1237 1267 1297 1327 1357 1387 1417 1447 1477 1507 1537 1567 1597 1627 1657 1687 1717 1747 1777 1807 1837 1867 1897 1927 1957 1987 2017 2047 2077 2107 2137 2167 2197 2227 2257 2287

 13 43 73 103 133 163 193 223 253 283 313 343 373 403 433 463 493 523 553 583 613 643 673 703 733 763 793 823 853 883 913 943 973 1003 1033 1063 1093 1123 1153 1183 1213 1243 1273 1303 1333 1363 1393 1423 1453 1483 1513 1543 1573 1603 1633 1663 1693 1723 1753 1783 1813 1843 1873 1903 1933 1963 1993 2023 2053 2083 2113 2143 2173 2203 2233 2263 2293

 19 49 79 109 139 169 199 229 259 289 319 349 379 409 439 469 499 529 559 589 619 649 679 709 739 769 799 829 859 889 919 949 979 1009 1039 1069 1099 1129 1159 1189 1219 1249 1279 1309 1339 1369 1399 1429 1459 1489 1519 1549 1579 1609 1639 1669 1699 1729 1759 1789 1819 1849 1879 1909 1939 1969 1999 2029 2059 2089 2119 2149 2179 2209 2239 2269 2299

 25 55 85 115 145 175 205 235 265 295 325 355 385 415 445 475 505 535 565 595 625 655 685 715 745 775 805 835 865 895 925 955 985 1015 1045 1075 1105 1135 1165 1195 1225 1255 1285 1315 1345 1375 1405 1435 1465 1495 1525 1555 1585 1615 1645 1675 1705 1735 1765 1795 1825 1855 1885 1915 1945 1975 2005 2035 2065 2095 2125 2155 2185 2215 2245 2275 2305

And as before, note that we can rearrange these to make the planes clearer, for instance:

 5n+1 5n+2 5n+3 5n+4 5n+5 7n+1 1 127 43 169 85 7n+2 121 37 163 79 205 7n+3 31 157 73 199 115 7n+4 151 67 193 109 25 7n+5 61 187 103 19 145 7n+6 181 97 13 139 55 7n+7 91 7 133 49 175

 7n+1 7n+2 7n+3 7n+4 7n+5 7n+6 7n+7 11n+1 1 331 661 991 1321 1651 1981 11n+2 211 541 871 1201 1531 1861 2191 11n+3 421 751 1081 1411 1741 2071 91 11n+4 631 961 1291 1621 1951 2281 301 11n+5 841 1171 1501 1831 2161 181 511 11n+6 1051 1381 1711 2041 61 391 721 11n+7 1261 1591 1921 2251 271 601 931 11n+8 1471 1701 2131 151 481 811 1141 11n+9 1681 2011 31 361 691 1021 1351 11n+10 1891 2211 241 571 901 1231 1561 11n+11 2101 121 451 781 1111 1441 1771

 7n+1 7n+2 7n+3 7n+4 7n+5 7n+6 7n+7 11n+1 463 793 1123 1453 1783 2113 133 11n+2 673 1003 1333 1663 1993 13 343 11n+3 883 1213 1543 1873 2203 223 553 11n+4 1093 1423 1753 2083 103 433 763 11n+5 1303 1633 1963 2293 313 643 973 11n+6 1513 1843 2173 193 523 853 1183 11n+7 1723 2053 73 403 733 1063 1393 11n+8 1933 2263 283 613 943 1273 1603 11n+9 2143 163 493 823 1153 1483 1813 11n+10 43 373 703 1033 1363 1693 2023 11n+11 253 583 913 1243 1573 1903 2233

For the proof above, these diagrams would need to be extended to 25n x 49n grids and 49n x 121n grids, but it is easier just to imagine those…

Appendix B

Just a quick thought on modulo residues:

I find it easiest to think of modulo residues this way – any prime number p has a modulo residue of p for every other prime. For instance, 19 is defined by the infinite set of pn + 19 residues. However, for numbers below 19, you need to convert this into numbers in the appropriate modulo range, so this is equivalent to 2n+1, 3n+1, 5n+4, 7n+5, 11n+8, 13n+6, 17n+2, 19n (or 19n+19) and every pn+19 thereafter.

Appendix C

Some people might not like the "merging" stage of the attempted proof. (it's certainly a bit dodgy as things stand!) An alternative method would be to simply remove the pn+2 planes from the grid at the same time as repeating other green planes. This would leave us with an incomplete number chain, with repetitions. It would still be topologically equivalent to the prime number hypercuboid. And when it comes to the p^2n+p and p^2n+p+2 planes, we could simply ignore one of them since both will contain the same numbers (>3) in locations where they are not intersected by black planes.