Back to reduction process    Irreducibles in sequence

IRREDUCIBLES [I] WITH ASSOCIATED PRIMES P (in prime sequence)

To evaluate I, given P (>2), it is necessary to know the unique values of x and y (<x) in the expression P = (x²+y²), and then to obtain integers a and b for which (ax-by) = ±1. Then Z = (ay+bx) has the desired property that (Z²+1) is divisible by P (another valid, if less helpful, way of obtaining a value with this property is to equate Z to ((P-1)/2)! - for a proof of this, see below). If 2Z<P, set I = Z; otherwise, subtract a suitable multiple of P from Z so that I, the absolute value of the result, is as small as possible (i.e. so that 2I < P). [I] is then the irreducible associated with P.

Example 1: suppose P = 1193 = (32²+13²).  (5.13 - 2.32) = 1; so Z = (5.32 + 2.13) = 186.  2Z < P, so I = Z = 186.
Example 2: suppose P = 269 = (13²+10²).    (7.13 - 9.10) = 1; so Z = (7.10 + 9.13) = 187.  2Z > P, so (in this case) I = |Z-P| = 82.

Note that if y > 1 and x-y = 1 (e.g. for P=41, x=5 and y=4), I = x+y; and if y = 1, I = x.

  [1] P=2    (1²+1²)
  [2] P=5    (2²+1²)
  [5] P=13   (3²+2²)
  [4] P=17   (4²+1²)
 [12] P=29   (5²+2²)
  [6] P=37   (6²+1²)
  [9] P=41   (5²+4²)
 [23] P=53   (7²+2²)
 [11] P=61   (6²+5²)
 [27] P=73   (8²+3²)
 [34] P=89   (8²+5²)
 [22] P=97   (9²+4²)
 [10] P=101  (10²+1²)
 [33] P=109  (10²+3²)
 [15] P=113  (8²+7²)
 [37] P=137  (11²+4²)
 [44] P=149  (10²+7²)
 [28] P=157  (11²+6²)
 [80] P=173  (13²+2²)
 [19] P=181  (10²+9²)
 [81] P=193  (12²+7²)
 [14] P=197  (14²+1²)
[107] P=229  (15²+2²)
 [89] P=233  (13²+8²)
 [64] P=241  (15²+4²)
 [16] P=257  (16²+1²)
 [82] P=269  (13²+10²)
 [60] P=277  (14²+9²)
 [53] P=281  (16²+5²)
[138] P=293  (17²+2²)
 [25] P=313  (13²+12²)
[114] P=317  (14²+11²)
[148] P=337  (16²+9²)
[136] P=349  (18²+5²)
 [42] P=353  (17²+8²)
[104] P=373  (18²+7²)
[115] P=389  (17²+10²)
 [63] P=397  (19²+6²)
 [20] P=401  (20²+1²)
[143] P=409  (20²+3²)
 [29] P=421  (15²+14²)
[179] P=433  (17²+12²)
 [67] P=449  (20²+7²)
[109] P=457  (21²+4²)
 [48] P=461  (19²+10²)
[208] P=509  (22²+5²)
[235] P=521  (20²+11²)
 [52] P=541  (21²+10²)
[118] P=557  (19²+14²)
 [86] P=569  (20²+13²)
 [24] P=577  (24²+1²)
 [77] P=593  (23²+8²)
[125] P=601  (24²+5²)
 [35] P=613  (18²+17²)
[194] P=617  (19²+16²)
[154] P=641  (25²+4²)
[149] P=653  (22²+13²)
[106] P=661  (25²+6²)
 [58] P=673  (23²+12²)
 [26] P=677  (26²+1²)
[135] P=701  (26²+5²)
 [96] P=709  (22²+15²)
[353] P=733  (27²+2²)
 [87] P=757  (26²+9²)
 [39] P=761  (20²+19²)
 [62] P=769  (25²+12²)
[317] P=773  (22²+17²)
[215] P=797  (26²+11²)
[318] P=809  (28²+5²)
[295] P=821  (25²+14²)
[246] P=829  (27²+10²)
[333] P=853  (23²+18²)
[207] P=857  (29²+4²)
[151] P=877  (29²+6²)
[387] P=881  (25²+16²)
[324] P=929  (23²+20²)
[196] P=937  (24²+19²)
 [97] P=941  (29²+10²)
[442] P=953  (28²+13²)
[252] P=977  (31²+4²)
[161] P=997  (31²+6²)
[469] P=1009 (28²+15²)
 [45] P=1013 (23²+22²)
[374] P=1021 (30²+11²)
[355] P=1033 (32²+3²)
[426] P=1049 (32²+5²)
[103] P=1061 (31²+10²)
[249] P=1069 (30²+13²)
[530] P=1093 (33²+2²)
[341] P=1097 (29²+16²)
[354] P=1109 (25²+22²)
[214] P=1117 (26²+21²)
[168] P=1129 (27²+20²)
[140] P=1153 (33²+8²)
[243] P=1181 (34²+5²)
[186] P=1193 (32²+13²)
 [49] P=1201 (25²+24²)
[495] P=1213 (27²+22²)
 [78] P=1217 (31²+16²)
[597] P=1229 (35²+2²)
[546] P=1237 (34²+9²)
[585] P=1249 (32²+15²)
[113] P=1277 (34²+11²)
[479] P=1289 (35²+8²)
 [36] P=1297 (36²+1²)
 [51] P=1301 (26²+25²)
[257] P=1321 (36²+5²)
[614] P=1361 (31²+20²)
[668] P=1373 (37²+2²)
[366] P=1381 (34²+15²)
[452] P=1409 (28²+25²)
[620] P=1429 (30²+23²)
[542] P=1433 (37²+8²)
[497] P=1453 (38²+3²)
[465] P=1481 (35²+16²)
[225] P=1489 (33²+20²)
[432] P=1493 (38²+7²)
 [88] P=1549 (35²+18²)
[339] P=1553 (32²+23²)
[610] P=1597 (34²+21²)
etc.

Proof that if P is a prime of the form 4m+1, and Z = ((P-1)/2)!, then (Z²+1) is divisible by P.

By Wilson's theorem, P divides ((P-1)! + 1) = ((4m)! + 1). Consider the first 2m terms in (4m)!, i.e. 4m, 4m-1, 4m-2, ...., 2m+1; these can be rewritten as: P-1, P-2, P-3, ...., P-2m, respectively. Therefore their product is congruent (modulo P) to (2m)!. The remaining terms in (4m)! are 2m, 2m-1, 2m-2, ...., 2, 1, and their product is exactly (2m)!. It follows that ((4m)! + 1) is congruent (modulo P) to (((2m)!)²+1), i.e. that if Z = ((P-1)/2)! = (2m)!, (Z²+1) is divisible by P.

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