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Another case of divisibility (Posted on 2009-02-15) Difficulty: 2 of 5
Determine all pairs (a, b) of positive integers such that ab2 + b + 7 divides evenly a2b + a + b.

See The Solution Submitted by pcbouhid    
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Some Thoughts partial list -- possible spoiler. | Comment 1 of 3
 a   b      a^2b+a+b    ab^2+b+7        a/7  b/7
 11  1   133  19         11/7  1/7
 7  7    357  357        1  1
 28  14          11018  5509     4  2
 49  1   2451  57        7  1/7
 63  21          83433  27811    9  3
 112  28         351372  87843   16  4
 175  35         1072085  214417         25  5
 252  42         2667462  444577         36  6
 343  49         5765193  823599         49  7
 448  56         11239928  1404991       64  8
 567  63         20254437  2250493       81  9
 700  70         34300770  3430077       100  10
 847  77         55241417  5021947       121  11
 1008  84        85350468  7112539       144  12
 1183  91        127354773  9796521      169  13
 1372  98        184475102  13176793     196  14
 1575  105       260467305  17364487     225  15
 1792  112       359663472  22478967     256  16
 2023  119       487013093  28647829     289  17
 2268  126       648124218  36006901     324  18
 2527  133       849304617  44700243     361  19
 2800  140       1097602940  54880147    400  20
 3087  147       1400849877  66707137    441  21
 3388  154       1767699318  80349969    484  22
 3703  161       2207669513  95985631    529  23
 4032  168       2731184232  113799343   576  24
 4375  175       3349613925  133984557   625  25
 4732  182       4075316882  156742957   676  26
 5103  189       4921680393  182284459   729  27
 5488  196       5903161908  210827211   784  28
 5887  203       7035330197  242597593   841  29
 6300  210       8334906510  277830217   900  30
 6727  217       9819805737  316767927   961  31
 7168  224       11509177568  359661799          1024  32
 7623  231       13423447653  406771141          1089  33
 8092  238       15584358762  458363493          1156  34
 8575  245       18015011945  514714627          1225  35
 9072  252       20739907692  576108547          1296  36
 9583  259       23784987093  642837489          1369  37
 10108  266      27177672998  715201921          1444  38
 10647  273      30946911177  793510543          1521  39
 11200  280      35123211480  878080287          1600  40
 11767  287      39738688997  969236317          1681  41
 12348  294      44827105218  1067312029         1764  42
 12943  301      50423909193  1172649051         1849  43
10   for T=1 to 1000000
20   for A=1 to T-1
30     B=T-A
40     R=(A*A*B+A+B)@(A*B*B+B+7)
50     if R=0 then print A;B,A*A*B+A+B;A*B*B+B+7;A*B*B+B+7,A//7;B//7
60   next
70   next

was stopped when the DOS screen became full.

Aside from (11,1) and (49,1) any (7x^2,7x) seems to work.

Doing the division (7^3*x^5 + 7*x^2 + 7*x) / (7^3*x^4 + 7*x + 7) gives x with no remainder, verifying the sufficiency of the latter part of the solution.

Probably there are no other sporadics like (11,1) and (49,1).

Edited on February 15, 2009, 2:20 pm
  Posted by Charlie on 2009-02-15 14:19:34

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