there doesn't appear to any for 15a = 1+651n
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Correct. Did you find any for 16b = 1 + 651n?
No solution for 15a=1+651n because if you rewrite it as 15a-651n=1 you have on your left a multiple of 3 which will never be one.
About 16b=1+651m I can say that the solutions are (k integer):
b=-122+651k
m=-3+16k
Jul 13, 2013
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#6
As Sam_math_88 noted, 15a is a multiple of three and so is 651n=3*217n, so adding the one makes one side a multiple of three but the other side not, so no solution.
Upon division by 16, 651 has remainder 11. If we can find a multiple of 651 having remainder 15, then adding one will give us a multiple of 16. Cycling through the remainders: 11, 6, 1, 12, 7, 2, 13, 8, 3, 14, 9, 4, 15, so let n = 13. Then 1 + 651(13) = 1 + 8463 = 8464 = 16(529). So, we will be able to find a b so that 16b = 1 + 651(13 + 16k), i.e., m = 13 + 16k (or, as Sam_math_88 puts it, for m = -3 + 16k).
Then 16b = 1 + 8463 + 10416k = 8464 + 10416k, so b = 529 + 651k (or, as Sam_math_88 puts it, -122 + 651k).
Are there integers a,n such that 15a = 1 + 651n? If so, specify them. Same question for integers b,m such that 16b = 1 + 651m.