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1) Re: y^2 = x^n + 4
Posted: Aug 13, 2015 11:22 AM, by: robersi

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2) Re: (abc)^p | p[1 + SUM_{i=1}^{(p-1)/2}[ p!/((p-i)!i!)(-1)^i] -
(p-1)!/(((p-1)/2)!)^2 * (-1)^((p-1)/2)]

Posted: Aug 9, 2015 2:49 PM, by: robersi

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3) Re: (abc)^p | p[1 + (p-1)!/(((p-1)/2)!)^2 * (-1)^((p-1)/2)]
Posted: Aug 9, 2015 2:49 PM, by: robersi

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4) Re: Close to FLT...Seriously!
Posted: Aug 9, 2015 2:48 PM, by: robersi

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5) Re: (abc)^p | p[1 + (p-1)!/(((p-1)/2)!)^2 * (-1)^((p-1)/2)]
Posted: Aug 9, 2015 3:10 AM, by: robersi

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6) Re: So much for meds.
Posted: Aug 8, 2015 4:09 AM, by: robersi

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7) Re: third time is a charm, not really.
Posted: Aug 6, 2015 6:07 PM, by: robersi

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8) Re: (x+1)^p = x^p + 1 (mod p^3), number of solutions given one
solution, x = y_1

Posted: Aug 6, 2015 12:43 AM, by: robersi

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9) Re: can g^2=1 (mod p) when 1<g<p-1?
Posted: Aug 6, 2015 12:09 AM, by: robersi

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10) Re: A very incomplete outline of a proof for the number of solutions
to the equation (x+1)^p = x^p + 1 (mod p^3).

Posted: Aug 5, 2015 7:09 PM, by: robersi

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11) Re: Fighting Infinity
Posted: Jul 13, 2015 3:50 PM, by: robersi

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12) Re: z_1 + ... + z_5 = 0
Posted: Jul 13, 2015 1:01 PM, by: robersi

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13) Re: a + b + c + d = 0
Posted: Jul 13, 2015 12:51 PM, by: robersi

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14) Re: TROLL SPOTTING: A beginner's guide
Posted: Jul 12, 2015 11:30 PM, by: robersi

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15) (a-b)^3 divides [2(a^n-b^n) - 2n(a-b)(ab)^((n-1)/2)] if a!=b
Posted: Apr 23, 2015 3:19 AM, by: robersi

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