■(素数)^2−1(その15)

以上より,p>3のとき

  p^2−1=24k=k(2^2−1)(3^2−1)

    3=2^2−1

    8=3^2−1

   24=5^2−1=(2^2−1)(3^2−1)

   48=7^2−1=2(2^2−1)(3^2−1)

  120=11^2−1=5(2^2−1)(3^2−1)

  168=13^2−1=7(2^2−1)(3^2−1)

  288=17^2−1=4(2^2−1)^2(3^2−1)

  360=19^2−1=5(2^2−1)^2(3^2−1)

  528=23^2−1=22(2^2−1)(3^2−1)

  840=29^2−1=35(2^2−1)(3^2−1)

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 kがΠ(pi^2−1)の形に表されることはあると思われる.確かめてみよう.15=4^2−1,35=6^2−1であるが,p^2−1の形ではない.続行.

  960=31^2−1=5(2^2−1)(3^2−1)^2

 1368=37^2−1=57(2^2−1)(3^2−1)

 1680=41^2−1=70(2^2−1)(3^2−1)

 1848=43^2−1=77(2^2−1)(3^2−1)

 2208=47^2−1=92(2^2−1)(3^2−1)

 2808=53^2−1=13(2^2−1)^3(3^2−1)

 3480=59^2−1=145(2^2−1)(3^2−1)

 3720=61^2−1=155(2^2−1)(3^2−1)

 4488=67^2−1=187(2^2−1)(3^2−1)

 5040=71^2−1=70(2^2−1)^2(3^2−1)

 5328=73^2−1=111(2^2−1)^2(3^2−1)

 6240=79^2−1=260(2^2−1)(3^2−1)

 6888=83^2−1=287(2^2−1)(3^2−1)

 7920=89^2−1=110(2^2−1)^2(3^2−1)

 9408=97^2−1=49(2^2−1)(3^2−1)^2

10200=101^2−1=425(2^2−1)(3^2−1)

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