■キャノンボール問題(その10)

m=24k,24k+6,24k+12,24k+18

m=24k+1,24k+7,24k+13,24k+19

の場合まで絞られた.

m=3k→m^2=3(3k^2)

m=3k+1→m^2=3(k^2+2k)+1

m=3k−1→m^2=3(k^2−2k)+1

m=4k→m^2=4(4k^2)

m=4k±1→m^2=4(4k^2±2k)+1

m=4k±2→m^2=4(4k^2±4k+1)

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m=24k+6

m(m+1)(2m+1)/6=(4k+1)(24k+7)(48k+13)

24k+7=3  (mod4)

m=24k+12

m(m+1)(2m+1)/6=(4k+2)(24k+13)(48k+25)

4k+2=2  (mod4)

m=24k+18

m(m+1)(2m+1)/6=(4k+3)(24k+19)(48k+37)

4k+3=3  (mod4)

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m=24k+7

m(m+1)(2m+1)/6=(24k+7)(8k+1)(24k+7)

24k+7=3  (mod4)

m=24k+13

m(m+1)(2m+1)/6=(24k+13)(12k+7)(16k+9)

12k+7=3  (mod4)

m=24k+19

m(m+1)(2m+1)/6=(24k+19)(12k+10)(16k+13)

12k+10=2  (mod4)

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