[Q]所与の自然数nについて,
[1]1/(a-b)(a-c)+1/(b-a)(b-c)+1/(c-a)(c-b)=n
[2]a/(a-b)(a-c)+b/(b-a)(b-c)+c/(c-a)(c-b)=n
[3]a^2/(a-b)(a-c)+b^2/(b-a)(b-c)+c^2/(c-a)(c-b)=n
[4]a^3/(a-b)(a-c)+b^3/(b-a)(b-c)+c^3/(c-a)(c-b)=n
の自然数解(a,b,c)を求めよ.
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[A]
[1]1/(a-b)(a-c)+1/(b-a)(b-c)+1/(c-a)(c-b)=0
[2]a/(a-b)(a-c)+b/(b-a)(b-c)+c/(c-a)(c-b)=0
[3]a^2/(a-b)(a-c)+b^2/(b-a)(b-c)+c^2/(c-a)(c-b)=1
[4]a^3/(a-b)(a-c)+b^3/(b-a)(b-c)+c^3/(c-a)(c-b)=a+b+c
より,[1][2]はn=0,[3]はn=1,[4]はn=a+b+cのとき,無限に整数解をもちます.
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