※ 引述《oxs77 (安)》之銘言:
: 1.已知n為正整數
: 且(n^2-9n-1)^1/2 亦為整數
: 求 n=?
設(n^2-9n-1)^1/2=p (p為正整數) => n^2-9n-1=p^2
9 + (85+4p^2)^0.5
=> n^2-9n-(p^2+1)=0 => n= ----------------- ,又n為正整數
2
=> (85+4p^2)^0.5為正整數 => 設 85+4p^2=q^2 ( q為正整數 => q^2 > 4p^2 )
=> q^2-4p^2=85 => (q+2p)(q-2p)=85*1=17*5
=> (n,p,q)=(26,21,43) or (10,3,11) => n=10 or 26
有誤請更正
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