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設n為正整數,且n4次方-6n平方+25為質數,是求此質數及n?

2005-10-10 16:11:34 · 3 個解答 · 發問者 Anonymous in 科學 數學

3 個解答

設質數P=n^4-6n^2+25
那可化成P=(n^2-3)^2+16
再細心地慢慢找,找到質數P=17,n=2這一組解
若有其它解也歡迎指正!

2005-10-10 16:46:46 · answer #1 · answered by 金錢遊戲 3 · 0 0

n4-6n2+25=(n4+10n2+25)-16n2=(n2+5)2-(4n)2=(n2+4n+5)(n2-4n+5)為質數因為必然n2-4n+5

2005-10-11 18:50:04 · answer #2 · answered by ? 7 · 0 0

令n^4-6^2+25=k → n^4-6^2+(25-k)=0
十字交乘:
n^2 \/ p
n^2 /\ q
─────
得p+q=-6, p×q = 25-k

p+q = -6 = -1 + -5 = -2 + -4 = -3 + -3
p×q =1× 5 = 5 = 25-k k=20(非質數,不合)
p×q = 2 × 4 = 8 = 25-k k=17(質數,合)
p×q = 3 × 3 = 9 = 25-k k=16(非質數,不合)

故此質數為17,n=2

2005-10-14 18:49:15 補充:
怎麼可能一個慢慢找阿,那如果N很大怎辦?不同意...

2005-10-10 21:42:36 · answer #3 · answered by Gangster 2 · 0 0

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