设a.b,c,d∈N,证明(a的(4b+d)次方)减去(a的(4c+d)次方)能被240整除a4b+d-a4c+d
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设a.b,c,d∈N,证明(a的(4b+d)次方)减去(a的(4c+d)次方)能被240整除a4b+d-a4c+d
设a.b,c,d∈N,证明(a的(4b+d)次方)减去(a的(4c+d)次方)能被240整除
a4b+d-a4c+d
设a.b,c,d∈N,证明(a的(4b+d)次方)减去(a的(4c+d)次方)能被240整除a4b+d-a4c+d
240=3*5*16,3,5,16两两互质.
只要证明
A=a^(4b+d)-a^(4c+d)分别被3,5,16整除即可.
如果你知道费马小定理那就很容易了.否则需要用带余数除法慢慢整.怕你不知道费马小定理,就慢慢来吧.
预备知识:
1.3不整除x,那么3整除x^2-1,或者说x^2除以3余1.
这是因为x^2-1=(x-1)(x+1),x除以3的余数是1的话,x-1是3的倍数,x除以3的余数是2的话,x+1是3的倍数,总之,x^2-1是3的倍数.
2.5不整除x,那么5整除x^4-1,或者说x^4除以5余1.
这是因为
x^4-1=(x^2-1)(x^2+1)=(x^2-1)(x^2-4+5)
=(x-2)(x-1)(x+1)(x+2)+5(x^2-1)
无论x除以5余1,2,3,4上式都是5的倍数.
3.偶数的4次方是16的倍数,奇数的平方被8除余1,奇数的4次方被16除余1.
第一句话显然正确,第三句话可由第二句推出.
只证明:奇数的平方被8除余1.
因为(2k+1)^2=4k(k+1)+1,k和k+1一定有一个偶数,所以4k(k+1)是8的倍数.
有了这些准备,就很容易了.
看式子 A=a^(4b+d)-a^(4c+d)=a^d[(a^b)^4-(a^c)^4]
=a^d[(a^4)^b-(a^4)^c]
如果a是2,3,5的倍数,显然A就是16,3,5的倍数.下面设a不是2,3,5的倍数.
3不整除a,根据预备知识1(a^b)^4和(a^c)^4除以3都余1,二者之差就是3的倍数.5不整除a,(a^b)^4和(a^c)^4除以5都余1,二者之差是5的倍数.2不整除a,
(a^b)^4和(a^c)^4除以16都余1,二者之差是16的倍数.
证毕.
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This past May, our own Lovely Kicks shop held a vault release of the Air Jordan 10 “Steel Grey.” The event was a great success & lots of had the chance to buy a pair of these classic Jordans from 2005. Carrying on with the success of the colorway & design, Jordan Brand has decided to generate a fusion version.
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因为 a的4次方为偶数 so(a的(4b)次方)减去(a的(4c)次方)依然为偶数 若d为奇数 a也为奇数 则(a的(4b+d)次方) 也为奇数 (a的(4c+d)次方) 也为奇数 而奇数减奇数为偶数 又因为a为偶数时 (a的(4b+d)次方)减去(a的(4c+d)次方)定为偶数 注【题目 a.b,c,d∈N,应为整数集】 若符合整数及则 定能被240整除...
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因为 a的4次方为偶数 so(a的(4b)次方)减去(a的(4c)次方)依然为偶数 若d为奇数 a也为奇数 则(a的(4b+d)次方) 也为奇数 (a的(4c+d)次方) 也为奇数 而奇数减奇数为偶数 又因为a为偶数时 (a的(4b+d)次方)减去(a的(4c+d)次方)定为偶数 注【题目 a.b,c,d∈N,应为整数集】 若符合整数及则 定能被240整除
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