show that a number of the form 3^m*5^N*7^K can never be a perfect number

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show that a number of the form 3^m*5^N*7^K can never be a perfect number
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show that a number of the form 3^m*5^N*7^K can never be a perfect number
show that a number of the form 3^m*5^N*7^K can never be a perfect number

show that a number of the form 3^m*5^N*7^K can never be a perfect number
题:show that a number of the form 3^m*5^N*7^K can never be a perfect number
译:形如3^m*5^n*7^k的数不可能是完全数.补充:m,n,k为正整数.
等价于证明以下不定方程无
(3^(m+1)-1)/2 * (5^(n+1)-1)/4 * (7^(k+1)-1)/6 = 2 * 3^m * 5^n *7^k
(3^(m+1)-1) * (5^(n+1)-1) * (7^(k+1)-1) = 32 * 3^(m+1) * 5^n *7^k
其中m,n,k均>=1.
尝试一下,以抛砖引玉:
两边mod3:
-1*((-1)^(n+1)-1)*0==0,无矛盾.
mod 4,无矛盾.
两边mod5,7,==0,同时成立.
即 (3^(m+1)-1) * (-1) * (2^(k+1)-1) ==0 mod 5
其解为:m==3mod4或k==3mod4
(3^(m+1)-1) * (5^(n+1)-1) * (-1) ==0 mod 7
其解为 m==5 mod 7 或n==5 mod 6.
故有:
m==3 mod 4且n==5 mod6(##1)

m==19 mod 28(##2)

k==3 mod 4且n==5 mod6(##3)
以下再考虑原等式:
(3^(m+1)-1)/2 * (5^(n+1)-1)/4 * (7^(k+1)-1)/6 = 2 * 3^m * 5^n *7^k
即(3^(m+1)-1) * (5^(n+1)-1) * (7^(k+1)-1) = 32 * 3^(m+1) * 5^n *7^k
即(3-3^(-m))(5-5^(-n))(7-7^(-k))=96
情况##1:
左边>=(3-3^(-3))(5-5^(-5))(7-1/7)>96,等式不成立.
情况##2:
左边>=(3-3^(-19))(5-1/5)(7-1/7)>96,等式不成立.
情况##3:
左边>=(3-3^(-1))(5-5^(-5))(7-7^(-3))=2,则左边>=(3-3^(-2))(5-5^(-5))(7-7^(-3))>96,
至此,只有一种情况没有排除:
m=1且k==3 mod 4且n==5 mod6(##3##)
即原数形如:
3*5^(3+4k)*7^(5+6t),k,t>=0
相信亦不难解决.我不做了,抛砖引玉吧.

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