Question

Given the sequence 10001,100010001,1000100010001,..., prove each term of the sequence is a composite number

Collected in the board: Number Theory

Steven Zheng posted 3 hours ago

Answer


a_1 = 10001 =10^{4\times1}+1= 73\times 137

a_2=100010001 = 10^{4\times2}+10^{4\times1}+1

a_3 = 1000100010001 =10^{4\times3}+ 10^{4\times2}+10^{4\times2}+10^{4\times1}+1

...

a_n = \underbrace{100010001\dots 0001}_{\text{n 0001}}= 10^{4n}+10^{4n-1}+\dots+10^4+1

Using the Sum formula for a geometric sequence

S_n=\dfrac{a_1(1-r^n)}{1-r}

a_n = \dfrac{1-10^{4(n+1)}}{1-10^4} =\dfrac{1-(10^{2n+2})^2}{1-10^4} =\dfrac{(10^{2n+2}-1)(10^{2n+2}+1) }{10^4-1}

Since 10^4-1< 10^{2n+2}-1 < 10^{2n+2}+1

a_n is not a prime number but a composite number


Steven Zheng posted 3 hours ago

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