Prove that (10101)b is a composite integer for any natural number base b>=2?

(10101)b = 10101 to the base b.

ie, if b=2, then 10101 = 2^0 + 2^2 + 2^4.

Update:

Stephen m, great answer. Could you please show me how you went about factorising b^4 + b^2 + 1. Step by step. I would really appreciate it. Thanks.

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  • 1 decade ago
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    10101 in base b is b^4 + b^2 + 1.

    That factorises as (b^2 + b + 1)(b^2 - b + 1)

    Since b >= 2, both of those are > 1, so it is composite.

    Re your additional question..

    Firstly, I've done far too much maths to know have already seen the polynomial b^4 + b^2 + 1 before, so I knew what it factorised to from memory :P

    Option B would be just playing around with factors and seeing if they work. Theres not going to be a linear factor, as substituting b=1 and -1 in don't give you 0. So you're going to have something like:

    (b^2 + mb ± 1)(b^2 + nb ± 1).

    That expands to b^4 + (m+n)b^3 + (mn ± 2)b^2 ± (m+n)b + 1.

    So you want m+n = 0, and mn ± 2 to be 1: That tells you that m and n are 1 and -1.

    Finally, theres a nice trick, which is what you're really wanting. Complete the square: b^4 + b^2 + 1 = b^4 + 2b^2 + 1 - b^2 = (b^2+1)^2 - b^2 = (b^2 + 1 - b)(b^2 + 1 + b) since its a difference of two squares.

  • Anonymous
    1 decade ago

    Stephen+

    Assume a=b^2, then a^2+a+1=0; thus a1=-1/2 – sqrt(-3)/2 and a2= -1/2 + sqrt(-3)/2, isn’t it?

    So a1 = cos(2pi/3) - i*sin(2pi/3) = exp(-i*2pi/3) and a2 = cos(2pi/3) + i*sin(2pi/3) = exp(+i*2pi/3); hence

    b1=-sqrt(a1)=-exp(-i*pi/3)

    b2=+sqrt(a1)=+exp(-i*pi/3)

    b3=-sqrt(a2)=-exp(i*pi/3)

    b4=+sqrt(a2)=+exp(i*pi/3)

    and (b-b1)*(b-b3) * (b-b2)*(b-b4) = (b^2-(b1+b3)*b+1) * (b-(b2+b4)*b+1), where b1+b3 = -cos(pi/3)+i*sin(pi/3) - cos(pi/3)-i*sin(pi/3) = -1, and b2+b4 = +1; thus = (b^2-b+1) * (b^2+b+1);

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