Whose sum is this?
Let and be the roots of the equation . And denote . What is the smallest positive integer such that are all divisible by 8?
The answer is 4.
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1 solution
Look, for this problem, we don't require to solve this equation and find its roots(in fact we won't even get real roots), all we got to do is substitute values of sum and product of the roots. Given equation, x^4+4x^3+4x^2+4x+4 =(x-a)(x-b)(x-c)(x-d).... Equating, coefficients of various powers of x from LHS and RHS of the equation, we have, a+b+c+d=-4 ab+ac+ad+bc+bd+cd=4 abc+bcd+acd+abd=-4 abcd=4, Now, since a satisfies above equation, a^4=-4(a^3+a^2+a+1) =>a^(n+4)= -4(a^(n+3)+a^(n+2)+a^(n+1)+a^(n)). So do the same for b,c,d. Thus,S(n+4)=-4(S(n+3)+S(n+2)+S(n+1)+S(n)), for all integer n(not just positive). Hence always divisible by 8 if the rest are divisible. S1=a+b+c+d=-4,S0=4, S2=(a+b+c+d)^2-(ab+bc+ca+ad+bd+cd)=8 Hence, S3=28 and S4 onwards divisible by 8. Therefore m=8.