9a = 1 0r 2
sub and see
14 Answers
62x13+x+90= x(x12+1) + 90
= x(x4+1)(x8-x4+1) + 90
= x(x4+1){(x4+x2)(x4-x2)+1} + 90
x2=x-a
x4=(x-a)2=x2-2ax+a2=x-2ax-a+a2
x4=(x-a)2=x2-2ax+a2
x4-x2=-2ax+a2
hence the above expression is
x(x-2ax-a+a2+1){(x-2ax-a+a2+x-a)(-2ax+a2)+1}+90
x(x-2ax-a+a2+1){(2x-2ax-2a+a2)(-2ax+a2)+1}+90
{x2(1-2a)-a+a2+1)}{(2x-2ax-2a+a2)(-2ax+a2)+1}+90
{(x-a)(1-2a)-a+a2+1)}{(2x-2ax-2a+a2)
{x(1-2a)-a+2a2-a+a2+1)}{(2x-2ax-2a+a2)
{x(1-2a)-2a+3a2+1)}{(2x(1-a)-2a+a2)
=x2(1-2a)(1-a) + x{(-2a+a2)(1-2a)+2(1-a)(-2a+3a2+1)} + (-2a+3a2+1)(-2a+a2)
= (x-a)(1-2a)(1-a) + x{(-2a+a2)(1-2a)+2(1-a)(-2a+3a2+1)} + (-2a+3a2+1)(-2a+a2)
Now wee need the condition that this expression is zero for all values of x.
For that take x=0 and x=1
we get two simultaneous eqtns in a. We solve them for common roots.
I hope Prophet sir does not feel bad because I have a feeling that i have made an elegant question look so badly messed up :(
341If x2-x+1 were a divisor, then we must have (-\omega)^{13} - \omega + 90 = 0 where \omega is a non-real cube root of 1
9sir i dint yet post the full solution i jus thought in my mind that a can be 1,2 or -1 ,-2
so offered a clue for juniors wrkin on it
ill post the method by working out soon
9in the mean time can u pls look at my Riemann hypothesis thread ??
http://targetiit.com/iit_jee_forum/posts/prophet_sir_pls_help_9801.html
9x13 +x+90 =(x2-x+a)(f(x))
putx= 0 and 1 and subtract
2=a(f(1) - f(0) )
so a = 1,2,-1,-2 are the contestants !
9oops i think first i need to prove f(x) is a polynomial of integer coefficients
341i dont know abt the riemann hypothesis, but the wiki article gives a functional eqn with a sine term in it which explains the -ve even zeros. I am puzzled u did not agonise over the value at s = 0 being -1/2
9yup i cudnt even understand the statement , never knew maths cud go this weird !
341Post 8: by Horner's detached coefficients it should be obvious that f(x) has integer coefficients.
Otherwise , you can at least prove that the coefficients will be rational and then use Gauss's lemma - http://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial) (see the last sentence)
Actually, if i remember right, the proof will simply extend the argument you will use to prove that the coeffs are rational.
9sir i have another proof that straight away tells a=2 is only feasible integer
( it involves complex no in euler form )
afterwards i had verified a=2 is exactly dividing manually
but is there a derivation that gives straight away that a=2 ??
341This giving of links will irritate b555 :D, but I am short of time
http://www.goiit.com/posts/list/algebra-question-101690.htm
9waw i thought nishant sirs solution was very inelegant but now it seems to be the most direct proof of this Q
excellent sir