\hspace{-16}$Let $\mathbf{x^n+a_{1}.x^{n-1}+a_{2}.x^{n-2}+a_{3}.x^{n-3}+.......+a_{n}=2011}$\\\\ Where $\mathbf{a_{i}\in \mathbb{Z}\;\forall \;i\in \left\{1,2,3,....,n\right\}}$ has $\mathbf{4}$ Integrals Roots.\;\\\\ Then the no. of Integral Roots of the equation\\\\ $\mathbf{x^n+a_{1}.x^{n-1}+a_{2}.x^{n-2}+a_{3}.x^{n-3}+.......+a_{n}=2013}$ is
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1 Answers
341Let P(x) = xn+a1xn-1+...+an-2011.
Then P(x) = (x-a)(x-b)(x-c)(x-d) Q(x) where a,b,c, and d are integers and Q(x) is a monic polynomial with integer coefficients.
Suppose P(z)=2 for some integer z, then
(z-a)(z-b)(z-c)(z-d) Q(z) = 2
z-a, z-b,z-c, z-d are all distinct. But we know that 2 can be factored into at most three distinct factors
i.e. as (-2) X (-1) X (1). Hence the above equation is not possible.
So no integer roots are possible
