1For which exam you study man!!! i have never seen such sums in jee !!!
9 Answers
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341Since by Fermat's Little Theorem x^3 \equiv x \pmod{3} we have
x^3 +6x^2+2x-6 \equiv 6x^2+3x-6 \equiv 0 \pmod{3}
Hence we only need to solve the cubics x^3 +6x^2+2x-6 = \pm 3 for integer solutions
which have their only integer roots as -1 and 1 respectively
1708Thanks hsbhatt Sir.
Sir I want Some explanation on Fermat,s Little theorem for Polynomial Modulo.
1708Thanks Pritish., Next time I will not post Such Type of Questions.
21@piritish , when u know that this is not for jee (and this is true) why do u waste ur time over here? In stead of coercing others to stop their discussion why can't u simply ignore this type of threads ?
341in a way jagdish cannot be faulted. If the polynomial could have been factorised, then the question would have fallen in jee syllabus! He couldnt have known beforehand that the solution would use FLT.
62The proof that can be used withing IIT JEE syllabus is by just changing the idea of congruence to divisibility
First observe that x is not a multiple of both 2 and 3..
So x has to be of the form 6k+1 or 6k-1
Then we can analyze these two cases seperately... to end up at prophet sir's result.. [1]
1remember having seen it earlier http://www.goiit.com/posts/list/algebra-prime-no-1101041.htm