Pre RMO 2016
x 2 0 1 6 + ( 2 0 1 6 ! + 1 ! ) x 2 0 1 5 + ( 2 0 1 5 ! + 2 ! ) x 2 0 1 4 + ⋯ + ( 1 ! + 2 0 1 6 ! ) = 0
Find the number of integral roots of the equation above.
The answer is 0.
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3 solutions
Let f(x) be the given polynomial. Note that f(1) and f(0) are both odd. Thus it cannot have any integer root
why it can't have any integer root ?
Consider the cases separately when x is even and odd. In both of these, evaluate f(x) (mod 2).
Why don't you show me how you obtained the answer of 0 first?
Note that if x is an integral solution, then, x k + x 2 0 1 5 − k will divide x 2 0 1 6 for all k = 0 , ⋯ , 2 0 1 5 . Since x 2 0 1 5 − k + x k = x k ( x 2 0 1 5 − 2 k + 1 ) , ∀ k = 0 , 1 , ⋯ , 2 0 1 5 , x + 1 is a factor, and thus x + 1 ∣ x 2 0 1 6 ⟹ x + 1 ∣ x , which can happen only if x = 0 , but clearly, that is not a solution to the equation.
Suppose α is an integral solution of the equation p ( x ) = 0 , i.e. p ( α ) = 0 , where the given polynomial is denoted by p ( x ) .
Then, α − 0 ∣ p ( α ) − p ( 0 ) ⇒ α ∣ p ( 0 )
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Again, α − 1 ∣ p ( α ) − p ( 1 ) ⇒ α − 1 ∣ p ( 1 )
But, α and α − 1 are two consecutive integers, so one of them is even and the other is odd. But, note that both p ( 0 ) and p ( 1 ) are odd.
Hence, contradiction!! So, no such integer α exists.
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Please proof this in the solution box. @Raushan Sharma
It is always 0 .