De monics
Let P ( x ) be a monic third degree polynomial. P ( x ) = − 4 has three distinct nonzero integer solutions whose sum is 2 . If there exists a positive integer n , such that P ( n ) = 2 0 1 7 , find n .
The answer is 31.
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1 solution
You also need to show that, that is the only possible value for n. For any pair of negative factors in(-47,-43,-1) there is no possible integral solution for n.
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( n − a ) = − 4 3 ( n − b ) = − 4 7 ( n − c ) = 1 then 3 n − ( a + b + c ) = − 4 3 − 4 7 + 1 = − 8 9 = > 3 n = − 8 7 = > n = − 2 9
Why this is not allowable ?
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It is given in the question that n is a positive integer.
O yeah, I may have miscalculated. But has the question changed since to see if n is only positive?
Nice problem.
Let G ( x ) = P ( x ) + 4 , such that G ( x ) = 0 will have three distinct integer solutions which add up to 2. That means, we can assign values a , b , and c such that G ( x ) = ( x − a ) ( x − b ) ( x − c ) = 0 .
Thus, G ( n ) = 2 0 1 7 + 4 = 2 0 2 1 , which in turn equates to
2 0 2 1 = ( n − a ) ( n − b ) ( n − c )
2 0 2 1 has two factors 4 7 and 4 3 . So we can assign these to any of the three factors of G ( n ) and assign the other one with 1 .
n − a = 4 7
n − b = 4 3
n − c = 1
adding them, we get
3 n − ( a + b + c ) = 9 1
3 n − 2 = 9 1
3 n = 9 3
and voila! n = 3 1 as desired.