x → 1 lim ( x m + 1 − m x + ( m − 1 ) 1 − x n + 1 − n x + ( n − 1 ) 1 ) = 5 2 1
Given that m and n are positive integers satisfying the above limit. Find the minimum value of n .
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Let f L ( x ) = x L + 1 − L x + ( L − 1 ) . Then, f L ′ ( x ) = ( L + 1 ) x L − L ⇒ f L ′ ′ ( x ) = L ( L + 1 ) x L − 1 .
And f L ( 1 ) = 0 , f L ′ ( 1 ) = 1 , f L ′ ′ ( 1 ) = L ( L + 1 ) .
Given, 5 2 1 = = = = = = 5 2 1 × 8 = 5 2 1 × 2 = x → 1 lim ( x m + 1 − m x + ( m − 1 ) 1 − x n + 1 − n x + ( n − 1 ) 1 ) x → 1 lim ( f m ( x ) 1 − f n ( x ) 1 ) x → 1 lim ( f m ( x ) ⋅ f n ( x ) f n ( x ) − f m ( x ) ) x → 1 lim ( f m ′ ( x ) ⋅ f n ( x ) + f m ( x ) ⋅ f n ′ ( x ) f n ′ ( x ) − f m ′ ( x ) ) x → 1 lim ( f m ′ ′ ( x ) ⋅ f n ( x ) + f m ′ ( x ) ⋅ f n ′ ( x ) + f m ′ ( x ) ⋅ f n ′ ( x ) + f m ( x ) ⋅ f n ′ ′ ( x ) f n ′ ′ ( x ) − f m ′ ′ ( x ) ) 0 + 2 × 1 2 + 0 n ( n + 1 ) − m ( m + 1 ) 4 n ( n + 1 ) − 4 m ( m + 1 ) = ( 2 n + 1 ) 2 − ( 2 m + 1 ) 2 = ( 2 n − 2 m ) ( 2 n + 2 m + 2 ) ( n − m ) ( n + m + 1 ) Apply L’hopital rule because 0 / 0 Apply L’hopital rule again because 0 / 0
Then, ( n − m , n + m + 1 ) = ( 2 , 5 2 1 ) , ( 1 , 5 2 1 × 1 ) ⇒ ( n , m ) = ( 5 2 1 , 5 2 0 ) , ( 2 6 1 , 2 5 9 ) . The answer is 2 6 1 .
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Note that x m + 1 − m x + ( m − 1 ) = ( x − 1 ) ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) .
Then
x → 1 lim ( x m + 1 − m x + ( m − 1 ) 1 − x n + 1 − n x + ( n − 1 ) 1 ) = x → 1 lim ( x − 1 ) ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) − ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) .
Let f ( x ) = ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) − ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) and g ( x ) = ( x − 1 ) ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) .
As f ( 1 ) = g ( 1 ) = 0 , we may use L'Hôpital's Rule.
It is not difficult to see that f ′ ( 1 ) = 2 n ( n + 1 ) − 2 m ( m + 1 ) .
For g ( x ) , let g ( x ) = ( x − 1 ) h ( x ) where h ( x ) = ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) . Now g ′ ( x ) = ( x − 1 ) h ′ ( x ) + h ( x ) and so g ′ ( 1 ) = h ( 1 ) = 1 as we know that h ′ ( 1 ) exists.
Now,
x → 1 lim ( x m + 1 − m x + ( m − 1 ) 1 − x n + 1 − n x + ( n − 1 ) 1 ) = x → 1 lim ( x − 1 ) ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) ( x n + x n − 1 + … + x 2 + x − ( n − 1 ) ) − ( x m + x m − 1 + … + x 2 + x − ( m − 1 ) ) = x → 1 lim g ( x ) f ( x ) = x → 1 lim g ′ ( x ) f ′ ( x ) = g ′ ( 1 ) f ′ ( 1 ) = 2 n ( n + 1 ) − 2 m ( m + 1 )
This means that 2 n ( n + 1 ) − 2 m ( m + 1 ) = 5 2 1 . As 521 is a prime number, there are only two pairs of positive integer solutions ( m , n ) to this Diophantine Equation: ( n − m ) ( n + m + 1 ) = 2 × 5 2 1 = 1 × 1 0 4 2
Namely: n − m = 2 , n + m + 1 = 5 2 1 or n − m = 1 , n + m + 1 = 1 0 4 2 . Solve the equations and we obtain ( m , n ) = ( 2 5 9 , 2 6 1 ) or ( 5 2 0 , 5 2 1 ) . Hence the minimum value of n is 2 6 1 .