( 6 1 5 5 ) x + ( 6 1 5 6 ) x + ⋯ + ( 6 1 6 0 ) x = ( 6 1 6 2 ) x + ( 6 1 6 3 ) x + ⋯ + ( 6 1 6 5 ) x + 1
If x satisfies the equation above, find the value of ( x 2 + 1 ) 4 + ( x + 2 3 ) 2 + ( x + 1 ) 6 + 3 6 .
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Wow. This identity is the first of its kind I've ever seen.
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Read the book. You'll see it there ;)
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OMG now I remember Titu Andreescu was the one who signed the certificates for the purple comet competition...
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To make things a bit nicer, multiply both sides of the equation by 6 1 x . Hence, the equation becomes,
5 5 x + 5 6 x + . . . + 6 0 x = 6 1 x + 6 2 x + . . . + 6 5 x
By doing a little bit of Trial and Error, we get that x = 2 is a solution. (Well, that is if you're not lazy.)
But how do we get x = 2 ?
Consider the identity,
a 2 + ( a + 1 ) 2 + . . . + ( a + k ) 2 = ( a + k + 1 ) 2 + . . . + ( a + 2 k ) 2
for a = k ( 2 k + 1 ) , k ∈ N
The equation can be rewritten as,
5 5 x + ( 5 5 + 1 ) x + . . . + ( 5 5 + 5 ) x = ( 5 5 + 6 ) x + ( 5 5 + 7 ) x + . . . + ( 5 5 + 1 0 ) x
Then, k = 5 and indeed, a = 5 × ( 2 × 5 + 1 ) = 5 5
Since the equation follows the identity, we know now (at least not by Trial and Error) that x = 2 is a solution.
To prove that only x = 2 is the solution, notice in the original equation that the left-hand side is a decreasing function in x and the right-hand side is an increasing function in x , thus, it has at most one solution, which is x = 2 .
Therefore, the value of ( x 2 + 1 ) 4 + ( x + 2 3 ) 2 + ( x + 1 ) 6 + 3 6 = ( 4 + 1 ) 4 + ( 2 + 2 3 ) 2 + ( 2 + 1 ) 6 + 3 6 = 6 2 5 + 6 2 5 + 7 2 9 + 3 6 = 2 0 1 5
This problem is modified from Zuming Feng and Titu Andreescu's book, 101 Problems in Algebra From the Training of the USA IMO Team