Is this one hard?!
Algebra
Level
pending
Is is the only integer-valued function defined on the integers that satisfies the following conditions?
- for
- for
No
Yes
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1 solution
By condition ( 3 ) , the first point is f ( 0 ) = 1 .
If we now plug in n = 0 into ( 1 ) , we get f ( f ( 0 ) ) = 0 ⇔ f ( 1 ) = 0 . In general, if f ( n ) = m , then ( 1 ) gives f ( f ( n ) ) = n ⇔ f ( m ) = n . Let's call this fact ( 4 ) .
So now we know the values for n = 0 , 1 . If we can find some n so that n + 2 = 0 , 1 , we can simplify ( 2 ) to get a new point. Let's pick n = − 1 , so by ( 2 ) , f ( f ( − 1 + 2 ) + 2 ) = − 1 ⇔ f ( 2 ) = − 1 . If we pick n = − 2 , we get f ( f ( − 2 + 2 ) + 2 ) = − 2 ⇔ f ( 3 ) = − 2 .
By ( 4 ) , f ( 2 ) = − 1 ⇔ f ( − 1 ) = 2 and f ( 3 ) = − 2 ⇔ f ( − 2 ) = 3 .
With these four new point, we can use ( 2 ) to find f ( f ( − 3 + 2 ) ) = − 3 ⇔ f ( 4 ) = − 3 and f ( f ( − 4 + 2 ) ) = − 4 ⇔ f ( 5 ) = − 4 .
Now we can use ( 4 ) to generate two more points and then plug these into ( 2 ) , and then this result in ( 4 ) and so on. So we can uniquely define all values of this function and therefore its equation is also unique.
Note: A function like f ( n ) = 1 − n + sin ( π n ) would also give the same points because sin ( π n ) = 0 , n ∈ N , but because of the constraint integer valued I supposed that it was about functions like polynomials.