Let f:N∪{0}→N∪{0} be a function which satisfies f(x2+y2)=f(x)2+f(y)2 for all x,y∈N∪{0}. It' easy to see that f(0)=0 and f(1)=0 or f(1)=1. Suppose let's assume that f(1)=1. Then it's very easy to see that f(2)=2 and since f(2)=2, we can see that f(4)=4 and f(5)=5 because 5=22+12. To see f(3)=3, note that 25=f(52)=f(3)2+f(4)2⟹f(3)=3 One can even see that f(6)=6, f(7)=7 and so on. Is it generally true that f(n)=n for all n?
Edit. If anyone wondering how f(7)=7 here is the way. Note that f satisfies f(n2)=f(n)2. One can see that 252=242+72⟹252=f(24)2+f(7)2 We have to show f(24)=24. For this note that 262=242+102⟹f(26)2=f(24)2+f(10)2 But f(26)=26 because 26=52+12 and f(10)=10 since 10=32+12. In short if n=x2+y2 for some x,y and f(1)=1, then we can see that f(n)=n.
Answer
Yes.
Suppose you already know that f(n)=n for all n<N. You want to find a,b,c, all smaller than N, such that N2+a2=b2+c2. If you find such a triplet, you immediately conclude that f(N)=N, and you have the necessary inductive step.
For odd N>6, use
N2+(N−52)2=(N−2)2+(N+32)2
For even N>6, use
N2+(N2−5)2=(N−4)2+(N2+3)2
Since the question details already prove that f(n)=n for small values of n, this inductive step finishes the argument.
No comments:
Post a Comment