Wednesday, 10 September 2014

functional equations - If f(1)=1, then is it true that f(n)=n for all ninmathbbNcup0.




Let f:N{0}N{0} be a function which satisfies f(x2+y2)=f(x)2+f(y)2 for all x,yN{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)2f(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+72252=f(24)2+f(7)2 We have to show f(24)=24. For this note that 262=242+102f(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+(N52)2=(N2)2+(N+32)2



For even N>6, use



N2+(N25)2=(N4)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.


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