If f ( f ( x + y ) ) = f ( x 2 ) + f ( y 2 ) then f ( x ) = ? for all integers ( f : Z → Z )I know how to solve the following problem though: if f ( f ( x + y ) ) = f ( 2 x ) + 2 f ( y ) then f ( x ) = ? We can easily analyze that f(x) here (in the second problem) is a linear function. And hence solve it by using the linear equation.But, As for the main problem (the one I mentioned first) I don't know how to proceed. Should I use the quadratic equation? (my calculation says its a quadratic function)

Question

If   f  (  f  (  x  +  y  )  )  =  f  (      x    2    )  +  f  (      y    2    ) then   f  (  x  )  =  ? for all integers (  f  :      Z    →      Z  )I know how to solve the following problem though: if   f  (  f  (  x  +  y  )  )  =  f  (  2  x  )  +  2  f  (  y  ) then   f  (  x  )  =  ? We can easily analyze that f(x) here (in the second problem) is a linear function. And hence solve it by using the linear equation.But, As for the main problem (the one I mentioned first) I don&#039;t know how to proceed. Should I use the quadratic equation? (my calculation says its a quadratic function)

toliwask · Accepted Answer

Step 1Substituting   y  =  −  x grants   f  (  f  (  0  )  )  =  2  f  (      x    2    ). Thus   f  (      x    2    ) is constant for all x, i.e. f is constant for nonnegative numbers, the range of values for       x    2  .Step 2(If your f has domain       R   or something else, you&#039;ll need to specify to proceed further.)

Keyla Koch · Accepted Answer

Step 1in this answer, x and y are integers.let   y  =  −  x.then   2  f  (      x    2    )  =  f  (  f  (  0  )  )let   x  =  0then   2  f  (  0  )  =  f  (  f  (  0  )  )Let   a  =  f  (  0  )  f  (  a  )  =  2  a  f  (  f  (  x  +  y  )  )  =  f  (      x    2    )  +  f  (      y    2    )  =  2  a  f  (  f  (  x  )  )  =  2  a  f  (  2  a  )  =  f  (  f  (  a  )  )  =  2  af has only one fixed point 2f(0).also, the fixed point cannot be a square or twice a square.Step 2so the solutions f must satisfy 2 conditions:-   f  (  f  (  x  )  )  =  2  f  (  0  )-   f  (      x    2    )  =  f  (  0  )f cannot be a polynomial, because if it were, its second iterate is a polynomial and a constant at the same time, contradiction. in particular, f cannot be linear or quadratic. also the qn mentioned another functional eqn, but the solution of one won&#039;t help with solving the other.

If f(f(x+y))=f(x^2)+f(y^2) then f(x)=?

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